[Rd] Matrix memory layout R vs. C
Larissa Hauer
larissahauer at googlemail.com
Fri Dec 6 14:21:10 CET 2013
Hi everybody,
I'm trying to pass a matrix from R to C, where some computation is done
for performance reasons, and back to R for evaluation. But I've run into
the problem that R and C seem to have different ways of representing the
matrix in main memory. The C representation of a 2D matrix in linear
memory is concatenation of the rows whereas in R, it's a concatenation
of the columns. That leads to the problem. that an R-matrix, for example
123
456
789
is seen by C as
147
258
369
and vice versa.
Here's an example of C code that simply prints the matrix it gets from R:
#include <stdlib.h>
#include "R.h"
void printMatrix(int *mPtr, int *m, int *n) {
int (*matrix)[*n] = mPtr;
int j,k;
for(j = 0; j < *m; j++){
for(k = 0; k < *n; k++) {
printf("%d", matrix[j][k]);
}
printf("\n");
}
}
And here's what happens when I call the function in R:
> m <- 3; n <- 3
> mat <- matrix(c(1:9), nrow=m, ncol=n, byrow=TRUE)
> mat
[,1] [,2] [,3]
[1,] 1 2 3
[2,] 4 5 6
[3,] 7 8 9
> mat <- .C("printMatrix", mat, as.integer(m), as.integer(n))[[1]]
147
258
369
No matter if you create the matrix with byrow=TRUE or FALSE, C always
interprets it the other way round. Is there a way to avoid this? I've
read previous posts on passing a matrix from R to C, but the essence of
the answers was that "a matrix in R is just a vector with attributes",
but I don't see how this helps. Maybe someone can clarify.
Thanks a lot in advance!
Cheers
Larissa
Here's the C main function showing that the C code itself is correct:
#include <stdlib.h>
void printMatrix(int *mPtr, int *m, int *n);
int main(void) {
int m, n, i;
int *mPtr, *nPtr;
m = 3;
n = 3;
mPtr = &m;
nPtr = &n;
int *M = malloc(m * n * sizeof(int));
for (i = 0; i < m * n; i++){
M[i] = i + 1;
}
printMatrix(M, mPtr, nPtr);
return EXIT_SUCCESS;
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