Stata: Counting number of consecutive occurrences of a pre-defined length - stata

Observations in my data set contain the history of moves for each player. I would like to count the number of consecutive series of moves of some pre-defined length (2, 3 and more than 3 moves) in the first and the second halves of the game. The sequences cannot overlap, i.e. the sequence 1111 should be considered as a sequence of the length 4, not 2 sequences of length 2. That is, for an observation like this:
+-------+-------+-------+-------+-------+-------+-------+-------+
| Move1 | Move2 | Move3 | Move4 | Move5 | Move6 | Move7 | Move8 |
+-------+-------+-------+-------+-------+-------+-------+-------+
| 1 | 1 | 1 | 1 | . | . | 1 | 1 |
+-------+-------+-------+-------+-------+-------+-------+-------+
…the following variables should be generated:
Number of sequences of 2 in the first half =0
Number of sequences of 2 in the second half =1
Number of sequences of 3 in the first half =0
Number of sequences of 3 in the second half =0
Number of sequences of >3 in the first half =1
Number of sequences of >3 in the second half = 0
I have two potential options of how to proceed with this task but neither of those leads to the final solution:
Option 1: Elaborating on Nick’s tactical suggestion to use strings (Stata: Maximum number of consecutive occurrences of the same value across variables), I have concatenated all “move*” variables and tried to identify the starting position of a substring:
egen test1 = concat(move*)
gen test2 = subinstr(test1,"11","X",.) // find all consecutive series of length 2
There are several problems with Option 1:
(1) it does not account for cases with overlapping sequences (“1111” is recognized as 2 sequences of 2)
(2) it shortens the resulting string test2 so that positions of X no longer correspond to the starting positions in test1
(3) it does not account for variable length of substring if I need to check for sequences of the length greater than 3.
Option 2: Create an auxiliary set of variables to identify the starting positions of the consecutive set (sets) of the 1s of some fixed predefined length. Building on the earlier example, in order to count sequences of length 2, what I am trying to get is an auxiliary set of variables that will be equal to 1 if the sequence of started at a given move, and zero otherwise:
+-------+-------+-------+-------+-------+-------+-------+-------+
| Move1 | Move2 | Move3 | Move4 | Move5 | Move6 | Move7 | Move8 |
+-------+-------+-------+-------+-------+-------+-------+-------+
| 0 | 0 | 0 | 0 | 0 | 0 | 1 | 0 |
+-------+-------+-------+-------+-------+-------+-------+-------+
My code looks as follows but it breaks when I am trying to restart counting consecutive occurrences:
quietly forval i = 1/42 {
gen temprow`i' =.
egen rowsum = rownonmiss(seq1-seq`i') //count number of occurrences
replace temprow`i'=rowsum
mvdecode seq1-seq`i',mv(1) if rowsum==2
drop rowsum
}
Does anyone know a way of solving the task?

