使累积和更快 [英] Make cumulative sum faster

问题描述

我正在尝试对矩阵的每一列求累计和.这是我在 R 中的代码:

I'm trying to take cumulative sums for each column of a matrix. Here's my code in R:

testMatrix = matrix(1:65536, ncol=256);
microbenchmark(apply(testMatrix, 2, cumsum), times=100L);

Unit: milliseconds
                         expr      min       lq     mean  median       uq      max neval
 apply(testMatrix, 2, cumsum) 1.599051 1.766112 2.329932 2.15326 2.221538 93.84911 10000

我使用 Rcpp 进行比较:

I used Rcpp for comparison:

cppFunction('NumericMatrix apply_cumsum_col(NumericMatrix m) {
    for (int j = 0; j < m.ncol(); ++j) {
        for (int i = 1; i < m.nrow(); ++i) {
            m(i, j) += m(i - 1, j);
        }
    }
    return m;
}');
microbenchmark(apply_cumsum_col(testMatrix), times=10000L);

Unit: microseconds
                         expr     min      lq     mean  median      uq      max neval
 apply_cumsum_col(testMatrix) 205.833 257.719 309.9949 265.986 276.534 96398.93 10000

所以 C++ 代码的速度是原来的 7.5 倍.在纯 R 中是否有可能比 apply(testMatrix, 2, cumsum) 做得更好?感觉就像我无缘无故地有一个数量级的开销.

So the C++ code is 7.5 times as fast. Is it possible to do better than apply(testMatrix, 2, cumsum) in pure R? It feels like I have an order of magnitude overhead for no reason.

推荐答案

使用字节编译的 for 循环比在我的系统上调用 apply 稍快.我预计它会更快，因为它比 apply 做的工作更少.不出所料，R 循环仍然比您编写的简单 C++ 函数慢.

Using a byte-compiled for loop is slightly faster than the apply call on my system. I expected it to be faster because it does less work than apply. As expected, the R loop is still slower than the simple C++ function you wrote.

colCumsum <- compiler::cmpfun(function(x) {
  for (i in 1:ncol(x))
    x[,i] <- cumsum(x[,i])
  x
})

testMatrix <- matrix(1:65536, ncol=256)
m <- testMatrix
require(microbenchmark)
microbenchmark(colCumsum(m), apply_cumsum_col(m), apply(m, 2, cumsum), times=100L)
# Unit: microseconds
#                 expr      min        lq    median        uq       max neval
#      matrixCumsum(m) 1478.671 1540.5945 1586.1185 2199.9530 37377.114   100
#  apply_cumsum_col(m)  178.214  192.4375  204.3905  234.8245  1616.030   100
#  apply(m, 2, cumsum) 1879.850 1940.1615 1991.3125 2745.8975  4346.802   100
all.equal(colCumsum(m), apply(m, 2, cumsum))
# [1] TRUE

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