I'm trying to create a function, which creates an matrix of size nxm, with entries of the uniform distribution, with the Rcpp package. (I'm inexperienced using this package.)
library(Rcpp)
cppFunction('NumericMatrix rngCpp(const int n,const int m) {
NumericMatrix X(n, m);
X(_,0) = runif(n);
return X;
}')
set.seed(1)
rngCpp(4,5)
[,1] [,2] [,3] [,4] [,5]
[1,] 0.2655087 0 0 0 0
[2,] 0.3721239 0 0 0 0
[3,] 0.5728534 0 0 0 0
[4,] 0.9082078 0 0 0 0
Expected output
set.seed(1)
matrix(runif(4*5), nrow=4, ncol = 5)
[,1] [,2] [,3] [,4] [,5]
[1,] 0.2655087 0.2016819 0.62911404 0.6870228 0.7176185
[2,] 0.3721239 0.8983897 0.06178627 0.3841037 0.9919061
[3,] 0.5728534 0.9446753 0.20597457 0.7698414 0.3800352
[4,] 0.9082078 0.6607978 0.17655675 0.4976992 0.7774452
Well with
X(_,0) = runif(n);
you explicitly only assign to the first column. So just loop over all m column doing the same.
Another (inside-baseball) way is to request a vector of n*m, and then set the dim
attributes of (n,m) to make it a matrix.
Related, there is IIRC a constructor that would take that vector and re-dim it.
Edit: And that last approach is the simplest and works like this:
R> cppFunction('NumericMatrix mu(int n, int m) {
NumericVector v = runif(n*m);
return NumericMatrix(n, m, v.begin()); }')
R> set.seed(1); mu(4,5)
[,1] [,2] [,3] [,4] [,5]
[1,] 0.265509 0.201682 0.6291140 0.687023 0.717619
[2,] 0.372124 0.898390 0.0617863 0.384104 0.991906
[3,] 0.572853 0.944675 0.2059746 0.769841 0.380035
[4,] 0.908208 0.660798 0.1765568 0.497699 0.777445
R>
Edit 2: Additional variants as F.Privé insists on arguing:
We can add these two:
// [[Rcpp::export]]
NumericMatrix mu3(int n, int m) {
return NumericMatrix(n, m, runif(n*m).begin());
}
// [[Rcpp::export]]
NumericMatrix mu4(int n, int m) {
NumericVector v = runif(n * m);
v.attr("dim") = Dimension(n, m);
return as<NumericMatrix>(v);
}
and then get
R> N <- 1000; M <- 1000
R> microbenchmark::microbenchmark(
+ mu(N, M),
+ mu2(N, M),
+ mu3(N, M),
+ mu4(N, M),
+ rngCpp(N, M)
+ )
Unit: milliseconds
expr min lq mean median uq max neval
mu(N, M) 4.77894 4.98485 7.32315 5.18153 5.36468 33.2427 100
mu2(N, M) 3.99137 4.05308 5.43207 4.36296 4.57510 30.7335 100
mu3(N, M) 4.73176 5.01524 6.35186 5.17173 5.39541 31.7425 100
mu4(N, M) 3.99784 4.10052 4.72563 4.41176 4.60303 30.6166 100
rngCpp(N, M) 5.18726 5.60165 7.53171 5.83892 6.14315 34.5934 100
R>
In the OP's code, the assignment only happens for the 1st column index ie 0
. We can loop through the column index and assign the uniform distribution values
cppFunction('NumericMatrix rngCpp(const int n,const int m) {
NumericMatrix X(n, m);
for(int i = 0; i < m; i++){
X(_,i) = runif(n);
}
return X;
}')
set.seed(1)
rngCpp(4,5)
# [,1] [,2] [,3] [,4] [,5]
#[1,] 0.2655087 0.2016819 0.62911404 0.6870228 0.7176185
#[2,] 0.3721239 0.8983897 0.06178627 0.3841037 0.9919061
#[3,] 0.5728534 0.9446753 0.20597457 0.7698414 0.3800352
#[4,] 0.9082078 0.6607978 0.17655675 0.4976992 0.7774452
Generally, I would use the second idea of Dirk: changing the dimension of a vector so that it becomes a matrix.
#include <Rcpp.h>
using namespace Rcpp;
// [[Rcpp::export]]
NumericVector mu2(int n, int m) {
NumericVector v = runif(n * m);
v.attr("dim") = Dimension(n, m);
return v;
}
// [[Rcpp::export]]
NumericMatrix mu(int n, int m) {
NumericVector v = runif(n*m);
return NumericMatrix(n, m, v.begin());
}
// [[Rcpp::export]]
NumericMatrix rngCpp(const int n,const int m) {
NumericMatrix X(n, m);
for(int i = 0; i < m; i++){
X(_,i) = runif(n);
}
return X;
}
/*** R
set.seed(1); mu(4, 5)
set.seed(1); mu2(4, 5)
N <- 1000; M <- 1000
microbenchmark::microbenchmark(
mu(N, M),
mu2(N, M),
rngCpp(N, M)
)
*/
Microbenchmark:
Unit: milliseconds
expr min lq mean median uq max neval cld
mu(N, M) 7.627591 10.290691 16.05829 10.932174 11.416555 119.2110 100 a
mu2(N, M) 6.315160 6.590293 12.30125 8.510765 9.368282 117.8750 100 a
rngCpp(N, M) 8.330158 10.306584 16.14374 10.893767 11.398675 121.4233 100 a
So, mu2
is a bit faster because it doesn't make any copy.
The technical post webpages of this site follow the CC BY-SA 4.0 protocol. If you need to reprint, please indicate the site URL or the original address.Any question please contact:yoyou2525@163.com.