Assume a string variable concatenating all moves all (the name test1 is hardly evocative).
FIRST TRY: TAKING YOUR EXAMPLE LITERALLY
From your example with 8 moves, the first half of the game is moves 1-4 and the second half moves 5-8. Thus there is for each half only one way to have >3 moves, namely that there are 4 moves. In that case each substring will be "1111" and counting reduces to testing for the one possibility:
gen count_1_4 = substr(all, 1, 4) == "1111"
gen count_2_4 = substr(all, 5, 4) == "1111"
Extending this approach, there are only two ways to have 3 moves in sequence:
gen count_1_3 = inlist(substr(all, 1, 4), "111.", ".111")
gen count_2_3 = inlist(substr(all, 5, 4), "111.", ".111")
In similar style, there can't be two instances of 2 moves in sequence in each half of the game as that would qualify as 4 moves. So, at most there is one instance of 2 moves in sequence in each half. That instance must match either of two patterns, "11." or ".11". ".11." is allowed, so either includes both. We must also exclude any false match with a sequence of 3 moves, as just mentioned.
gen count_1_2 = (strpos(substr(all, 1, 4), "11.") | strpos(substr(all, 1, 4), ".11") ) & !count_1_3
gen count_2_2 = (strpos(substr(all, 5, 4), "11.") | strpos(substr(all, 5, 4), ".11") ) & !count_2_3
The result of each strpos() evaluation will be positive if a match is found and (arg1 | arg2) will be true (1) if either argument is positive. (For Stata, non-zero is true in logical evaluations.)
That's very much tailored to your particular problem, but not much worse for that.
P.S. I didn't try hard to understand your code. You seem to be confusing subinstr() with strpos(). If you want to know positions, subinstr() cannot help.
SECOND TRY
Your last code segment implies that your example is quite misleading: if there can be 42 moves, the approach above can not be extended without pain. You need a different approach.
Let's suppose that the string variable all can be 42 characters long. I will set aside the distinction between first and second halves, which can be tackled by modifying this approach. At its simplest, just split the history into two variables, one for the first half and one for the second and repeat the approach twice.
You can clone the history by
clonevar work = all
gen length1 = .
gen length2 = .
and set up your count variables. Here count_4 will hold counts of 4 or more.
gen count_4 = 0
gen count_3 = 0
gen count_2 = 0
First we look for move sequences of length 42, ..., 2. Every time we find one, we blank it out and bump up the count.
qui forval j = 42(-1)2 {
replace length1 = length(work)
local pattern : di _dup(`j') "1"
replace work = subinstr(work, "`pattern'", "", .)
replace length2 = length(work)
if `j' >= 4 {
replace count4 = count4 + (length1 - length2) / `j'
}
else if `j' == 3 {
replace count3 = count3 + (length1 - length2) / 3
}
else if `j' == 2 {
replace count2 = count2 + (length1 - length2) / 2
}
}
The important details here are
If we delete (repeated instances of) a pattern and measure the change in length, we have just deleted (change in length) / (length of pattern) instances of that pattern. So, if I look for "11" and found that the length decreased by 4, I just found two instances.
Working downwards and deleting what we found ensures that we don't find false positives, e.g. if "1111111" is deleted, we don't find later "111111", "11111", ..., "11" which are included within it.
Deletion implies that we should work on a clone in order not to destroy what is of interest.

Related

Finding the first occurence of 1-digit number in a list in Raku

I've got a number of lists of various lengths. Each of the lists starts with some numbers which are multiple digits but ends up with a recurring 1-digit number. For instance:
my #d = <751932 512775 64440 59994 9992 3799 423 2 2 2 2>;
my #e = <3750 3177 4536 4545 686 3 3 3>;
I'd like to find the position of the first occurence of the 1-digit number (for #d 7 and for #e 5) without constructing any loop. Ideally a lambda (or any other practical thing) should iterate over the list using a condition such as $_.chars == 1 and as soon as the condition is fulfilled it should stop and return the position. Instead of returing the position, it might as well return the list up until the 1-digit number; changes and improvisations are welcome. How to do it?
You want the :k modifier on first:
say #d.first( *.chars == 1, :k ) # 7
say #e.first( *.chars == 1, :k ) # 5
See first for more information.
To answer your second part of the question:
say #d[^$_] with #d.first( *.chars == 1, :k );
# (751932 512775 64440 59994 9992 3799 423)
say #e[^$_] with #e.first( *.chars == 1, :k );
# (3750 3177 4536 4545 686)
Make sure that you use the with to ensure you only show the slice if first actually found an entry.
See with for more information.

Coding dichotomous variables in Stata

I have a set of dichotomous variables for firm size:
emp1_2 (i.e. firm with 1 or 2 employed people, including the owner), emp3_9, emp10_19, emp20_49, emp50_99, emp100_249, emp250_499, emp500, plus I do not have information on 27 firms size but I have an educated guess that they are large firms.
I want to create a dichotomous variable for a firm being a "small firm"; therefore, this variable equals 1 when emp1_2==1 | emp3_9==1 | emp10_19==1 equals 1, and 0 otherwise.
To my understanding of Stata, of which I am a bare user, the two following methods to construct dichotomous variables should be equivalent.
Method 1)
gen lar_firm = 0
replace lar_firm = 1 if emp1_2==1 | emp3_9==1 | emp10_19==1
Method 2)
gen lar_firm = (emp1_2 | emp3_9 | emp10_19)
Instead I have found out that with method 2) lar_firm equals 1 for firms for which emp1_2 | emp3_9 | emp10_19 and for firms that do not enter in any of the categories (i.e. emp1_2, emp3_9, emp10_19, emp20_49, emp50_99, emp100_249, emp250_499, emp500) but for which I have an educated guess that they are large firms.
I am wondering whether there is some subtle difference between the two methods. I though they should lead to equal outcomes.
When you do
gen lar_firm = emp1_2 | emp3_9 | emp10_19
you're testing if
(emp1_2 != 0) | (emp3_9 != 0) |(emp10_19 != 0)
In particular, missing values . are different from 0: they are greater in fact.
For more information:
http://www.stata.com/support/faqs/data-management/logical-expressions-and-missing-values/

