Obtain matrices by switch a one and a zero-Local search - r

Question

Let's start with the following matrix.
M <- matrix(c(0,0,0,1,0,0,1,1,
0,0,1,0,0,1,1,0,
0,0,0,0,0,1,1,1,
0,0,0,1,1,0,1,0,
0,0,0,1,1,1,0,0,
0,0,1,0,1,0,0,1),nrow = 8,ncol = 6)
[,1] [,2] [,3] [,4] [,5] [,6]
[1,] 0 0 0 0 0 0
[2,] 0 0 0 0 0 0
[3,] 0 1 0 0 0 1
[4,] 1 0 0 1 1 0
[5,] 0 0 0 1 1 1
[6,] 0 1 1 0 1 0
[7,] 1 1 1 1 0 0
[8,] 1 0 1 0 0 1
I want to obtain set of matrices by switching ones and zeros. For each column, starting from column 1, I wanna obtain set of matrices by switching 1 in (4,1) with 0 in (1,1), (2,1), (3,1), (5,1), (6,1) and then do the same for 1s in (7,1) and (8,1). Then continue to the other columns. There are altogether
90 matrices (15 for each column, 15*6) after switching. This is just an example. I have bigger size matrices. How do I generalize for other cases?

Answer 1

Here's a solution. You could wrap the whole thing up into a function. It produces a list of lists of matrices, results, where results[[i]] is a list of matrices with the ith column switched.
column_switcher = function(x) {
ones = which(x == 1)
zeros = which(x == 0)
results = matrix(rep(x, length(ones) * length(zeros)), nrow = length(x))
counter = 1
for (one in ones) {
for (zero in zeros) {
results[one, counter] = 0
results[zero, counter] = 1
counter = counter + 1
}
}
return(results)
}
switched = lapply(1:ncol(M), function(col) column_switcher(M[, col]))
results = lapply(seq_along(switched), function(m_col) {
lapply(1:ncol(switched[[m_col]]), function(i) {
M[, m_col] = switched[[m_col]][, i]
return(M)
})
})
results[[1]]
# [[1]]
# [,1] [,2] [,3] [,4] [,5] [,6]
# [1,] 1 0 0 0 0 0
# [2,] 0 0 0 0 0 0
# [3,] 0 1 0 0 0 1
# [4,] 0 0 0 1 1 0
# [5,] 0 0 0 1 1 1
# [6,] 0 1 1 0 1 0
# [7,] 1 1 1 1 0 0
# [8,] 1 0 1 0 0 1
#
# [[2]]
# [,1] [,2] [,3] [,4] [,5] [,6]
# [1,] 0 0 0 0 0 0
# [2,] 1 0 0 0 0 0
# [3,] 0 1 0 0 0 1
# [4,] 0 0 0 1 1 0
# [5,] 0 0 0 1 1 1
# [6,] 0 1 1 0 1 0
# [7,] 1 1 1 1 0 0
# [8,] 1 0 1 0 0 1
#
# ...
Checking the length of the list and the lengths of the sublists, they're all there.
length(results)
# [1] 6
lengths(results)
# [1] 15 15 15 15 15 15