StataCorp now provides free tutorial videos on StataCorp’s YouTube channel,

There are 24 videos providing 1 hour 51 minutes of instructional entertainment:

Stata Quick Tour (5:47)
Stata Quick Help (2:47)
Stata PDF Documentation (6:37)

Stata SEM Builder (8:09)
Stata One-way ANOVA (5:15)
Stata Two-way ANOVA (5:57)

Stata Box Plots (4:04)
Stata Basic Scatterplots (5:19)
Stata Bar Graphs (4:15)
Stata Histograms (4:50)
Stata Pie Charts (5:32)

And more are forthcoming.

The inside story

Alright, that’s the official announcement.

Last Friday, 21 September 2012, was an exciting day here at StataCorp. After a couple of years of “wouldn’t it be cool if”, and a couple of months of “we’re almost there”, Stata’s YouTube channel was finally ready for prime time.

Stata’s YouTube Channel was the brainchild of Karen Strope, StataCorp’s Director of Marketing, but I had something to do with it, too. Well, maybe more than something, but I’m a modest guy. Anyway, I thought it sounded like fun and recorded a few prototype videos. Annette Fett, StataCorp’s Graphic Designer, added the cool splash-screen and after a few experiments, we soon had 24 Blu-ray resolution videos. We’ve kicked off with videos covering topics such as a tour of Stata’s interface, how to create basic graphs, how to conduct many common statistical analyses, and more.

My personal favorite is the video entitled Combining Crosstabs and Descriptives because it’s relevant to nearly all Stata users and works well as a video demonstration.

Stata has over 9,000 pages of documentation included in PDF format, a built-in Help system, and a collection of books on general and special topics published by Stata Press, and an extensive collection of dialog boxes that make even the most complex graphs and analyses easy to perform.

So aren’t the videos, ahh, unnecessary?

The problem is, it’s cumbersome to describe how to use all of Stata’s features, especially dialog boxes, in a manual, even when you have 9,000 pages, and 9,000 pages tries even the most dedicated user’s patience.

In a 3-7 minutes video, we can show you how to create complicated graphs or a sophisticated structural equation model.

We have three audiences in mind.

1. Videos for non-Stata users, whom we call future Stata users; videos intended to provide a loosely guided tour of Stata’s features.
2. Videos for new Stata users, such as the person who might simply want to know “How do I calculate a twoway ANOVA in Stata?” or “How do I create a Pie Chart?”. These videos will get them up and running quickly and painlessly.
3. Videos for experienced Stata users who want to learn new tips and tricks.

There’s actually a fourth group that’s of interest, too; experienced Stata users teaching statistics or data analysis classes, who don’t want to spend valuable class time showing their students how to use Stata. They can refer their students to the relevant videos as homework and thus free class time for the teaching of statistics.

One of the fun things about working at StataCorp is that management doesn’t much use the word “no”. New ideas are more often met with the phrase, “well, let’s try it and see what happens”. So I’m trying this. My plan is to add a couple of videos to the channel every week or two as time permits. I have a list of topics I’d like to cover including things like multiple imputation, survey analysis, mixed models, Stata’s “immediate” commands (tabi, ttesti, csi, cci, etc…), and more examples using the SEM Builder.

However, I will take requests. If you have a suggested topic or a future video, leave a comment.

I’d like to keep the videos brief, between 3-7 minutes, so please don’t request feature-length films like “How to do survival analysis in Stata”. Similarly, topics that are only interesting to you and your two post-docs such as “Please describe the difference between the Laplacian Approximation and Adaptive Gauss-Hermite Quadrature in the xtmepoisson command” are not likely to see the light of day. But I am very interested in your ideas for small, bite-sized topics that will work in a video format.

Categories: Company Tags:

## Using Stata’s random-number generators, part 3, drawing with replacement

The topic for today is drawing random samples with replacement. If you haven’t read part 1 and part 2 of this series on random numbers, do so. In the series we’ve
discussed that

1. Stata’s runiform() function produces random numbers over the range [0,1). To produce such random numbers, type
. generate double u = runiform()


2. To produce continuous random numbers over [a,b), type
. generate double u = (b-a)*runiform() + a


3. To produce integer random numbers over [a,b], type
. generate ui = floor((b-a+1)*runiform() + a)


If b > 16,777,216, type

. generate long ui = floor((b-a+1)*runiform() + a)


4. To place observations in random order — to shuffle observations — type
. set seed #
. generate double u = runiform()
. sort u


5. To draw without replacement a random sample of n observations from a dataset of N observations, type
. set seed #
. sort variables_that_put_dta_in_unique_order
. generate double u = runiform()
. sort u
. keep in 1/n


If N>1,000, generate two random variables u1 and u2 in place of u, and substitute sort u1 u2 for sort u.

6. To draw without replacement a P-percent random sample, type
. set seed #
. keep if runiform() <= P/100


I’ve glossed over details, but the above is the gist of it.

Today I’m going to tell you

1. To draw a random sample of size n with replacement from a dataset of size N, type
. set seed #
. drop _all
. set obs n
. generate long obsno = floor(N*runiform()+1)
. sort obsno
. save obsnos_to_draw

. use your_dataset, clear
. generate long obsno = _n
. merge 1:m obsno using obsnos_to_draw, keep(match) nogen


2. You need to set the random-number seed only if you care about reproducibility. I’ll also mention that if N ≤ 16,777,216, it is not necessary to specify that new variable obsno be stored as long; the default float will be sufficient.

The above solution works whether n<N, n=N, or n>N.

Drawing samples with replacement

The solution to sampling with replacement n observations from a dataset of size N is

1. Draw n observation numbers 1, …, N with replacement. For instance, if N=4 and n=3, we might draw observation numbers 1, 3, and 3.

2. Select those observations from the dataset of interest. For instance, select observations 1, 3, and 3.

As previously discussed in part 1, to generate random integers drawn with replacement over the range [a, b], use the formula

generate varname = floor((b-a+1)*runiform() + a)

In this case, we want a=1 and b=N, and the formula reduces to,

generate varname = floor(N*runiform() + 1)

So the first half of our solution could read

. drop _all
. set obs n
. generate obsno = floor(N*runiform() + 1)


Now we are merely left with the problem of selecting those observations from our dataset, which we can do using merge by typing

. sort obsno
. save obsnos_to_draw
. use dataset_of_interest, clear
. generate obsno = _n
. merge 1:m obsno using obsnos_to_draw, keep(match) nogen


Let’s do an example. In part 2 of this series, I had a dataset with observations corresponding to playing cards:

. use cards

. list in 1/5

+-------------+
| rank   suit |
|-------------|
1. |  Ace   Club |
2. |    2   Club |
3. |    3   Club |
4. |    4   Club |
5. |    5   Club |
+-------------+


There are 52 observations in the dataset; I’m showing you just the first five. Let’s draw 10 cards from the deck, but with replacement.

The first step is to draw the observation numbers. We have N=52 cards in the deck, and we want to draw n=10, so we generate 10 random integers from the integers [1, 52]:

. drop _all

. set obs 10                            // we want n=10
obs was 0, now 10

. gen obsno = floor(52*runiform()+1)    // we draw from N=52

. list obsno                            // let's see what we have

+-------+
| obsno |
|-------|
1. |    42 |
2. |    52 |
3. |    16 |
4. |     9 |
5. |    40 |
|-------|
6. |    11 |
7. |    34 |
8. |    20 |
9. |    49 |
10. |    42 |
+-------+

If you look carefully at the list, you will see that observation number 42 repeats. It will be easier to see the duplicate if we sort the list,

. sort obsno
. list

+-------+
| obsno |
|-------|
1. |     9 |
2. |    11 |
3. |    16 |
4. |    20 |
5. |    34 |
|-------|
6. |    40 |
7. |    42 |     <- Obs. 42 repeats
8. |    42 |     <- See?
9. |    49 |
10. |    52 |
+-------+

An observation didn’t have to repeat, but it’s not surprising that one did because in drawing n=10 from N=52, we would expect one or more repeated cards about 60% of the time.

Anyway, we now know which cards we want, namely cards 9, 11, 16, 20, 34, 40, 42, 42 (again), 49, and 52.

The final step is to select those observations from cards.dta. The way to do that is to perform a one-to-many merge of cards.dta with the list above and keep the matches. Before we can do that, however, we must (1) save the list of observation numbers as a dataset, (2) load cards.dta, and (3) add a variable called obsno to it. Then we will be able to perform the merge. So let’s get that out of the way,

. save obsnos_to_draw                // 1. save the list above
file obsnos_to_draw.dta saved

. use cards                          // 2. load cards.dta

. gen obsno = _n                     // 3.  Add variable obsno to it

Now we can perform the merge:

. merge 1:m obsno using obsnos_to_draw, keep(matched) nogen

Result                           # of obs.
-----------------------------------------
not matched                             0
matched                                10
-----------------------------------------


I’ll list the result, but let me first briefly explain the command

merge 1:m obsno using obsnos_to_draw, keep(matched) nogen

merge …, we are performing the merge command,

1:m …, the merge is one-to-many,

using obsnos_to_draw …, we merge data in memory with obsnos_todraw.dta,

, keep(matched) …, we keep observations that appear in both datasets,

nogen, do not add variable _merge to the resulting dataset; _merge reports the source of the resulting observations; we said keep(matched) so we know each came from both sources.