How to populate missing values for string variable in a column based on fixed criteria

To populate missing data with a fixed range of values
I would like to check how to populate column aktype with a range of values (the range of values for the same pidlink are always fixed at 11 types of values listed below) for those cells with missing values. I have about 17,000+ observations that are missing.
The range of values are as follows:
A
B
C
D
E
G
H
I
J
K
L
I have tried the following command but it does not work:-
foreach x of varlist aktype=1/11 {
replace aktype = "A" in 1 if aktype==""
replace aktype = "B" in 2 if aktype==""
replace aktype = "C" in 3 if aktype==""
replace aktype = "D" in 4 if aktype==""
replace aktype = "E" in 5 if aktype==""
replace aktype = "G" in 6 if aktype==""
replace aktype = "H" in 7 if aktype==""
replace aktype = "I" in 8 if aktype==""
replace aktype = "J" in 9 if aktype==""
replace aktype = "K" in 10 if aktype==""
replace aktype = "L" in 11 if aktype==""
}
Would appreciate it if you could advise on the right command to use. Many thanks!
I would generate a variable AK that has letters A-K in positions 1-11 (and 12-22, and 23-33, and so on). The replace missing values with the value of this variable AK.
* generate data
clear
set obs 20
generate aktype = ""
replace aktype = "foo" in 1/1
replace aktype = "bar" in 10/12
* generate variable with letters A-K
generate AK = char(65 + mod(_n - 1, 11))
* fill missing values
replace aktype = AK if missing(aktype)
list
This yields the following.
. list
+-------------+
| aktype AK |
|-------------|
1. | foo A |
2. | B B |
3. | C C |
4. | D D |
5. | E E |
|-------------|
This first addresses the comment "it does not work".
Generally, in this kind of forum you should always be specific and say exactly what happens, namely where the code breaks down and what the result is (e.g. what error message you get). If necessary, add why that is not what is wanted.
Specifically, in this case Stata would get no further than
foreach x of varlist aktype=1/11
which is illegal (as well as unclear to Stata programmers).
You can loop over a varlist. In this case looping over a single variable aktype is legal. (It is usually pointless, but that's style, not syntax.) So this is legal:
foreach x of varlist aktype
By the way, you define x as the loop argument, but never refer to it inside the loop. That isn't illegal, but it is unusual.
You can also loop over a numlist, e.g.
foreach x of numlist 1/11
although
forval x = 1/11
is a more direct way of doing that. All this follows from the syntax diagrams for the commands concerned, where whatever is not explicitly allowed is forbidden.
On occasions when you need to loop over a varlist and a numlist you will need to use different syntax, but what is best depends on the precise problem.
Now second to the question: I can't see any kind of rule in the question for which values get assigned A through L, so can't advise positively.

