I face the following problem: I need many subsets of a big matrix. Actually I just need views as input for another function f(), so I don't need to change the values. However it seems, that R is terribly slow for this task, or I'm doing something wrong (which seems more likely). The toy example illustrates how much time it takes just to select the columns, and then use them in another function (in this toy example the primitive function sum()). As 'benchmark' I also test the calculation time against summing up the whole matrix, which is surprisingly faster. I also experimented with the package ref, however could not achieve any performance gain. So the key question is how to subset the matrix without copying it? I appreciate any help, Thanks!
library(microbenchmark)
library(ref)
m0 <- matrix(rnorm(10^6), 10^3, 10^3)
r0 <- refdata(m0)
microbenchmark(m0[, 1:900], sum(m0[, 1:900]), sum(r0[,1:900]), sum(m0))
Unit: milliseconds expr min lq mean median uq m0[, 1:900] 10.087403 12.350751 16.697078 18.307475 19.054157 sum(m0[, 1:900]) 11.067583 13.341860 17.286514 19.123748 19.990661 sum(r0[, 1:900]) 11.066164 13.194244 16.869551 19.204434 20.004034 sum(m0) 1.015247 1.040574 1.059872 1.049513 1.067142 max neval 58.238217 100 25.664729 100 23.505308 100 1.233617 100
The benchmark task of summing the whole matrix takes 1.059872 milliseconds and is about 16 times faster than the other functions.
Subsetting matricesA matrix is subset with two arguments within single brackets, [] , and separated by a comma. The first argument specifies the rows, and the second the columns.
Similar to vectors, you can use the square brackets [ ] to select one or multiple elements from a matrix. Whereas vectors have one dimension, matrices have two dimensions. You should therefore use a comma to separate the rows you want to select from the columns.
Something not mentioned by @Michal is that not only is a matrix smaller than the equivalent data frame, using matrices can make your code far more efficient than using data frames, often considerably so. That is one reason why internally, a lot of R functions will coerce to matrices data that are in data frames.
The problem with your solution is that the subsetting is allocating another matrix, which takes times.
You have two solutions:
If the time taken with sum
on the whole matrix is okay with you, you could use colSums
on the whole matrix and subset the result:
sum(colSums(m0)[1:900])
Or you could use Rcpp to compute the sum
with subsetting without copying the matrix.
#include <Rcpp.h>
using namespace Rcpp;
// [[Rcpp::export]]
double sumSub(const NumericMatrix& x,
const IntegerVector& colInd) {
double sum = 0;
for (IntegerVector::const_iterator it = colInd.begin(); it != colInd.end(); ++it) {
int j = *it - 1;
for (int i = 0; i < x.nrow(); i++) {
sum += x(i, j);
}
}
return sum;
}
microbenchmark(m0[, 1:900], sum(m0[, 1:900]), sum(r0[,1:900]), sum(m0),
sum(colSums(m0)[1:900]),
sumSub(m0, 1:900))
Unit: milliseconds
expr min lq mean median uq max neval
m0[, 1:900] 4.831616 5.447749 5.641096 5.675774 5.861052 6.418266 100
sum(m0[, 1:900]) 6.103985 6.475921 7.052001 6.723035 6.999226 37.085345 100
sum(r0[, 1:900]) 6.224850 6.449210 6.728681 6.705366 6.943689 7.565842 100
sum(m0) 1.110073 1.145906 1.175224 1.168696 1.197889 1.269589 100
sum(colSums(m0)[1:900]) 1.113834 1.141411 1.178913 1.168312 1.201827 1.408785 100
sumSub(m0, 1:900) 1.337188 1.368383 1.404744 1.390846 1.415434 2.459361 100
You could use unrolling optimization to further optimize the Rcpp version.
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