And here is the result:

. list

+-------------------------+
|  rank      suit   obsno |
|-------------------------|
1. |     8      Club       9 |
2. |  Jack      Club      11 |
3. |   Ace     Spade      16 |
4. |     2   Diamond      20 |
5. |     6     Spade      34 |
|-------------------------|
6. |     8     Spade      40 |
7. |     9     Heart      42 |   <- Obs. 42 is here ...
8. | Queen     Spade      49 |
9. |  King     Spade      52 |
10. |     9     Heart      42 |   <- and here
+-------------------------+


We drew 10 cards — those are the observation numbers on the left. Variable obsno in our dataset records the original observation (card) number and really, we no longer need the variable. Anyway, obsno==42 appears twice, in real observations 7 and 10, and thus we drew the 9 of Hearts twice.

What could go wrong?

Not much can go wrong, it turns out. At this point, our generic solution is

. drop _all
. set obs n
. generate obsno = floor(n*runiform()+1)
. sort obsno
. save obsnos_to_draw

. use our_dataset
. gen obsno = _n
. merge 1:m obsno using obsnos_to_draw, keep(matched) nogen


If you study this code, there are two lines that might cause problems,

. generate obsno = floor(N*runiform()+1)


and

. generate obsno = _n


When you are looking for problems and see a generate or replace, think about rounding.

Let’s look at the right-hand side first. Both calculations produce integers over the range [1, N]. generate performs all calculations in double and the largest integer that can be stored without rounding is 9,007,199,254,740,992 (see previous blog post on precision). Stata allows datasets up to 2,147,483,646, so we can be sure that N is less than the maximum precise-integer double. There are no rounding issues on the right-hand side.

Next let’s look at the left-hand side. Variable obsno is being stored as a float because we did not instruct otherwise. The largest integer value that can be stored without rounding as a float (also covered in previous blog post on precision) is 16,777,216, and that is less than Stata’s 2,147,483,646 maximum observations. When N exceeds 16,777,216, the solution is to store obsno as a long. We could remember to use long on the rare occasion when dealing with such large datasets, but I’m going to change the generic solution to use longs in all cases, even when it’s unnecessary.

What else could go wrong? Well, we tried an example with n<N and that seemed to work. We should now try examples with n=N and n>N to verify there’s no hidden bug or assumption in our code. I’ve tried examples of both and the code works fine.

We’re done for today

That’s it. Drawing samples with replacement turns out to be easy, and that shouldn’t surprise us because we have a random-number generator that draws with replacement.

We could complicate the discussion and consider solutions that would run a bit more efficiently when n=N, which is of special interest in statistics because it is a key ingredient in bootstrapping, but we will not. The above solution works fine in the n=N case, and I always advise researchers to favor simple-even-if-slower solutions because they will probably save you time. Writing complicated code takes longer than writing simple code, and testing complicated code takes even longer. I know because that’s what we do at StataCorp.

Categories: Numerical Analysis Tags:

## Using Stata’s random-number generators, part 2, drawing without replacement

Last time I told you that Stata’s runiform() function generates rectangularly (uniformly) distributed random numbers over [0, 1), from 0 to nearly 1, and to be precise, over [0, 0.999999999767169356]. And I gave you two formulas,

1. To generate continuous random numbers between a and b, use

generate double u = (b-a)*runiform() + a

The random numbers will not actually be between a and b: they will be between a and nearly b, but the top will be so close to b, namely 0.999999999767169356*b, that it will not matter.

2. To generate integer random numbers between a and b, use

generate ui = floor((b-a+1)*runiform() + a)

I also mentioned that runiform() can solve a variety of problems, including

• shuffling data (putting observations in random order),
• drawing random samples without replacement (there’s a minor detail we’ll have to discuss because runiform() itself produces values drawn with replacement),
• drawing random samples with replacement (which is easier to do than most people realize),
• drawing stratified random samples (with or without replacement),
• manufacturing fictional data (something teachers, textbook authors, manual writers, and blog writers often need to do).

Today we will cover shuffling and drawing random samples without replacement — the first two topics on the list — and we will leave drawing random samples with replacement for next time. I’m going to tell you

1. To place observations in random order — to shuffle the observations — type
. generate double u = runiform()
. sort u

2. To draw without replacement a random sample of n observations from a dataset of N observations, type
. set seed #
. generate double u = runiform()
. sort u
. keep in 1/n


I will tell you that there are good statistical reasons for setting the random-number seed even if you don’t care about reproducibility.

If you do care about reproducibility, I will mention (1) that you need to use sort to put the original data in a known, reproducible order, before you generate the random variate u, and I will explain (2) a subtle issue that leads us to use different code for N≤1,000 and N>1,000. The code for for N≤1,000 is

. set seed #
. sort variables_that_put_data_in_unique_order
. generate double u = runiform()
. sort u
. keep in 1/n


and the code for N>1,000 is

. set seed #
. sort variables_that_put_data_in_unique_order
. generate double u1 = runiform()
. generate double u2 = runiform()
. sort u1 u2
. keep in 1/n


You can use the N>1,000 code for the N≤1,000 case.

3. To draw without replacement a P-percent random sample, type
. set seed #
. keep if runiform() <= P/100


There’s no issue in this case when N is large.

As I mentioned, we’ll discuss drawing random samples with replacement next time. Today, the topic is random samples without replacement. Let’s start.

Shuffling data

I have a deck of 52 cards, in order, the first four of which are

. list in 1/4

+-------------+
| rank   suit |
|-------------|
1. |  Ace   Club |
2. |    1   Club |
3. |    2   Club |
4. |    3   Club |
+-------------+


Well, actually I just have a Stata dataset with observations corresponding to playing cards. To shuffle the deck — to place the observations in random order — type

. generate double u = runiform()

. sort u


Having done that, here’s your hand,

. list in 1/5

+----------------------------+
|  rank      suit          u |
|----------------------------|
1. | Queen      Club   .0445188 |
2. |     5   Diamond   .0580662 |
3. |     7      Club   .0610638 |
4. |  King     Heart   .0907986 |
5. |     6     Spade   .0981878 |
+----------------------------+


and here’s mine:

. list in 6/10

+---------------------------+
| rank      suit          u |
|---------------------------|
6. |    8   Diamond   .1024369 |
7. |    5      Club   .1086679 |
8. |    8     Spade   .1091783 |
9. |    2     Spade   .1180158 |
10. |  Ace      Club   .1369841 |
+---------------------------+


All I did was generate random numbers — one per observation (card) — and then place the observations in ascending order of the random values. Doing that is equivalent to shuffling the deck. I used runiform() random numbers, meaning rectangularly distributed random numbers over [0, 1), but since I’m only exploiting the random-numbers’ ordinal properties, I could have used random numbers from any continuous distribution.

This simple, elegant, and obvious solution to shuffling data will play an important part of the solution to drawing observations without replacement. I have already more than hinted at the solution when I showed you your hand and mine.

Drawing n observations without replacement

Drawing without replacement is exactly the same problem as dealing cards. The solution to the physical card problem is to shuffle the cards and then draw the top cards. The solution to randomly selecting n from N observations is to put the N observations in random order and keep the first n of them.

. use cards, clear

. generate double u = runiform()

. sort u

. keep in 1/5
(47 observations deleted)

. list

+---------------------------+
| rank      suit          u |
|---------------------------|
1. |  Ace   Diamond   .0064866 |
2. |    6     Heart   .0087578 |
3. | King     Spade    .014819 |
4. |    3     Spade   .0955155 |
5. | King   Diamond   .1007262 |
+---------------------------+

. drop u


Reproducibility

You might later want to reproduce the analysis, meaning you do not want to draw another random sample, but you want to draw the same random sample. Perhaps you informally distributed some preliminary results and, of course, then discovered a mistake. You want to redistribute updated results and show that your mistake didn’t change results by much, and to drive home the point, you want to use the same samples as you used previously.

Part of the solution is to set the random-number seed. You might type

. set seed 49983882

. use cards, clear

. generate double u = runiform()

. sort u

. keep in 1/5


See help set seed in Stata. As a quick review, when you set the random-number seed, you set Stata’s random-number generator into a fixed, reproducible state, which is to say, the sequence of random numbers that runiform() produces is a function of the seed. Set the seed today to the same value as yesterday, and runiform() will produce the same sequence of random numbers today as it did yesterday. Thus, after setting the seed, if you repeat today exactly what you did yesterday, you will obtain the same results.

So imagine that you set the random number seed today to the value you set it to yesterday and you repeat the above commands. Even so, you might not get the same results! You will not get the same results if the observations in cards.dta are in a different order yesterday and today. Setting the seed merely ensures that if yesterday the smallest value of u was in observation 23, it will again be in observation 23 today (and it will be the same value). If yesterday, however, observation 23 was the 6 of Clubs, and today it’s the 7 of Hearts, then today you will select the 7 of Hearts in place of the 6 of Clubs.

So make sure the data are in the same order. One way to do that is put the dataset in a known order before generating the random values on which you will sort. For instance,

. set seed 49983882

. use cards, clear

. sort rank suit

. generate double u = runiform()

. sort u

. keep in 1/5


An even better solution would add the line

. by rank suit: assert _N==1


just before the generate. That line would check whether sorting on variables rank and suit uniquely orders the observations.

