CODEWITHSUNDEEP
rmatrixindexing
Ophiothrix
Ophiothrix

Reputation: 831

Preserve structure, when indexing a matrix with another matrix in R

Dear StackOverflowers,

I have an integer matrix in R and I would like to subset it so that I remove 1 specified cell in each column. So that, for instance, a 4x3 matrix becomes a 3x3 matrix. I have tried doing it by creating the second logical matrix of the same dimensions.

(subject.matrix <- matrix(1:12, nrow = 4))
     [,1] [,2] [,3]
[1,]    1    5    9
[2,]    2    6   10
[3,]    3    7   11
[4,]    4    8   12
(query.matrix <- matrix(c(T, T, F, T, T, F, T, T, T, T, T, F), nrow = 4))
      [,1]  [,2]  [,3]
[1,]  TRUE  TRUE  TRUE
[2,]  TRUE FALSE  TRUE
[3,] FALSE  TRUE  TRUE
[4,]  TRUE  TRUE FALSE

The problem is that, when I index the first matrix by the second one, it is simplified to an integer vector.

subject.matrix[query.matrix]
[1]  1  2  4  5  7  8  9 10 11

I've tried adding drop=F, but to no avail. I know, I can just wrap the resulting vector into a 3x3 matrix. So the expected outcome would be:

matrix(subject.matrix[query.matrix], nrow = 3)
     [,1] [,2] [,3]
[1,]    1    5    9
[2,]    2    7   10
[3,]    4    8   11

But I wonder if there's a more elegant/direct solution. I'm also not attached to using a logical matrix as the index, if that means a simpler solution. Perhaps, I could subset it with a vector of indices for the rows to be removed in each column, which in this case would translate into c(3, 2, 4).

Many thanks!

Edit based on @LyzandeR suggestion: My final goal was to take column sums of the resulting matrix. So replacing the redundant values with NA's seems to be the best way to go.

Upvotes: 3

Views: 699

Answers (2)

LyzandeR
LyzandeR

Reputation: 37889

I think that the only way you can preserve the matrix structure would be to use a more general way of your question edit i.e.:

matrix(subject.matrix[query.matrix], ncol = ncol(subject.matrix))

You could even convert it into a function if you plan on using it multiple times:

subset.mat <- function(mat, index, cols=ncol(mat)) {
  matrix(mat[index], ncol = cols)
}

Output:

> subset.mat(subject.matrix, query.matrix)
     [,1] [,2] [,3]
[1,]    1    5    9
[2,]    2    7   10
[3,]    4    8   11

Also (sorry just read your updated comment) you might consider using NAs in the matrix instead of subsetting them out, which will allow you to calculate the column sums as you say:

subject.matrix[!query.matrix] <- NA
subject.matrix
#    [,1] [,2] [,3]
#[1,]    1    5    9
#[2,]    2   NA   10
#[3,]   NA    7   11
#[4,]    4    8   NA

Upvotes: 3

Nancy
Nancy

Reputation: 4109

This is a little brute-forceish, but I think you'll be able to extrapolate it into something more general:

new.matrix = matrix(ncol = ncol(subject.matrix), nrow = nrow(subject.matrix) - 1)
for(i in 1:ncol(subject.matrix)){
  new.matrix[,i] = subject.matrix[,i][query.matrix[,i] == TRUE]
}
new.matrix

      [,1] [,2] [,3]
[1,]    1    5    9
[2,]    2    7   10
[3,]    4    8   11

Essentially, I just initialized an empty matrix, and then iterated through each column of subject.matrix taking only the TRUE values for query.matrix.

Upvotes: 1

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