Grabbing columns with special characters and upper case letters

I have a data frame and I'm trying to loop through the data frame to identify those columns which contain a special character or which are all capital letters.
I have tried a few things but nothing where I'm apple to catch the column names within the loop.
data = data.frame(one=c(1,3,5,1,3,5,1,3,5,1,3,5), two=c(1,3,5,1,3,5,1,3,5,1,3,5),
thr=c("A","B","D","E","F","G","H","I","J","H","I","J"),
fou=c("A","B","D","A","B","D","A","B","D","A","B","D"),
fiv=c(1,3,5,1,3,5,1,3,5,1,3,5),
six=c("A","B","D","E","F","G","H","I","J","H","I","J"),
sev=c("A","B","D","A","B","D","A","B","D","A","B","D"),
eig=c("A","B","D","A","B","D","A","B","D","A","B","D"),
nin=c(1.24,3.52,5.33,1.44,3.11,5.33,1.55,3.66,5.33,1.32,3.54,5.77),
ten=c(1:12),
ele=rep(1,12),
twe=c(1,2,1,2,1,2,1,2,1,2,1,2),
thir=c("THiS","THAT34","T(&*(", "!!!","#$#","$Q%J","who","THIS","this","this","this","this"),
stringsAsFactors = FALSE)
data
colls <- c()
spec=c("$","%","&")
for( col in names(data) ) {
if( length(strings[stringr::str_detect(data[,col], spec)]) >= 1 ){
print("HORRAY")
colls <- c(collls, col)
}
else print ("NOOOOOOOOOO")
}
for( col in names(data) ) {
if( any(data[,col]) %in% spec ){
print("HORRAY")
colls <- c(collls, col)
}
else print ("NOOOOOOOOOO")
}
Can anyone shed light on a good way to tackle this problem.
EDIT:
The end goal is to have a vector with a name of column names which meet that criteria. Sorry for my poor SO question, but hopefully this will help with what I'm trying to do
I would use grep() to search for the pattern you are interested in. See here.
[:upper:] Matches any upper case letters.
Combining it with anchors (^,$) and match one or more times (+) gives ^[[:upper:]]+$ and should only match entries completely in capitals.
The following would match the special characters in your toy data set (but is not guaranteed to match all special characters in your real data set i.e form feeds, carriage returns)
[:punct:] #Matches punctuation - ! " # $ % & ' ( ) * + , - . / : ; < = > ? # [ \ ] ^ _ ` { | } ~.
Note that rather than use [:punct:] you could define your special characters manually.
We can try the resultant code on the first row of your data set:
#Using grepl() rather than grep() so that we return a list of logical values.
grepl(x= data[1,], pattern = "^[[:upper:]]+$|[[:punct:]]")
[1] FALSE FALSE TRUE TRUE FALSE TRUE TRUE TRUE TRUE FALSE FALSE FALSE FALSE
This gives us our expected response except for column nine which has the value 1.24. Here the decimal point is being recognised as punctuation and is being flagged as a match.
We can add a "negative lookahead assertion" - (?!\\.) - to remove any periods from consideration, before they are even tested for being punctuation characters. Note we use \ to escape the period.
grepl(x= data[1,], perl = TRUE, pattern = "(?!\\.)(^[[:upper:]]+$|[[:punct:]])")
[1] FALSE FALSE TRUE TRUE FALSE TRUE TRUE TRUE FALSE FALSE FALSE FALSE TRUE
This returns a better response - it now no longer matches decimal places. NOTE: This might not be what you want as this pattern also won't match any fullstops in character fields. You would need to refine the pattern further.