With cards.dta, you can argue that the assert is unnecessary, but not because you know each rank-suit combination occurs once. You have only my assurances about that. I recommend you never trust anyone’s assurances about data. In this case, however, you can argue that the assert is unnecessary because we sorted on all the variables in the dataset and thus uniqueness is not required. Pretend there are two Ace of Clubs in the deck. Would it matter that the first card was Ace of Clubs followed by Ace of Clubs as opposed to being the other way around? Of course it would not; the two states are indistinguishable.

So let’s assume there is another variable in the dataset, say whether there was a grease spot on the back of the card. Yesterday, after sorting, the ordering might have been,

     +---------------------------------+
| rank   suit   grease          u |
|---------------------------------|
1. |  Ace   Club      yes   .6012949 |
2. |  Ace   Club       no   .1859054 |
+---------------------------------+


and today,

     +---------------------------------+
| rank   suit   grease          u |
|---------------------------------|
1. |  Ace   Club       no   .6012949 |
2. |  Ace   Club      yes   .1859054 |
+---------------------------------+


If yesterday you selected the Ace of Clubs without grease, today you would select the Ace of Clubs with grease.

My recommendation is (1) sort on whatever variables put the data into a unique order, and then verify that, or (2) sort on all the variables in the dataset and then don’t worry whether the order is unique.

Ensuring a random ordering

Included in our reproducible solution but omitted from our base solution was setting the random-number seed,

. set seed 49983882


Setting the seed is important even if you don’t care about reproducibility. Each time you launch Stata, Stata sets the same random-number seed, namely 123456789, and that means that runiform() generates the same sequence of random numbers, and that means that if you generated all your random samples right after launching Stata, you would always select the same observations, at least holding N constant.

So set the seed, but don’t set it too often. You set the seed once per problem. If I wanted to draw 10,000 random samples from the same data, I could code:

  use dataset, clear
set seed 1702213
sort variables_that_put_data_in_unique_order
preserve
forvalues i=1(1)10000 {
generate double u = runiform()
sort u
keep in 1/n
drop u
save samplei', replace
restore, preserve
}


In the example I save each sample in a file. In real life, I seldom (never) save the samples; I perform whatever analysis on the samples I need and save the results, which I usually append into a single dataset. I don’t need to save the individual samples because I can recreate them.

And the result still might not be reproducible …

runiform() draws random-numbers with replacement. It is thus possible that two or more observations could have the same random values associated with them. Well yes, you’re thinking, I see that it’s possible, but surely it’s so unlikely that it just doesn’t happen. But it does happen:

. clear all

. set obs 100000
obs was 0, now 100000

. generate double u = runiform()

. by u, sort: assert _N==1
assertion is false
r(9);


In the 100,000-observation dataset I just created, I got a duplicate! By the way, I didn’t have to look hard for such an example, I got it the first time I tried.

I have three things I want to tell you:

1. Duplicates happen more often than you might guess.

2. Do not panic about the duplicates. Because of how Stata is written, duplicates do not lower the quality of the sample selected. I’ll explain.

3. Duplicates do interfere with reproducibility, however, and there is an easy way around that problem.

Let’s start with the chances of observing duplicates. I mentioned in passing last time that runiform() is a 32-bit random-number generator. That means runiform() can return any of 232 values. Their values are, in order,

          0  =  0
1/232  =  2.32830643654e-10
2/232  =  4.65661287308e-10
3/232  =  6.98491930962e-10
.
.
.
(232-2)/232  =  0.9999999995343387
(232-1)/232  =  0.9999999997671694


So what are the chances that in N draws with replacement from an urn containing these 232 values, that all values are distinct? The probability p that all values are distinct is

      232 * (232-1) * ... *(232-N)
p  =  ----------------------------
N*232


Here are some values for various values of N. p is the probability that all values are unique, and 1-p is the probability of observing one or more repeated values.

------------------------------------
N         p            1-p
------------------------------------
50   0.999999715    0.000000285
500   0.999970955    0.000029045
1,000   0.999883707    0.000116293
5,000   0.997094436    0.002905564
10,000   0.988427154    0.011572846
50,000   0.747490440    0.252509560
100,000   0.312187988    0.687812012
200,000   0.009498117    0.990501883
300,000   0.000028161    0.999971839
400,000   0.000000008    0.999999992
500,000   0.000000000    1.000000000
------------------------------------


In shuffling cards we generated N=52 random values. The probability of a repeated values is infinitesimal. In datasets of N=10,000, I expect to see repeated values 1% of the time. In datasets of N=50,000, I expect to see repeated values 25% of the time. By N=100,000, I expect to see repeated values more often than not. By N=500,000, I expect to see repeated value in virtually all sequences.

Even so, I promised you that this problem does not affect the randomness of the ordering. It does not because of how Stata’s sort command is written. Remember the basic solution,


. use dataset, clear

. generate double u = runiform()

. sort u

. keep in 1/n


Did you know sort has its own, private random-number generator built into it? It does, and sort uses its random-number generator to determine the order of tied observations. In the manuals we at StataCorp are fond of writing, “the ties will be ordered randomly” and a few sophisticated users probably took that to mean, “the ties will be ordered in a way that we at StataCorp do not know and even though they might be ordered in a way that will cause a bias in the subsequent analysis, because we don’t know, we’ll ignore the possibility.” But we meant it when wrote that the ties will be ordered randomly; we know that because we put a random number generator into sort to ensure the result. And that is why I can now write that repeated values of the runiform() function cause a reproducibility issue, but not a statistical issue.

The solution to the reproducibility issue is to draw two random numbers and use the random-number pair to order the observations:


. use dataset, clear

. sort varnames

. set seed #

. generate double u1 = runiform()

. generate double u2 = runiform()

. sort u1 u2

. keep in 1/n


You might wonder if we would ever need three random numbers. It is very unlikely. p, the probability of no problem, equals 1 to at least 5 digits for N=500,000. Of course, the chances of duplication are always nonzero. If you are concerned about this problem, you could add an assert to the code to verify that the two random numbers together do uniquely identify the observations:

. use dataset, clear

. sort varnames

. set seed #

. generate double u1 = runiform()

. generate double u2 = runiform()

. sort u1 u2

. by u1 u2: assert _N==1            // added line

. keep in 1/n


I do not believe that doing that is necessary.

Is using doubles necessary?

In the generation of random numbers in all of the above, note that I am storing them as doubles. For the reproducibility issue, that is important. As I mentioned in part 1, the 32-bit random numbers that runiform() produces will be rounded if forced into 23-bit floats.

Above I gave you a table of probabilities p that, in creating

. generate double u = runiform()


the values of u would be distinct. Here is what would happen if you instead stored u as a float:

                               u stored as
---------------------------------------------------------
-------- double ----------     ----------float ----------
N         p            1-p               p           1-p
-------------------------------------------------------------------
50   0.999999715    0.000000285     0.999853979    0.000146021
500   0.999970955    0.000029045     0.985238383    0.014761617
1,000   0.999883707    0.000116293     0.942190868    0.057809132
5,000   0.997094436    0.002905564     0.225346930    0.774653070
10,000   0.988427154    0.011572846     0.002574145    0.997425855
50,000   0.747490440    0.252509560     0.000000000    1.000000000
100,000   0.312187988    0.687812012     0.000000000    1.000000000
200,000   0.009498117    0.990501883     0.000000000    1.000000000
300,000   0.000028161    0.999971839     0.000000000    1.000000000
400,000   0.000000008    0.999999992     0.000000000    1.000000000
500,000   0.000000000    1.000000000     0.000000000    1.000000000
-------------------------------------------------------------------


Drawing without replacement P-percent random samples

We have discussed drawing without replacement n observations from N observations. The number of observations selected has been fixed. Say instead we wanted to draw a 10% random sample, meaning that we independently allow each observation to have a 10% chance of appearing in our sample. In that case, the final number of observations is expected to be 0.1*N, but it may (and probably will) vary from that. The basic solution for drawing a 10% random sample is

. keep if runiform() <= 0.10


and the basic solution for drawing a P% random sample is

. keep if runiform() <= P/100


It is unlikely to matter whether you code <= or < in the comparison. As you now know, runiform() produces values drawn from 232 possible values, and thus the chance of equality is 2-32 or roughly 0.000000000232830644. If you want a P% sample, however, theory says you should code <=.

If you care about reproducibility, you should expand the basic solution to read,

. set seed #

. use data, clear

. sort variables_that_put_data_in_unique_order

. keep if runiform() <= P/100


Below I draw a 10% sample from the card.dta:

. set seed 838

. use cards, clear

. sort rank suit

. keep if runiform() <= 10/100
(46 observations deleted)

. list

+-----------------+
|  rank      suit |
|-----------------|
1. |     2   Diamond |
2. |     2     Heart |
3. |     3      Club |
4. |     5     Heart |
5. |  Jack   Diamond |
|-----------------|
+-----------------+


We’re not done, but we’re done for today

In part 3 of this series I will discuss drawing random samples with replacement.

