有一个向量,我想通过基于序列获取其元素来创建一个新向量:
set.seed(0)
n <- 1000
ncval1 <- as.integer(n)
ncval2 <- ncval1:1L
ncval3 <- sequence(ncval2, from = 1L, by = 1L)
x <- as.double(runif(n))
y <- x[ncval3]
这大约需要 2.2 毫秒。也许可以通过采用重复元素的属性来加快速度。
您可以使用
Rcpp
。
Rcpp::sourceCpp(code='
#include <Rcpp.h>
// [[Rcpp::export]]
Rcpp::NumericVector foo(int n) {
// draw from standard normal
Rcpp::NumericVector r(n);
r = Rcpp::runif(n);
// length of result
int l = 0;
for (int i = 0; i <= n; i++) {
l = l + i;
}
// subset and concatenate
Rcpp::NumericVector a(l);
int p = 0;
for (int i = 0; i < n; i++) {
for (int j = 0; j < n - i; j++) {
a[p] = r[j];
p = p + 1;
}
}
return a;
}
')
> set.seed(0)
> foo(10)
[1] 0.8966972 0.2655087 0.3721239 0.5728534 0.9082078 0.2016819 0.8983897
[8] 0.9446753 0.6607978 0.6291140 0.8966972 0.2655087 0.3721239 0.5728534
[15] 0.9082078 0.2016819 0.8983897 0.9446753 0.6607978 0.8966972 0.2655087
[22] 0.3721239 0.5728534 0.9082078 0.2016819 0.8983897 0.9446753 0.8966972
[29] 0.2655087 0.3721239 0.5728534 0.9082078 0.2016819 0.8983897 0.8966972
[36] 0.2655087 0.3721239 0.5728534 0.9082078 0.2016819 0.8966972 0.2655087
[43] 0.3721239 0.5728534 0.9082078 0.8966972 0.2655087 0.3721239 0.5728534
[50] 0.8966972 0.2655087 0.3721239 0.8966972 0.2655087 0.8966972
n <- 1e3
microbenchmark::microbenchmark(
OP={
set.seed(0)
ncval1 <- as.integer(n)
ncval2 <- ncval1:1L
ncval3 <- sequence(ncval2, from = 1L, by = 1L)
x <- as.double(runif(n))
x[ncval3]
},
foo={set.seed(0); foo(n)},
check='identical'
)
$ Rscript --vanilla foo.R
Unit: milliseconds
expr min lq mean median uq max neval cld
OP 2.109090 2.199845 3.119882 2.294714 4.213308 7.297789 100 a
foo 1.055756 1.190470 1.983916 1.318557 2.741124 6.850159 100 b
根据中位数,
foo()
只需要 57% 的时间。