Rather than use a 'for loop' to reiterate this code across every row in your dataframe I would use vectorization instead which is 'more R like'.
To do this we must convert our script into a function which we will call with apply()
myFunction <- function(x){
matches <- grepl(x= x, perl = TRUE, pattern = "(?!\\.)(^[[:upper:]]+$|[[:punct:]])")
#Given a set of logical vectors 'matches', is at least one of the values true? using any()
return(any(matches))
}
apply(X = data, 1, myFunction)
The 1 above instructs apply() to reiterate across rows rather than columns.
[1] TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE
In your example data set all rows have an entry containing a special character or a string of all capital letters. This is unsurprising as many columns in your example data set are a list of single capital letters.
If you are just interested in which values in column thirteen fit the stated criteria you can use:
matches <- grepl(x= data$thir, perl = TRUE, pattern = "(?!\\.)(^[[:upper:]]+$|[[:punct:]])")
matches
[1] FALSE FALSE TRUE TRUE TRUE TRUE FALSE TRUE FALSE FALSE FALSE FALSE
To subset your dataframe on matching rows:
data[matches,]
one two thr fou fiv six sev eig nin ten ele twe thir
3 5 5 D D 5 D D D 5.33 3 1 1 T(&*(
4 1 1 E A 1 E A A 1.44 4 1 2 !!!
5 3 3 F B 3 F B B 3.11 5 1 1 #$#
6 5 5 G D 5 G D D 5.33 6 1 2 $Q%J
8 3 3 I B 3 I B B 3.66 8 1 2 THIS
To subset your dataframe on non-matching rows:
data[!matches,]
one two thr fou fiv six sev eig nin ten ele twe thir
1 1 1 A A 1 A A A 1.24 1 1 1 THiS
2 3 3 B B 3 B B B 3.52 2 1 2 THAT34
7 1 1 H A 1 H A A 1.55 7 1 1 who
9 5 5 J D 5 J D D 5.33 9 1 1 this
10 1 1 H A 1 H A A 1.32 10 1 2 this
11 3 3 I B 3 I B B 3.54 11 1 1 this
12 5 5 J D 5 J D D 5.77 12 1 2 this
Note that the regular expression used doesn't match THAT34 as it isn't composed wholly of capitalised letters, having the number 34 at the end.
EDIT:
To get a list of column names identifying columns that fulfill the criteria in your edit use myFunction described above with:
colnames(data)[apply(X = data, 2, myFunction)]
"thr" "fou" "six" "sev" "eig" "thir"
The number in apply() changes from 1 to 2 to reiterate across columns rather than rows. We pass the output from apply(), a list of logical matches (TRUE or FALSE), to colnames(data) - this returns the matching column names via subsetting.
I would collapse the data into strings (one string per row)
strings = apply(data, 1, paste, collapse = "")
contains_only_caps = strings == toupper(strings)
strings[contains_only_caps]
# [1] "33BB3BBB3.52 212THAT34" "55DD5DDD5.33 311T(&*(" "11EA1EAA1.44 412!!!" "33FB3FBB3.11 511#$#"
# [5] "55GD5GDD5.33 612$Q%J" "33IB3IBB3.66 812THIS"
# escaping special characters
spec=c("\\$","%","\\&")
contains_spec = stringr::str_detect(strings, pattern = paste(spec, collapse = "|"))
strings[contains_spec]
# [1] "55DD5DDD5.33 311T(&*(" "33FB3FBB3.11 511#$#" "55GD5GDD5.33 612$Q%J"
You could also use which on contains_spec or contains_only_caps to get the corresponding row numbers for the original data frame. I think that using strings rather than row-wise data frame elements will by much faster - as long as you want to search the whole strings, not certain columns for certain conditions.