Categories: Numerical Analysis Tags:

## Using Stata’s random-number generators, part 1

I want to start a series on using Stata’s random-number function. Stata in fact has ten random-number functions:

1. runiform() generates rectangularly (uniformly) distributed random number over [0,1).
2. rbeta(a, b) generates beta-distribution beta(a, b) random numbers.
3. rbinomial(n, p) generates binomial(n, p) random numbers, where n is the number of trials and p the probability of a success.
4. rchi2(df) generates χ2 with df degrees of freedom random numbers.
5. rgamma(a, b) generates Γ(a, b) random numbers, where a is the shape parameter and b, the scale parameter.
6. rhypergeometric(N, K, n) generates hypergeometric random numbers, where N is the population size, K is the number of in the population having the attribute of interest, and n is the sample size.
7. rnbinomial(n, p) generates negative binomial -- the number of failures before the nth success -- random numbers, where p is the probability of a success. (n can also be noninteger.)
8. rnormal(μ, σ) generates Gaussian normal random numbers.
9. rpoisson(m) generates Poisson(m) random numbers.
10. rt(df) generates Student's t(df) random numbers.

You already know that these random-number generators do not really produce random numbers; they produce pseudo-random numbers. This series is not about that, so we'll be relaxed about calling them random-number generators.

You should already know that you can set the random-number seed before using the generators. That is not required but it is recommended. You set the seed not to obtain better random numbers, but to obtain reproducible random numbers. In fact, setting the seed too often can actually reduce the quality of the random numbers! If you don't know that, then read help set seed in Stata. I should probably pull out the part about setting the seed too often, expand it, and turn it into a blog entry. Anyway, this series is not about that either.

This series is about the use of random-number generators to solve problems, just as most users usually use them. The series will provide practical advice. I'll stay away from describing how they work internally, although long-time readers know that I won't keep the promise. At least I'll try to make sure that any technical details are things you really need to know. As a result, I probably won't even get to write once that if this is the kind of thing that interests you, StataCorp would be delighted to have you join our development staff.

runiform(), generating uniformly distributed random numbers

Mostly I'm going to write about runiform() because runiform() can solve such a variety of problems. runiform() can be used to solve,

• shuffling data (putting observations in random order),
• drawing random samples without replacement (there's a minor detail we'll have to discuss because runiform() itself produces values drawn with replacement),
• drawing random samples with replacement (which is easier to do than most people realize),
• drawing stratified random samples (with or without replacement),
• manufacturing fictional data (something teachers, textbook authors, manual writers, and blog writers often need to do).

runiform() generates uniformly, a.k.a. rectangularly distributed, random numbers over the interval, I quote from the manual, "0 to nearly 1".

Nearly 1? "Why not all the way to 1?" you should be asking. "And what exactly do you mean by nearly 1?"

The answer is that the generator is more useful if it omits 1 from the interval, and so we shaved just a little off. runiform() produces random numbers over [0, 0.999999999767169356].

Here are two useful formulas you should commit to memory.

1. If you want to generate continuous random numbers between a and b, use

generate double u = (b-a)*runiform() + a

The random numbers will not actually be between a and b, they will be between a and nearly b, but the top will be so close to b, namely 0.999999999767169356*b, that it will not matter.

Remember to store continuous random values as doubles.

2. If you want to generate integer random numbers between a and b, use

generate ui = floor((b-a+1)*runiform() + a)

In particular, do not even consider using the formula for continuous values but rounded to integers, which is to say, round(u) = round((b-a)*runiform() + a). If you use that formula, and if b-a>1, then a and b will be under represented by 50% each in the samples you generate!

I stored ui as a default float, so I am assuming that -16,777,216 ≤ a < b ≤ 16,777,216. If you have integers outside of that range, however, store as a long or double.

I’m going to spend the rest of this blog entry explaining the above.

First, I want to show you how I got the two formulas and why you must use the second formula for generating integer uniform deviates.

Then I want explain why we shaved a little from the top of runiform(), namely (1) while it wouldn’t matter for formula 1, it made formula 2 a little easier, (2) the code would run more quickly, (3) we could more easily prove that we had implemented the random-number generator correctly, and (4) anyone digging deeper into our random numbers would not be misled into thinking they had more than 32 bits of resolution. That last point will be important in a future blog entry.

Continuous uniforms over [a, b)

runiform() produces random numbers over [0, 1). It therefore obviously follows that (b-a)*runiform()+a produces number over [a, b). Substitute 0 for runiform() and the lower limit is obtained. Substitute 1 for runiform() and the upper limit is obtained.

I can tell you that in fact, runiform() produces random numbers over [0, (232-1)/232].

Thus (b-a)*runiform()+a produces random numbers over [a, ((232-1)/232)*b].

(232-1)/232) approximately equals 0.999999999767169356 and exactly equals 1.fffffffeX-01 if you will allow me to use %21x format, which Stata understands and which you can understand if you see my previous blog posting on precision.

Thus, if you are concerned about results being in the interval [a, b) rather than [a, b], you can use the formula

generate double u = ((b-a)*runiform() + a) / 1.fffffffeX-01

There are seven f’s followed by e in the hexadecimal constant. Alternatively, you could type

generate double u = ((b-a)*runiform() + a) * ((2^32-1)/2^32)

but multiplying by 1.fffffffeX-01 is less typing so I’d type that. Actually I wouldn’t type either one; the small difference between values lying in [a, b) or [a, b] is unimportant.

Integer uniforms over [a, b]

Whether we produce real, continuous random numbers over [a, b) or [a, b] may be unimportant, but if we want to draw random integers, the distinction is important.

runiform() produces continuous results over [0, 1).

(b-a)*runiform()+a produces continuous results over [a, b).

To produce integer results, we might round continuous results over segments of the number line:

           a    a+.5  a+1  a+1.5  a+2  a+2.5       b-1.5  b-1  b-.5    b
real line  +-----+-----+-----+-----+-----+-----------+-----+-----+-----+
int  line  |<-a->|<---a+1--->|<---a+2--->|           |<---b-1--->|<-b->|


In the diagram above, think of the numbers being produced by the continuous formula u=(b-a)*runiform()+a as being arrayed along the real line. Then imagine rounding those values, say by using Stata's round(u) function. If you rounded in that way, then

• Values of u between a and a+0.5 will be rounded to a.
• Values of u between a+0.5 and a+1.5 will be rounded to a+1.
• Values of u between a+1.5 and a+2.5 will be rounded to a+2.
• ...
• Values of u between b-1.5 and b-0.5 will be rounded to b-1.
• Values of u between b-0.5 and b-1 will be rounded to b.

Note that the width of the first and last intervals is half that of the other intervals. Given that u follows the rectangular distribution, we thus expect half as many values rounded to a and to b as to a+1 or a+2 or ... or b-1.

And indeed, that is exactly what we would see:

. set obs 100000
obs was 0, now 100000

. gen double u = (5-1)*runiform() + 1

. gen i = round(u)

. summarize u i

Variable |       Obs        Mean    Std. Dev.       Min        Max
-------------+--------------------------------------------------------
u |    100000    3.005933    1.156486   1.000012   4.999983
i |    100000     3.00489    1.225757          1          5

. tabulate i

i |      Freq.     Percent        Cum.
------------+-----------------------------------
1 |     12,525       12.53       12.53
2 |     24,785       24.79       37.31
3 |     24,886       24.89       62.20
4 |     25,284       25.28       87.48
5 |     12,520       12.52      100.00
------------+-----------------------------------
Total |    100,000      100.00


To avoid the problem we need to make the widths of all the intervals equal, and that is what the formula floor((b-a+1)*runiform() + a) does.

           a          a+1         a+2                     b-1          b          b+1
real line  +-----+-----+-----+-----+-----------------------+-----+-----+-----+-----+
int  line  |<--- a --->|<-- a+1 -->|                       |<-- b-1 -->|<--- b --->)


Our intervals are of equal width and thus we expect to see roughly the same number of observations in each:

. gen better = floor((5-1+1)*runiform() + 1)

. tabulate better

better |      Freq.     Percent        Cum.
------------+-----------------------------------
1 |     19,808       19.81       19.81
2 |     20,025       20.02       39.83
3 |     19,963       19.96       59.80
4 |     20,051       20.05       79.85
5 |     20,153       20.15      100.00
------------+-----------------------------------
Total |    100,000      100.00


So now you know why we shaved a little off the top when we implemented runiform(); it made the formula

floor((b-a+1)*runiform() + a):

easier. Our integer [a, b] formula did not have to concern itself that runiform() would sometimes — rarely — return 1. If runiform() did return the occasional 1, the simple formula above would produce the (correspondingly occasional) b+1.

How Stata calculates continuous random numbers

I’ve said that we shaved a little off the top, but the fact was that it was easier for us to do the shaving than not.

runiform() is based on the KISS random number generator. KISS produces 32-bit integers, meaning integers the range [0, 232-1], or [0, 4,294,967,295]. You might wonder how we converted that range to being continuous over [0, 1).

Start by thinking of the number KISS produces in its binary form:

b31b30b29b28b27b26b25b24b23b22b21b20b19b18b17b16b15b14b13b12b11b10b9b8b7b6b5b4b3b2b1b0

The corresponding integer is b31*231 + b31*230 + ... + b0*20. All we did was insert a binary point out front:

. b31b30b29b28b27b26b25b24b23b22b21b20b19b18b17b16b15b14b13b12b11b10b9b8b7b6b5b4b3b2b1b0

making the real value b31*2-1 + b30*2-2 + ... + b0*2-32. Doing that is equivalent to dividing by 2-32, except insertion of the binary point is faster. Nonetheless, if we had wanted runiform() to produce numbers over [0, 1], we could have divided by 232-1.

