Jere Jere - 2 months ago 10
R Question

Replacing matrix elements indexed by another matrix

After several hours of searching, I am turning to your expertise. Beginner in R, I try to speed up my code. My goal is to replace the values in a matrix

A
. However, I want to replace values based on two vectors of another matrix
B
.
B[, 1]
is the name of row
i
of the matrix
A
. The second column,
B[, 2]
corresponds to the name of column of the matrix
A
.

The first version of my code was to use the match function in a loop.

for(k in 1:L){
i <- B[k,1]
j <- B[k,2]
d <- match(i,rownames(A))
e <- match(j,colnames(A))
A[d, e] <- 0
}


The second version allowed me to speed a little bit:

for( k in 1:L) {
A[match(B[k,1],rownames(A)), match(B[k,2],colnames(A))] <- 0
}


However, the processing time is long, too long. So I thought to use the
apply
function. For this, I have to use
apply
in each row vectors of
B
.

Is Using
apply
function a great way? Or I am going in the wrong way?

Answer

It appears to me that you can simply do A[B[, 1:2]] <- 0, by using the power of matrix indexing.

For example, A[cbind(1:4, 1:4)] <- 0 will replace A[1,1], A[2,2], A[3,3] and A[4,4] to 0. In fact, if A has "dimnames" attributes (the "rownames" and "colnames" you refer to), we can also use the character strings as index.


Reproducible example

A <- matrix(1:16, 4, 4, dimnames = list(letters[1:4], LETTERS[1:4]))
#  A B  C  D
#a 1 5  9 13
#b 2 6 10 14
#c 3 7 11 15
#d 4 8 12 16

set.seed(0); B <- cbind(sample(letters[1:4])), sample(LETTERS[1:4]))
#     [,1] [,2]
#[1,] "d"  "D" 
#[2,] "a"  "A" 
#[3,] "c"  "B" 
#[4,] "b"  "C" 

## since `B` has just 2 columns, we can use `B` rather than `B[, 1:2]`
A[B] <- 0

#  A B  C  D
#a 0 5  9 13
#b 2 6  0 14
#c 3 0 11 15
#d 4 8 12  0
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