R: Choosing specific number of combinations from all possible combinations - r

Question

Let's say we have the following dataset
set.seed(144)
dat <- matrix(rnorm(100), ncol=5)
The following function creates all possible combinations of columns and removes the first
(cols <- do.call(expand.grid, rep(list(c(F, T)), ncol(dat)))[-1,])
# Var1 Var2 Var3 Var4 Var5
# 2 TRUE FALSE FALSE FALSE FALSE
# 3 FALSE TRUE FALSE FALSE FALSE
# 4 TRUE TRUE FALSE FALSE FALSE
# ...
# 31 FALSE TRUE TRUE TRUE TRUE
# 32 TRUE TRUE TRUE TRUE TRUE
My question is how can I calculate single, binary and triple combinations only ?
Choosing the rows including no more than 3 TRUE values using the following function works for this vector: cols[rowSums(cols)<4L, ]
However, it gives following error for larger vectors mainly because of the error in expand.grid with long vectors:
Error in rep.int(seq_len(nx), rep.int(rep.fac, nx)) :
invalid 'times' value
In addition: Warning message:
In rep.fac * nx : NAs produced by integer overflow
Any suggestion that would allow me to compute single, binary and triple combinations only ?

Answer 1

You could try either
cols[rowSums(cols) < 4L, ]
Or
cols[Reduce(`+`, cols) < 4L, ]

Answer 2

You can use this solution:
col.i <- do.call(c,lapply(1:3,combn,x=5,simplify=F))
# [[1]]
# [1] 1
#
# [[2]]
# [1] 2
#
# <...skipped...>
#
# [[24]]
# [1] 2 4 5
#
# [[25]]
# [1] 3 4 5
Here, col.i is a list every element of which contains column indices.
How it works: combn generates all combinations of the numbers from 1 to 5 (requested by x=5) taken m at a time (simplify=FALSE ensures that the result has a list structure). lapply invokes an implicit cycle to iterate m from 1 to 3 and returns a list of lists. do.call(c,...) converts a list of lists into a plain list.
You can use col.i to get certain columns from dat using e.g. dat[,col.i[[1]],drop=F] (1 is an index of the column combination, so you could use any number from 1 to 25; drop=F makes sure that when you pick just one column from dat, the result is not simplified to a vector, which might cause unexpected program behavior). Another option is to use lapply, e.g.
lapply(col.i, function(cols) dat[,cols])
which will return a list of data frames each containing a certain subset of columns of dat.
In case you want to get column indices as a boolean matrix, you can use:
col.b <- t(sapply(col.i,function(z) 1:5 %in% z))
# [,1] [,2] [,3] [,4] [,5]
# [1,] TRUE FALSE FALSE FALSE FALSE
# [2,] FALSE TRUE FALSE FALSE FALSE
# [3,] FALSE FALSE TRUE FALSE FALSE
# ...
[UPDATE]
More efficient realization:
library("gRbase")
coli <- function(x=5,m=3) {
col.i <- do.call(c,lapply(1:m,combnPrim,x=x,simplify=F))
z <- lapply(seq_along(col.i), function(i) x*(i-1)+col.i[[i]])
v.b <- rep(F,x*length(col.i))
v.b[unlist(z)] <- TRUE
matrix(v.b,ncol=x,byrow = TRUE)
}
coli(70,5) # takes about 30 sec on my desktop