Anyway, if the KISS random number generator produced 3190625931, which in binary is

10111110001011010001011010001011

we converted that to

0.10111110001011010001011010001011

which equals 0.74287549 in base 10.

The largest number the KISS random number generator can produce is, of course,

11111111111111111111111111111111

and 0.11111111111111111111111111111111 equals 0.999999999767169356 in base 10. Thus, the runiform() implementation of KISS generates random numbers in the range [0, 0.999999999767169356].

I could have presented all of this mathematically in base 10: KISS produces integers in the range [0, 232-1], and in runiform() we divide by 232 to thus produce continuous numbers over the range [0, (232-1)/232]. I could have said that, but it loses the flavor and intuition of my longer explanation, and it would gloss over the fact that we just inserted the binary point. If I asked you, a base-10 user, to divide 232 by 10, you wouldn’t actually divide in the same way that they would divide by, say 9. Dividing by 9 is work. Dividing by 10 merely requires shifting the decimal point. 232 divided by 10 is obviously 23.2. You may not have realized that modern digital computers, when programmed by “advanced” programmers, follow similar procedures.

Oh gosh, I do get to say it! If this sort of thing interests you, consider a career at StataCorp. We’d love to have you.

Is it important that runiform() values be stored as doubles?

Sometimes it is important. It’s obviously not important when you are generating random integers using floor((b-a+1)*runiform() + a) and -16,777,216 ≤ a < b ≤ 16,777,216. Integers in that range fit into a float without rounding.

When creating continuous values, remember that runiform() produces 32 bits. floats store 23 bits and doubles store 52, so if you store the result of runiform() as a float, it will be rounded. Sometimes the rounding matters, and sometimes it does not. Next time, we will discuss drawing random samples without replacement. In that case, the rounding matters. In most other cases, including drawing random samples with replacement — something else for later — the rounding
does not matter. Rather than thinking hard about the issue, I store all my non-integer
random values as doubles.

Tune in for the next episode

Yes, please do tune in for the next episode of everything you need to know about using random-number generators. As I already mentioned, we’ll discuss drawing random samples without replacement. In the third installment, I’m pretty sure we’ll discuss random samples with replacement. After that, I’m a little unsure about the ordering, but I want to discuss oversampling of some groups relative to others and, separately, discuss the manufacturing of fictional data.

Am I forgetting something?

Categories: Numerical Analysis Tags:

## Using import excel with real world data

Stata 12′s new import excel command can help you easily import real-world Excel files into Stata. Excel files often contain header and footer information in the first few and last few rows of a sheet, and you may not want that information loaded. Also, the column labels used in the sheet are invalid Stata variable names and therefore cannot be loaded. Both of these issues can be easily solved using import excel.

Let’s start by looking at an Excel spreadsheet, metro_gdp.xls, that is downloaded from the Bureau of Economic Analysis website.

As you can see, the first five rows of the Excel file contain a description of the data, and rows 374 through 381 contain footer notes. We don’t want to load these rows into Stata. import excel has a cellrange() option that can help us avoid unwanted information being loaded.

With cellrange(), you specify the upper left cell and the lower right cell (using standard Excel notation) of the area of data you want loaded. In the file metro_gdp.xls, we want all the data from column A row 6 (upper left cell) to column L row 373 (lower right cell) loaded into Stata. To do this, we type

. import excel metro_gdp.xls, cellrange(A6:L373) clear


In Stata, we open the Data Editor to inspect the loaded data.

The first row of the data we loaded contained column labels. Because of these labels, import excel loaded all the data as strings. import excel again has an easy fix. We need to specify the firstrow option to tell import excel that the first row of data contains the variable names.

. import excel metro_gdp.xls, cellrange(A6:L373) firstrow clear


We again open the Data Editor to inspect the data.

The data are now in the correct format, but we are missing the year column labels. Stata does not accept numeric variable names, so import excel has to use the Excel column name (C, D, …) for the variable names instead of 2001, 2002, …. The simple solution is to rename the column headers in Excel to something like y2001, y2002, etc., before loading. You can also use Stata to rename the column headers. import excel saves the values in the first row of data as variable labels so that the information is not lost. If we describe the data, we will see all the column labels from the Excel file saved as variable labels.

. describe

Contains data
obs:           367
vars:            12
size:        37,067
-------------------------------------------------------------------------------
storage  display     value
variable name   type   format      label      variable label
-------------------------------------------------------------------------------
Fips            str5   %9s                    Fips
Area            str56  %56s                   Area
C               long   %10.0g                 2001
D               long   %10.0g                 2002
E               long   %10.0g                 2003
F               long   %10.0g                 2004
G               long   %10.0g                 2005
H               long   %10.0g                 2006
I               long   %10.0g                 2007
J               long   %10.0g                 2008
K               long   %10.0g                 2009
L               long   %10.0g                 2010
-------------------------------------------------------------------------------
Sorted by:
Note:  dataset has changed since last saved


We want to grab the variable label for each variable by using the extended macro function :variable label varname, create a valid lowercase variable name from that label by using the strtoname() and lower() functions, and rename the variable to the new name by using rename. We can do this with a foreach loop.

foreach var of varlist _all {
local label : variable label var'
local new_name = lower(strtoname("label'"))
rename var' new_name'
}


Now when we describe our data, they look like this:

. describe

Contains data
obs:           367
vars:            12
size:        37,067
-------------------------------------------------------------------------------
storage  display     value
variable name   type   format      label      variable label
-------------------------------------------------------------------------------
fips            str5   %9s                    Fips
area            str56  %56s                   Area
_2001           long   %10.0g                 2001
_2002           long   %10.0g                 2002
_2003           long   %10.0g                 2003
_2004           long   %10.0g                 2004
_2005           long   %10.0g                 2005
_2006           long   %10.0g                 2006
_2007           long   %10.0g                 2007
_2008           long   %10.0g                 2008
_2009           long   %10.0g                 2009
_2010           long   %10.0g                 2010
-------------------------------------------------------------------------------
Sorted by:
Note:  dataset has changed since last saved


One last thing we might want to do is to rename the year variables from _20## to y20##, which we can easily accomplish with rename:

. rename (_*) (y*)

. describe

Contains data
obs:           367
vars:            12
size:        37,067
-------------------------------------------------------------------------------
storage  display     value
variable name   type   format      label      variable label
-------------------------------------------------------------------------------
fips            str5   %9s                    Fips
area            str56  %56s                   Area
y2001           long   %10.0g                 2001
y2002           long   %10.0g                 2002
y2003           long   %10.0g                 2003
y2004           long   %10.0g                 2004
y2005           long   %10.0g                 2005
y2006           long   %10.0g                 2006
y2007           long   %10.0g                 2007
y2008           long   %10.0g                 2008
y2009           long   %10.0g                 2009
y2010           long   %10.0g                 2010
-------------------------------------------------------------------------------
Sorted by:
Note:  dataset has changed since last saved

Categories: Data Management Tags:

## The Penultimate Guide to Precision

There have recently been occasional questions on precision and storage types on Statalist despite all that I have written on the subject, much of it posted in this blog. I take that as evidence that I have yet to produce a useful, readable piece that addresses all the questions researchers have.

So I want to try again. This time I’ll try to write the ultimate piece on the subject, making it as short and snappy as possible, and addressing every popular question of which I am aware — including some I haven’t addressed before — and doing all that without making you wade with me into all the messy details, which I know I have a tendency to do.

I am hopeful that from now on, every question that appears on Statalist that even remotely touches on the subject will be answered with a link back to this page. If I succeed, I will place this in the Stata manuals and get it indexed online in Stata so that users can find it the instant they have questions.

What follows is intended to provide everything scientific researchers need to know to judge the effect of storage precision on their work, to know what can go wrong, and to prevent that. I don’t want to raise expectations too much, however, so I will entitle it …

THE PENULTIMATE GUIDE TO PRECISION

1. Contents

2. Numeric types

1.1 Stata provides five numeric types for storing variables, three of them integer types and two of them floating point.

1.2 The floating-point types are float and double.

1.3 The integer types are byte, int, and long.

1.4 Stata uses these five types for the storage of data.

1.5 Stata makes all calculations in double precision (and sometimes quad precision) regardless of the type used to store the data.

3. Floating-point types

2.1 Stata provides two IEEE 754-2008 floating-point types: float and double.

2.2 float variables are stored in 4 bytes.

2.3 double variables are stored in 8 bytes.

2.4 The ranges of float and double variables are

     Storage
type             minimum                maximum
-----------------------------------------------------
float     -3.40282346639e+ 38      1.70141173319e+ 38
double    -1.79769313486e+308      8.98846567431e+307
-----------------------------------------------------
In addition, float and double can record missing values
., .a, .b, ..., .z.

The above values are approximations. For those familiar with %21x floating-point hexadecimal format, the exact values are

     Storage
type                   minimum                maximum
-------------------------------------------------------
float   -1.fffffe0000000X+07f     +1.fffffe0000000X+07e
double  -1.fffffffffffffX+3ff     +1.fffffffffffffX+3fe
-------------------------------------------------------

Said differently, and less precisely, float values are in the open interval (-2128, 2127), and double values are in the open interval (-21024, 21023). This is less precise because the intervals shown in the tables are closed intervals.

