a simple loop with data.table

Question

I want to do a simple loop using data.table. I have 20 dichotomous (0,1) variables (from var_1 to var_20) and I would like to do a loop for this:

dat[var_1==1, newvar:=1]
dat[var_2==1, newvar:=2]
dat[var_3==1, newvar:=3]
...
dat[var_20==1, newvar:=21]

My main problem is I don't know how specify i (i.e. var_1==1, var_2==2...) using a loop. Below a short example:

var_1  <- c(1, rep(0,9))
var_2  <- c(0,1, rep(0,8))
var_3  <- c(0,0,1, rep(0,7))
dat  <- data.table(var_1, var_2, var_3)

dat[var_1==1, newvar:=1]
dat[var_2==1, newvar:=2]
dat[var_3==1, newvar:=3]

Any ideas about how to do this with a loop? Thanks!

Wojciech Sobala · Accepted Answer · 2012-12-02 19:21:09Z

4

To take advantage of data.table class it is better to set key.

dat[ ,newvar:= NA_integer_]
for(i in ncol(dat)) {
 setkeyv(dat, names(dat)[i])
 dat[J(1), newvar:=i]
}

edited Dec 2, 2012 at 19:21

answered Dec 2, 2012 at 13:37

Wojciech Sobala

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3 Comments

sdaza Over a year ago

Thank you. Do you know how to do that newvar:=1L increases from 1 to the number of variables (e.g., newvar should be equal to 2 for var_2, 3 for var_3, and so on). It is also possible that some variables have a value 1 simultaneously, in that case, I would like to impute just the bigger value (e.g., a case has var_1=1 and var_3=1, I would like to get newvar=3).

mnel Over a year ago

The code above should do this. I would be interested in the performance of repeatedly setting keys compared to sequential but single vector scans on large data.tables

Matt Dowle Over a year ago

@WojciechSobala Usually it is better to setkey but setkey has to read every value in the column (scan) in the process of sorting it. So in the special case of a single scan for a single value on a single column, a vector scan should be faster than setkey+join. Worth testing though, I haven't tested that myself.

mnel · Accepted Answer · 2012-12-02 05:12:34Z

4

Something like this will work.

nams <- names(dat)
for(n in seq_along(nams)){
  nam <- nams[n] 
  char <- sprintf('%s==1',nam)
  dat[eval(parse(text=char)), newvar := n]
}
dat
var_1 var_2 var_3 newvar
1:     1     0     0      1
2:     0     1     0      2
3:     0     0     1      3
4:     0     0     0     NA
5:     0     0     0     NA
6:     0     0     0     NA
7:     0     0     0     NA
8:     0     0     0     NA
9:     0     0     0     NA
10:    0     0     0     NA

answered Dec 2, 2012 at 5:12

mnel

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