基于列组的列的随机抽样

Question

I have a simple problem which can be solved in a dirty way, but I'm looking for a clean way using data.table

I have the following data.table with n columns belonging to m unequal groups. Here is an example of my data.table:

dframe   <- as.data.frame(matrix(rnorm(60), ncol=30))
cletters <- rep(c("A","B","C"), times=c(10,14,6))
colnames(dframe) <- cletters


           A           A          A           A           A          A
1 -0.7431185 -0.06356047 -0.2247782 -0.15423889 -0.03894069  0.1165187
2 -1.5891905 -0.44468389 -0.1186977  0.02270782 -0.64950716 -0.6844163
          A         A          A          A         B         B          B
1 -1.277307 1.8164195 -0.3957006 -0.6489105 0.3498384 -0.463272  0.8458673
2 -1.644389 0.6360258  0.5612634  0.3559574 1.9658743  1.858222 -1.4502839
           B          B          B         B          B           B          B
1  0.3167216 -0.2919079  0.5146733 0.6628149  0.5481958 -0.01721261 -0.5986918
2 -0.8104386  1.2335948 -0.6837159 0.4735597 -0.4686109  0.02647807  0.6389771
           B          B           B          B          C           C
1 -1.2980799  0.3834073 -0.04559749  0.8715914  1.1619585 -1.26236232
2 -0.3551722 -0.6587208  0.44822253 -0.1943887 -0.4958392  0.09581703
           C          C          C         C
1 -0.1387091 -0.4638417 -2.3897681 0.6853864
2  0.1680119 -0.5990310  0.9779425 1.0819789

What I want to do is to take a random subset of the columns (of a sepcific size), keeping the same number of columns per group (if the chosen sample size is larger than the number of columns belonging to one group, take all of the columns of this group).

I have tried an updated version of the method mentioned in this question:

sample rows of subgroups from dataframe with dplyr

but I'm not able to map the column names to the by argument.

Can someone help me with this?

Answer 1

Here's another approach, IIUC:

idx <- split(seq_along(dframe), names(dframe))
keep <- unlist(Map(sample, idx, pmin(7, lengths(idx))))

dframe[, keep]

Explanation:

The first step splits the column indices according to the column names:

idx
# $A
# [1]  1  2  3  4  5  6  7  8  9 10
# 
# $B
# [1] 11 12 13 14 15 16 17 18 19 20 21 22 23 24
# 
# $C
# [1] 25 26 27 28 29 30

In the next step we use

pmin(7, lengths(idx))
#[1] 7 7 6

to determine the sample size in each group and apply this to each list element (group) in idx using Map . We then unlist the result to get a single vector of column indices.

Answer 2

Not sure if you want a solution with dplyr , but here's one with just lapply :

dframe   <- as.data.frame(matrix(rnorm(60), ncol=30))
cletters <- rep(c("A","B","C"), times=c(10,14,6))
colnames(dframe) <- cletters

# Number of columns to sample per group
nc <- 8


res <- do.call(cbind,
       lapply(unique(colnames(dframe)),
              function(x){
                         dframe[,if(sum(colnames(dframe) == x) <= nc) which(colnames(dframe) == x) else sample(which(colnames(dframe) == x),nc,replace = F)]
                         }
))

It might look complicated, but it really just takes all columns per group if there's less than nc , and samples random nc columns if there are more than nc columns.

And to restore your original column-name scheme, gsub does the trick:

colnames(res) <- gsub('.[[:digit:]]','',colnames(res))