4. Integer types

3.1 Stata provides three integer storage formats: byte, int, and long. They are 1 byte, 2 bytes, and 4 bytes, respectively.

3.2 Integers may also be stored in Stata’s IEEE 754-2008 floating-point storage formats float and double.

3.3 Integer values may be stored precisely over the ranges

     storage
type                   minimum                 maximum
------------------------------------------------------
byte                      -127                     100
int                    -32,767                  32,740
long            -2,147,483,647           2,147,483,620
------------------------------------------------------
float              -16,777,216              16,777,216
double  -9,007,199,254,740,992   9,007,199,254,740,992
------------------------------------------------------
In addition, all storage types can record missing values
., .a, .b, ..., .z.

The overall ranges of float and double were shown in (2.4) and are wider than the ranges for them shown here. The ranges shown here are the subsets of the overall ranges over which no rounding of integer values occurs.

5. Integer precision

4.1 (Automatic promotion.) For the integer storage types — for byte, int, and long — numbers outside the ranges listed in (3.3) would be stored as missing (.) except that storage types are promoted automatically. As necessary, Stata promotes bytes to ints, ints to longs, and longs to doubles. Even if a variable is a byte, the effective range is still [-9,007,199,254,740,992, 9,007,199,254,740,992] in the sense that you could change a value of a byte variable to a large value and that value would be stored correctly; the variable that was a byte would, as if by magic, change its type to int, long, or double if that were necessary.

4.2 (Data input.) Automatic promotion (4.1) applies after the data are input/read/imported/copied into Stata. When first reading, importing, copying, or creating data, it is your responsibility to choose appropriate storage types. Be aware that Stata’s default storage type is float, so if you have large integers, it is usually necessary to specify explicitly the types you wish to use.

If you are unsure of the type to specify for your integer variables, specify double. After reading the data, you can use compress to demote storage types. compress never results in a loss of precision.

4.3 Note that you can use the floating-point types float and double to store integer data.

4.3.1 Integers outside the range [-2,147,483,647, 2,147,483,620] must be stored as doubles if they are to be precisely recorded.

4.3.2 Integers can be stored as float, but avoid doing that unless you are certain they will be inside the range [-16,777,216, 16,777,216] not just when you initially read, import, or copy them into Stata, but subsequently as you make transformations.

4.3.3 If you read your integer data as floats, and assuming they are within the allowed range, we recommend that you change them to an integer type. You can do that simply by typing compress. We make that recommendation so that your integer variables will benefit from the automatic promotion described in (4.1).

4.4 Let us show what can go wrong if you do not follow our advice in (4.3). For the floating-point types — for float and double — integer values outside the ranges listed in (3.3) are rounded.

Consider a float variable, and remember that the integer range for floats is [-16,777,216, 16,777,216]. If you tried to store a value outside the range in the variable — say, 16,777,221 — and if you checked afterward, you would discover that actually stored was 16,777,220! Here are some other examples of rounding:

     desired value                            stored (rounded)
to store            true value             float value
------------------------------------------------------
maximum             16,777,216              16,777,216
maximum+1           16,777,217              16,777,216
------------------------------------------------------
maximum+2           16,777,218              16,777,218
------------------------------------------------------
maximum+3           16,777,219              16,777,220
maximum+4           16,777,220              16,777,220
maximum+5           16,777,221              16,777,220
------------------------------------------------------
maximum+6           16,777,222              16,777,222
------------------------------------------------------
maximum+7           16,777,223              16,777,224
maximum+8           16,777,224              16,777,224
maximum+9           16,777,225              16,777,224
------------------------------------------------------
maximum+10          16,777,226              16,777,226
------------------------------------------------------

When you store large integers in float variables, values will be rounded and no mention will be made of that fact.

And that is why we say that if you have integer data that must be recorded precisely and if the values might be large — outside the range ±16,777,216 — do not use float. Use long or use double; or just use the compress command and let automatic promotion handle the problem for you.

4.5 Unlike byte, int, and long, float and double variables are not promoted to preserve integer precision.

Float values are not promoted because, well, they are not. Actually, there is a deep reason, but it has to do with the use of float variables for their real purpose, which is to store non-integer values.

Double values are not promoted because there is nothing to promote them to. Double is Stata’s most precise storage type. The largest integer value Stata can store precisely is 9,007,199,254,740,992 and the smallest is -9,007,199,254,740,992.

Integer values outside the range for doubles round in the same way that float values round, except at absolutely larger values.

6. Floating-point precision

5.1 The smallest, nonzero value that can be stored in float and double is

     Storage
type      value          value in %21x         value in base 10
-----------------------------------------------------------------
float     ±2^-127    ±1.0000000000000X-07f   ±5.877471754111e-039
double    ±2^-1022   ±1.0000000000000X-3fe   ±2.225073858507e-308
-----------------------------------------------------------------

We include the value shown in the third column, the value in %21x, for those who know how to read it. It is described in (9), but it is unimportant. We are merely emphasizing that these are the smallest values for properly normalized numbers.

5.2 The smallest value of epsilon such that 1+epsilon ≠ 1 is

     Storage
type      epsilon       epsilon in %21x        epsilon in base 10
-----------------------------------------------------------------
float      ±2^-23     ±1.0000000000000X-017    ±1.19209289551e-07
double     ±2^-52     ±1.0000000000000X-034    ±2.22044604925e-16
-----------------------------------------------------------------

Epsilon is the distance from 1 to the next number on the floating-point number line. The corresponding unit roundoff error is u = ±epsilon/2. The unit roundoff error is the maximum relative roundoff error that is introduced by the floating-point number storage scheme.

The smallest value of epsilon such that x+epsilon ≠ x is approximately |x|*epsilon, and the corresponding unit roundoff error is ±|x|*epsilon/2.

5.3 The precision of the floating-point types is, depending on how you want to measure it,

     Measurement                           float              double
----------------------------------------------------------------
# of binary digits                       23                  52
# of base 10 digits (approximate)         7                  16

Relative precision                   ±2^-24              ±2^-53
... in base 10 (approximate)      ±5.96e-08           ±1.11e-16
----------------------------------------------------------------

Relative precision is defined as

                       |x - x_as_stored|
± max   ------------------
x            x

performed using infinite precision arithmetic, x chosen from the subset of reals between the minimum and maximum values that can be stored. It is worth appreciating that relative precision is a worst-case relative error over all possible numbers that can be stored. Relative precision is identical to roundoff error, but perhaps this definition is easier to appreciate.

5.4 Stata never makes calculations in float precision, even if the data are stored as float.

Stata makes double-precision calculations regardless of how the numeric data are stored. In some cases, Stata internally uses quad precision, which provides approximately 32 decimal digits of precision. If the result of the calculation is being stored back into a variable in the dataset, then the double (or quad) result is rounded as necessary to be stored.

5.5 (False precision.) Double precision is 536,870,912 times more accurate than float precision. You may worry that float precision is inadequate to accurately record your data.

Little in this world is measured to a relative accuracy of ±2-24, the accuracy provided by float precision.

Ms. Smith, it is reported, made $112,293 this year. Do you believe that is recorded to an accuracy of ±2-24*112,293, or approximately ±0.7 cents? David was born on 21jan1952, so on 27mar2012 he was 21,981 days old, or 60.18 years old. Recorded in float precision, the precision is ±60.18*2-24, or roughly ±1.89 minutes. Joe reported that he drives 12,234 miles per year. Do you believe that Joe’s report is accurate to ±12,234*2-24, equivalent to ±3.85 feet? A sample of 102,400 people reported that they drove, in total, 1,252,761,600 miles last year. Is that accurate to ±74.7 miles (float precision)? If it is, each of them is reporting with an accuracy of roughly ±3.85 feet. The distance from the Earth to the moon is often reported as 384,401 kilometers. Recorded as a float, the precision is ±384,401*2-24, or ±23 meters, or ±0.023 kilometers. Because the number was not reported as 384,401.000, one would assume float precision would be accurate to record that result. In fact, float precision is more than sufficiently accurate to record the distance because the distance from the Earth to the moon varies from 356,400 to 406,700 kilometers, some 50,300 kilometers. The distance would have been better reported as 384,401 ±25,150 kilometers. At best, the measurement 384,401 has relative accuracy of ±0.033 (it is accurate to roughly two digits). Nonetheless, a few things have been measured with more than float accuracy, and they stand out as crowning accomplishments of mankind. Use double as required. 7. Advice concerning 0.1, 0.2, … 6.1 Stata uses base 2, binary. Popular numbers such as 0.1, 0.2, 100.21, and so on, have no exact binary representation in a finite number of binary digits. There are a few exceptions, such as 0.5 and 0.25, but not many. 6.2 If you create a float variable containing 1.1 and list it, it will list as 1.1 but that is only because Stata’s default display format is %9.0g. If you changed that format to %16.0g, the result would appear as 1.1000000238419. This scares some users. If this scares you, go back and read (5.5) False Precision. The relative error is still a modest ±2-24. The number 1.1000000238419 is likely a perfectly acceptable approximation to 1.1 because the 1.1 was never measured to an accuracy of less than ±2-24 anyway. 6.3 One reason perfectly acceptable approximations to 1.1 such as 1.1000000238419 may bother you is that you cannot select observations containing 1.1 by typing if x==1.1 if x is a float variable. You cannot because the 1.1 on the right is interpreted as double precision 1.1. To select the observations, you have to type if x==float(1.1). 6.4 If this bothers you, record the data as doubles. It is best to do this at the point when you read the original data or when you make the original calculation. The number will then appear to be 1.1. It will not really be 1.1, but it will have less relative error, namely, ±2-53. 6.5 If you originally read the data and stored them as floats, it is still sometimes possible to recover the double-precision accuracy just as if you had originally read the data into doubles. You can do this if you know how many decimal digits were recorded after the decimal point and if the values are within a certain range. If there was one digit after the decimal point and if the data are in the range [-1,048,576, 1,048,576], which means the values could be -1,048,576, -1,048,575.9, …, -1, 0, 1, …, 1,048,575.9, 1,048,576, then typing . gen double y = round(x*10)/10 will recover the full double-precision result. Stored in y will be the number in double precision just as if you had originally read it that way. It is not possible, however, to recover the original result if x is outside the range ±1,048,576 because the float variable contains too little information. You can do something similar when there are two, three, or more decimal digits:  # digits to right of decimal pt. range command ----------------------------------------------------------------- 1 ±1,048,576 gen double y = round(x*10)/10 2 ± 131,072 gen double y = round(x*100)/100 3 ± 16,384 gen double y = round(x*1000)/1000 4 ± 1,024 gen double y = round(x*10000)/10000 5 ± 128 gen double y = round(x*100000)/100000 6 ± 16 gen double y = round(x*1000000)/1000000 7 ± 1 gen double y = round(x*10000000)/10000000 ----------------------------------------------------------------- Range is the range of x over which command will produce correct results. For instance, range = ±16 in the next-to-the-last line means that the values recorded in x must be -16 ≤ x ≤ 16. 8. Advice concerning exact data, such as currency data 7.1 Yes, there are exact data in this world. Such data are usually counts of something or are currency data, which you can think of as counts of pennies ($0.01) or the smallest unit in whatever currency you are using.