Stata - assign different variables depending on the value within a variable

Sorry that title is confusing. Hopefully it's clear below.
I'm using Stata and I'd like to assign the value 1 to a variable that depends on the value within a different variable. I have 20 order variables and also 20 corresponding variables. For example if order1 = 3, I'd like to assign variable3 = 1. Below is a snippet of what the final dataset would look like if I had only 3 of each variable.
Right now I'm doing this with two loops but I have to another loop around this that goes through this 9 more times plus I'd doing this for a couple hundred data files. I'd like to make it more efficient.
forvalues i = 1/20 {
forvalues j = 1/20 {
replace variable`j' = 1 if order`i'==`j'
}
}
Is it possible to use the value of order'i' to assign the variable[order`i'VALUE] directly? Then I can get rid of the j loop above. Something like this.
forvalues i = 1/20 {
replace variable[`order`i'value] = 1
}
Thanks for your help!
***** CLARIFICATION ADDED Feb 2nd.**
I simplified my problem and the dataset too much bc the solutions suggested work for what I presented but, are not getting at what I'm really attempting to do. Thank you three for your solutions though. I was not clear enough in my post.
To clarify, my data doesn't have a one to one correspondence of each order# assigning variable# a 1 if it's not missing. For example, the first observation for order1=3, variable1 isn't supposed to get a 1, variable3 should get a 1. What I didn't include in my original post is that I'm actually checking for other conditions to set it equal to 1.
For more background, I'm counting up births of women by birth order(1st child, 2nd child, etc) that occurred at different ages of mothers. So in the data, each row is a woman, each order# is the number birth (order1=3, it's her third child). The corresponding variable#s are the counts (variable# means the woman has a child of birth order #). I mentioned in the post, that I do this 9 times bc I'm doing it for 5 year age groups (15-19; 20-24; etc). So the first set of variable# would be counts of birth by order when women were ages 15-19; the second set of variable# would be counts of births by order when women were 20-24. etc etc. After this, I sum up the counts in different ways (by woman's education, geography, etc).
So with the additional loop what I do is something more like this
forvalues k = 1/9{
forvalues i = 1/20 {
forvalues j = 1/20 {
replace variable`k'_`j' = 1 if order`i'==`j' & age`i'==`k' & birth_age`i'<36
}
}
}
Not sure if it's possible, but I wanted to simplify so I only need to cycle through each child once, without cycling through the birth orders and directly use the value in order# to assign a 1 to the correct variable. So if order1=3 and the woman had the child at the specific age group, assign variable[agegroup][3]=1; if order1=2, then variable[agegroup][2] should get a 1.
forvalues k=1/9{
forvalues i = 1/20 {
replace variable`k'_[`order`i'value] = 1 if age`i'==`k' & birth_age`i'<36
}
}
I would reshape twice. First reshape to long, then condition variable on !missing(order), then reshape back to wide.
* generate your data
clear
set obs 3
forvalues i = 1/3 {
generate order`i' = .
local k = (3 - `i' + 1)
forvalues j = 1/`k' {
replace order`i' = (`k' - `j' + 1) if (_n == `j')
}
}
list
*. list
*
* +--------------------------+
* | order1 order2 order3 |
* |--------------------------|
* 1. | 3 2 1 |
* 2. | 2 1 . |
* 3. | 1 . . |
* +--------------------------+
* I would rehsape to long, then back to wide
generate id = _n
reshape long order, i(id)
generate variable = !missing(order)
reshape wide order variable, i(id) j(_j)
order order* variable*
drop id
list
*. list
*
* +-----------------------------------------------------------+
* | order1 order2 order3 variab~1 variab~2 variab~3 |
* |-----------------------------------------------------------|
* 1. | 3 2 1 1 1 1 |
* 2. | 2 1 . 1 1 0 |
* 3. | 1 . . 1 0 0 |
* +-----------------------------------------------------------+
Using a simple forvalues loop with generate and missing() is orders of magnitude faster than other proposed solutions (until now). For this problem you need only one loop to traverse the complete list of variables, not two, as in the original post. Below some code that shows both points.
*----------------- generate some data ----------------------
clear all
set more off
local numobs 60
set obs `numobs'
quietly {
forvalues i = 1/`numobs' {
generate order`i' = .
local k = (`numobs' - `i' + 1)
forvalues j = 1/`k' {
replace order`i' = (`k' - `j' + 1) if (_n == `j')
}
}
}
timer clear
*------------- method 1 (gen + missing()) ------------------
timer on 1
quietly {
forvalues i = 1/`numobs' {
generate variable`i' = !missing(order`i')
}
}
timer off 1
* ----------- method 2 (reshape + missing()) ---------------
drop variable*
timer on 2
quietly {
generate id = _n
reshape long order, i(id)
generate variable = !missing(order)
reshape wide order variable, i(id) j(_j)
}
timer off 2
*--------------- method 3 (egen, rowmax()) -----------------
drop variable*
timer on 3
quietly {
// loop over the order variables creating dummies
forvalues v=1/`numobs' {
tab order`v', gen(var`v'_)
}
// loop over the domain of the order variables
// (may need to change)
forvalues l=1/`numobs' {
egen variable`l' = rmax(var*_`l')
drop var*_`l'
}
}
timer off 3
*----------------- method 4 (original post) ----------------
drop variable*
timer on 4
quietly {
forvalues i = 1/`numobs' {
gen variable`i' = 0
forvalues j = 1/`numobs' {
replace variable`i' = 1 if order`i'==`j'
}
}
}
timer off 4
*-----------------------------------------------------------
timer list
The timed procedures give
. timer list
1: 0.00 / 1 = 0.0010
2: 0.30 / 1 = 0.3000
3: 0.34 / 1 = 0.3390
4: 0.07 / 1 = 0.0700
where timer 1 is the simple gen, timer 2 the reshape, timer 3 the egen, rowmax(), and timer 4 the original post.
The reason you need only one loop is that Stata's approach is to execute the command for all observations in the database, from top (first observation) to bottom (last observation). For example, variable1 is generated but according to whether order1 is missing or not; this is done for all observations of both variables, without an explicit loop.
I wonder if you actually need to do this. For future questions, if you have a further goal in mind, I think a good strategy is to mention it in your post.
Note: I've reused code from other posters' answers.
Here's a simpler way to do it (that still requires 2 loops):
// loop over the order variables creating dummies
forvalues v=1/20 {
tab order`v', gen(var`v'_)
}
// loop over the domain of the order variables (may need to change)
forvalues l=1/3 {
egen variable`l' = rmax(var*_`l')
drop var*_`l'
}