通过“["快速矩阵子集:按行、按列还是无关紧要? [英] Fast matrix subsetting via '[': by rows, by columns or doesn't matter?

问题描述

想象一下，我们正在尝试翻转一个方阵并且不关心它是按行还是按列翻转.我们有几个选择:

Imagine we are trying to flip a square matrix and do not care if it will be flipped row- or columnwise. We have a couple of alternatives:

flip.by.col <- function(x) x[, rev(seq_len(ncol(x)))]
flip.by.row <- function(x) x[rev(seq_len(nrow(x))), ]

这些矩阵不相等，但正如我所说，可以忽略.

These matrices are not equal, but, as I said, that can be ignored.

问题是:从计算的角度来看有什么不同吗?

The question is: is there any difference from a computational perspective?

可以想到一个可能的论点:行或列子集是一种利用矩阵在内存中存储方式的低级操作，因此答案是大概是".但是，在使用以下代码进行多次测试后，我找不到任何系统差异:

One can think of a possible argument: row or column subsetting is a low-level operation that utilizes how the matrix is being stored in memory, so the answer is "presumably yes". However, I could not find any systematic difference after several tests with the following code:

require(rbenchmark)
N <- 5e3
x <- matrix(rnorm(N^2), N)
benchmark(flip.by.row(x), flip.by.col(x), replications=10)

一般来说，是否存在这可能成为问题的情况?

Generally speaking, is there a situation where this can be an issue?

UPD(见评论中的讨论)

澄清一下，替换一列可能比替换相同长度的一行更快:矩阵按列存储，要替换的元素按顺序定位.我不确定它是否可以观察到.

To clarify, replacing one column may be faster than replacing one row of the same length: matrices are stored by columns, elements to be replaced are located sequentially. I'm not sure if it's observable though.

推荐答案

无论是按行翻转还是按列翻转，都需要移动相同数量的点:仅中间行/列的元素(当N 是奇数)保持固定位置.因此，无论哪种方式，您都在做相同数量的工作.

In each case, flipped by row or flipping by column, the same number of points need to be moved: only elements in the middle row/column (when N is odd) stay fixed in position. So you are doing the same amount of work either way.

请注意，您可以通过使用 seq.int 获得微小的性能提升.

Note that you can get a tiny performance boost by using seq.int.

flip.by.col2 <- function(x) x[, seq.int(ncol(x), 1)]
flip.by.row2 <- function(x) x[seq.int(nrow(x), 1), ]

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