7.2 Just because the data are exact does not mean you need exact answers. It may still be that calculated answers are adequate if the data are recorded to a relative accuracy of ±2-24 (float). For most analyses — even of currency data — this is often adequate. The U.S. deficit in 2011 was $1.5 trillion. Stored as a float, this amount has a (maximum) error of ±2-24*1.5e+12 = ±$89,406.97. It would be difficult to imagine that ±$89,406.97 would affect any government decision maker dealing with the full$1.5 trillion.

7.3 That said, you sometimes do need to make exact calculations. Banks tracking their accounts need exact amounts. It is not enough to say to account holders that we have your money within a few pennies, dollars, or hundreds of dollars.

In that case, the currency data should be converted to integers (pennies) and stored as integers, and then processed as described in (4). Assuming the dollar-and-cent amounts were read into doubles, you can convert them into pennies by typing

. replace x = x*100

7.4 If you mistakenly read the currency data as a float, you do not have to re-read the data if the dollar amounts are between ±\$131,072. You can type

. gen double x_in_pennies = round(x*100)

This works only if x is between ±131,072.

8.1 Stata does all calculations in double (and sometimes quad) precision.

Float precision may be adequate for recording most data, but float precision is inadequate for performing calculations. That is why Stata does all calculations in double precision. Float precision is also inadequate for storing the results of intermediate calculations.

There is only one situation in which you need to exercise caution — if you create variables in the data containing intermediate results. Be sure to create all such variables as doubles.

8.2 The same quad-precision routines StataCorp uses are available to you in Mata; see the manual entries [M-5] mean, [M-5] sum, [M-5] runningsum, and [M-5] quadcross. Use them as you judge necessary.

10. How to interpret %21x format (if you care)

9.1 Stata has a display format that will display IEEE 754-2008 floating-point numbers in their full binary glory but in a readable way. You probably do not care; if so, skip this section.

9.2 IEEE 754-2008 floating-point numbers are stored as a pair of numbers (a, b) that are given the interpretation

z = a * 2b

where -2 < a < 2. In double precision, a is recorded with 52 binary digits. In float precision, a is recorded with 23 binary digits. For example, the number 2 is recorded in double precision as

a = +1.0000000000000000000000000000000000000000000000000000
b = +1

The value of pi is recorded as

a = +1.1001001000011111101101010100010001000010110100011000
b = +1

9.3 %21x presents a and b in base 16. The double-precision value of 2 is shown in %21x format as

+1.0000000000000X+001

and the value of pi is shown as

+1.921fb54442d18X+001

In the case of pi, the interpretation is

a = +1.921fb54442d18 (base 16)
b = +001             (base 16)

Reading this requires practice. It helps to remember that one-half corresponds to 0.8 (base 16). Thus, we can see that a is slightly larger than 1.5 (base 10) and b = 1 (base 10), so _pi is something over 1.5*21 = 3.

The number 100,000 in %21x is

+1.86a0000000000X+010

which is to say

a = +1.86a0000000000 (base 16)
b = +010             (base 16)

We see that a is slightly over 1.5 (base 10), and b is 16 (base 10), so 100,000 is something over 1.5*216 = 98,304.

9.4 %21x faithfully presents how the computer thinks of the number. For instance, we can easily see that the nice number 1.1 (base 10) is, in binary, a number with many digits to the right of the binary point:

. display %21x 1.1
+1.199999999999aX+000

We can also see why 1.1 stored as a float is different from 1.1 stored as a double:

. display %21x float(1.1)
+1.19999a0000000X+000

Float precision assigns fewer digits to the mantissa than does double precision, and 1.1 (base 10) in base 16 is a repeating hexadecimal.

9.5 %21x can be used as an input format as well as an output format. For instance, Stata understands

. gen x = 1.86ax+10

Stored in x will be 100,000 (base 10).

9.6 StataCorp has seen too many competent scientific programmers who, needing a perturbance for later use in their program, code something like

epsilon = 1e-8

It is worth examining that number:

. display %21x 1e-8
+1.5798ee2308c3aX-01b

That is an ugly number that can only lead to the introduction of roundoff error in their program. A far better number would be

epsilon = 1.0x-1b

Stata and Mata understand the above statement because %21x may be used as input as well as output. Naturally, 1.0x-1b looks just like what it is,

. display %21x 1.0x-1b
+1.0000000000000X-01b

and all those pretty zeros will reduce numerical roundoff error.

In base 10, the pretty 1.0x-1b looks like

. display %20.0g 1.0x-1b
7.4505805969238e-09

and that number may not look pretty to you, but you are not a base-2 digital computer.

Perhaps the programmer feels that epsilon really needs to be closer to 1e-8. In %21x, we see that 1e-8 is +1.5798ee2308c3aX-01b, so if we want to get closer, perhaps we use

epsilon = 1.6x-1b

9.7 %21x was invented by StataCorp.

11. Also see

How to read the %21x format

How to read the %21x format, part 2

Precision (yet again), Part I

Precision (yet again), Part II

Categories: Numerical Analysis Tags:

## Our users’ favorite commands

We recently had a contest on our Facebook page. To enter, contestants posted their favorite Stata command, feature, or just a post telling us why they love Stata. Contestants then asked their friends, colleagues, and fellow Stata users to vote for their entry by ‘Like’-ing the post. The prize, a copy of Stata/MP 12 (8-core).

The response was overwhelming! We enjoyed reading all the reasons why users love Stata so much, we wanted to share them with you.

The contest question was:

Do you have a favorite command or feature in Stata? What about a memorable experience when using the software? Post your favorite command, feature, or experience in the comments section of this post. Then, get your friends to “like” your comment. The person with the most “likes” by March 13, 2012, wins. The winner will receive a single-user copy of Stata/MP8 12 with PDF documentation.

We had many submissions with multiple “likes”. The winning submissions are:

 2,235 Likes,1st place: Rodrigo Briceno One of the most remarkable experiences with Stata was when I learned to use loops. Making repetitive procedures in so short amounts of time is really amazing! I LIKE STATA! 1,464 Likes,2nd place: Juan Jose Salcedo My Favorite STATA command is by far COLLAPSE! Getting descriptive statistics couldn’t be any easier! 140 Likes,3rd place: Tymon Sloczynski My favourite command is ‘oaxaca’, a user-written command (by Ben Jann from Zurich) which can be used to carry out the so-called Oaxaca-Blinder decomposition. I often use it in my research and it saves a lot of time – which easily makes it favourite!
Categories: Company Tags:

## Comparing predictions after arima with manual computations

Some of our users have asked about the way predictions are computed after fitting their models with arima. Those users report that they cannot reproduce the complete set of forecasts manually when the model contains MA terms. They specifically refer that they are not able to get the exact values for the first few predicted periods. The reason for the difference between their manual results and the forecasts obtained with predict after arima is the way the starting values and the recursive predictions are computed. While Stata uses the Kalman filter to compute the forecasts based on the state space representation of the model, users reporting differences compute their forecasts with a different estimator that is based on the recursions derived from the ARIMA representation of the model. Both estimators are consistent but they produce slightly different results for the first few forecasting periods.

When using the postestimation command predict after fitting their MA(1) model with arima, some users claim that they should be able to reproduce the predictions with

where

However, the recursive formula for the Kalman filter prediction is based on the shrunk error (See section 13.3 in Hamilton (1993) for the complete derivation based on the state space representation):

where

: is the estimated variance of the white noise disturbance

: corresponds to the unconditional mean for the error term

Let’s use one of the datasets available from our website to fit a MA(1) model and compute the predictions based on the Kalman filter recursions formulated above:

** Predictions with Kalman Filter recursions (obtained with -predict- **
use http://www.stata-press.com/data/r12/lutkepohl, clear
arima dlinvestment, ma(1)
predict double yhat

** Coefficient estimates and sigma^2 from ereturn list **
scalar beta = _b[_cons]
scalar theta = [ARMA]_b[L1.ma]
scalar sigma2 = e(sigma)^2

** pt and shrinking factor for the first two observations**
generate double pt=sigma2 in 1/2
generate double sh_factor=(sigma2)/(sigma2+theta^2*pt) in 2

** Predicted series and errors for the first two observations **
generate double my_yhat = beta
generate double myehat = sh_factor*(dlinvestment - my_yhat) in 2

** Predictions with the Kalman filter recursions **
quietly {
forvalues i = 3/91 {
replace my_yhat = my_yhat + theta*l.myehat in i'
replace pt= (sigma2*theta^2*L.pt)/(sigma2+theta^2*L.pt) in i'
replace sh_factor=(sigma2)/(sigma2+theta^2*pt)          in i'
replace myehat=sh_factor*(dlinvestment - my_yhat)       in i'
}
}


List the first 10 predictions (yhat from predict and my_yhat from the manual computations):

. list qtr yhat my_yhat pt sh_factor in 1/10

+--------------------------------------------------------+
|    qtr        yhat     my_yhat          pt   sh_factor |
|--------------------------------------------------------|
1. | 1960q1   .01686688   .01686688   .00192542           . |
2. | 1960q2   .01686688   .01686688   .00192542   .97272668 |
3. | 1960q3   .02052151   .02052151   .00005251   .99923589 |
4. | 1960q4   .01478403   .01478403   1.471e-06   .99997858 |
5. | 1961q1   .01312365   .01312365   4.125e-08    .9999994 |
|--------------------------------------------------------|
6. | 1961q2   .00326376   .00326376   1.157e-09   .99999998 |
7. | 1961q3   .02471242   .02471242   3.243e-11           1 |
8. | 1961q4   .01691061   .01691061   9.092e-13           1 |
9. | 1962q1   .01412974   .01412974   2.549e-14           1 |
10. | 1962q2   .00643301   .00643301   7.147e-16           1 |
+--------------------------------------------------------+


Notice that the shrinking factor (sh_factor) tends to 1 as t increases, which implies that after a few initial periods the predictions produced with the Kalman filter recursions become exactly the same as the ones produced by the formula at the top of this entry for the recursions derived from the ARIMA representation of the model.

Reference:

Hamilton, James. 1994. Time Series Analysis. Princeton University Press.

Categories: Statistics Tags:

## Building complicated expressions the easy way

Have you every wanted to make an “easy” calculation–say, after fitting a model–and gotten lost because you just weren’t sure where to find the degrees of freedom of the residual or the standard error of the coefficient? Have you ever been in the midst of constructing an “easy” calculation and was suddenly unsure just what e(df_r) really was? I have a solution.

It’s called Stata’s expression builder. You can get to it from the display dialog (Data->Other Utilities->Hand Calculator)

In the dialog, click the Create button to bring up the builder. Really, it doesn’t look like much:

I want to show you how to use this expression builder; if you’ll stick with me, it’ll be worth your time.

Let’s start over again and assume you are in the midst of an analysis, say,

. sysuse auto, clear
. regress price mpg length

Next invoke the expression builder by pulling down the menu Data->Other Utilities->Hand Calculator. Click Create. It looks like this:

Now click on the tree node icon (+) in front of “Estimation results” and then scroll down to see what’s underneath. You’ll see

Click on Scalars:

The middle box now contains the scalars stored in e(). N happens to be highlighted, but you could click on any of the scalars. If you look below the two boxes, you see the value of the e() scalar selected as well as its value and a short description. e(N) is 74 and is the “number of observations”.

It works the same way for all the other categories in the box on the left: Operators, Functions, Variables, Coefficients, Estimation results, Returned results, System parameters, Matrices, Macros, Scalars, Notes, and Characteristics. You simply click on the tree node icon (+), and the category expands to show what is available.

You have now mastered the expression builder!

Let’s try it out.

Say you want to verify that the p-value of the coefficient on mpg is correctly calculated by regress–which reports 0.052–or more likely, you want to verify that you know how it was calculated. You think the formula is

or, as an expression in Stata,

2*ttail(e(df_r), abs(_b[mpg]/_se[mpg]))

But I’m jumping ahead. You may not remember that _b[mpg] is the coefficient on variable mpg, or that _se[mpg] is its corresponding standard error, or that abs() is Stata’s absolute value function, or that e(df_r) is the residual degrees of freedom from the regression, or that ttail() is Stata’s Student’s t distribution function. We can build the above expression using the builder because all the components can be accessed through the builder. The ttail() and abs() functions are in the Functions category, the e(df_r) scalar is in the Estimation results category, and _b[mpg] and _se[mpg] are in the Coefficients category.

What’s nice about the builder is that not only are the item names listed but also a definition, syntax, and value are displayed when you click on an item. Having all this information in one place makes building a complex expression much easier.

Another example of when the expression builder comes in handy is when computing intraclass correlations after xtmixed. Consider a simple two-level model from Example 1 in [XT] xtmixed, which models weight trajectories of 48 pigs from 9 successive weeks:

. use http://www.stata-press.com/data/r12/pig
. xtmixed weight week || id:, variance

The intraclass correlation is a nonlinear function of variance components. In this example, the (residual) intraclass correlation is the ratio of the between-pig variance, var(_cons), to the total variance, between-pig variance plus residual (within-pig) variance, or var(_cons) + var(residual).

The xtmixed command does not store the estimates of variance components directly. Instead, it stores them as log standard deviations in e(b) such that _b[lns1_1_1:_cons] is the estimated log of between-pig standard deviation, and _b[lnsig_e:_cons] is the estimated log of residual (within-pig) standard deviation. So to compute the intraclass correlation, we must first transform log standard deviations to variances:

exp(2*_b[lns1_1_1:_cons])
exp(2*_b[lnsig_e:_cons])

The final expression for the intraclass correlation is then

exp(2*_b[lns1_1_1:_cons]) / (exp(2*_b[lns1_1_1:_cons])+exp(2*_b[lnsig_e:_cons]))

The problem is that few people remember that _b[lns1_1_1:_cons] is the estimated log of between-pig standard deviation. The few who do certainly do not want to type it. So use the expression builder as we do below:

In this case, we’re using the expression builder accessed from Stata’s nlcom dialog, which reports estimated nonlinear combinations along with their standard errors. Once we press OK here and in the nlcom dialog, we’ll see

. nlcom (exp(2*_b[lns1_1_1:_cons])/(exp(2*_b[lns1_1_1:_cons])+exp(2*_b[lnsig_e:_cons])))

_nl_1:  exp(2*_b[lns1_1_1:_cons])/(exp(2*_b[lns1_1_1:_cons])+exp(2*_b[lnsig_e:_cons]))

------------------------------------------------------------------------------
weight |      Coef.   Std. Err.      z    P>|z|     [95% Conf. Interval]
-------------+----------------------------------------------------------------
_nl_1 |   .7717142   .0393959    19.59   0.000     .6944996    .8489288
------------------------------------------------------------------------------`

The above could easily be extended to computing different types of intraclass correlations arising in higher-level random-effects models. The use of the expression builder for that becomes even more handy.

Categories: Statistics Tags:

## The next leap second will be on June 30th, maybe

Leap seconds are the extra seconds inserted every so often to keep precise atomic clocks better synchronized with the rotation of the Earth. Scheduled for June 30th is the extra second 23:59:60 inserted between 23:59:59 and 00:00:00. Or maybe not.

Tomorrow or Friday a vote may be held at the International Telecommuncation Union (ITU) meeting in Geneva to abolish the leap second from the definition of UTC (Coordinated Universial Time). Which would mean StataCorp would not have to post an update to Stata to keep the %tC format working correctly.

As I’ve blogged before — scroll down to “Why Stata has two datetime encodings” in Using dates and times from other software — Stata supports both UTC time (%tC) and constant-86,400-seconds/day time (%tc). Stata does that because some data are collected using leap-second corrected time, and some uncorrected. Stata is unique or nearly unique in providing both time formats.

I read that Google does something very clever: they strech the last second of the year out when a leap second occurs, so the data they collect does not end up with ugly times like 23:59:60, and so that it can be more easily processed by software that assumes a constant 86,400 seconds per day.

The IT industry and a number of others, I gather, are pretty united about the benefits of scrapping the leap second.

The vote is predicted to go against continuing the leap second, according to The Economist magazine. The United States and France are for abolishing the leap second. Britain, Canada, and China are believed to be for continuing it. Some 192 countries will get to vote.

Whichever way the vote goes, I would like to remind readers of advice I previously offered to help alleviate the need for leap seconds: Face west and throw rocks. As I previously noted, the benefit will be transitory if the rocks land back on Earth, so you need to throw the rocks really hard. Having now thought more about this issue, a less strenuous way occurs to me: Push rocks downhill or carry them toward the poles, and preferably do both. These suggestions are designed to attack the real problem, which is that the Earth is currently rotating too slowly.

Categories: Data Management Tags: