如何在每个矩阵元素的索引上应用函数 [英] How to apply function over each matrix element's indices

问题描述

我想知道 R 中是否有一个内置函数可以将函数应用于矩阵的每个元素(当然，该函数应该基于矩阵索引计算).等效的将是这样的:

I am wondering if there is a built-in function in R which applies a function to each element of the matrix (of course, the function should be computed based on matrix indices). The equivalent would be something like this:

matrix_apply <- function(m, f) {
  m2 <- m
  for (r in seq(nrow(m2)))
    for (c in seq(ncol(m2)))
      m2[[r, c]] <- f(r, c)
  return(m2)
}

如果没有这样的内置函数，初始化矩阵以包含通过计算以矩阵索引作为参数的任意函数获得的值的最佳方法是什么?

If there is no such built-in function, what is the best way to initialize a matrix to contain values obtained by computing an arbitrary function which has matrix indices as parameters?

推荐答案

我怀疑你想要outer:

> mat <- matrix(NA, nrow=5, ncol=3)

> outer(1:nrow(mat), 1:ncol(mat) , FUN="*")
     [,1] [,2] [,3]
[1,]    1    2    3
[2,]    2    4    6
[3,]    3    6    9
[4,]    4    8   12
[5,]    5   10   15

> outer(1:nrow(mat), 1:ncol(mat) , FUN=function(r,c) log(r+c) )
          [,1]     [,2]     [,3]
[1,] 0.6931472 1.098612 1.386294
[2,] 1.0986123 1.386294 1.609438
[3,] 1.3862944 1.609438 1.791759
[4,] 1.6094379 1.791759 1.945910
[5,] 1.7917595 1.945910 2.079442

这会产生一个很好的紧凑输出.但是 mapply 在其他情况下可能会有用.将 mapply 视为执行此页面上其他人使用 Vectorize 进行相同操作的另一种方式，这很有帮助.mapply 更通用，因为 Vectorize 无法使用原始"函数.

That yields a nice compact output. but it's possible that mapply would be useful in other situations. It is helpful to think of mapply as just another way to do the same operation that others on this page are using Vectorize for. mapply is more general because of the inability Vectorize to use "primitive" functions.

data.frame(mrow=c(row(mat)),   # straightens out the arguments
           mcol=c(col(mat)), 
           m.f.res= mapply(function(r,c) log(r+c), row(mat), col(mat)  ) )
#   mrow mcol   m.f.res
1     1    1 0.6931472
2     2    1 1.0986123
3     3    1 1.3862944
4     4    1 1.6094379
5     5    1 1.7917595
6     1    2 1.0986123
7     2    2 1.3862944
8     3    2 1.6094379
9     4    2 1.7917595
10    5    2 1.9459101
11    1    3 1.3862944
12    2    3 1.6094379
13    3    3 1.7917595
14    4    3 1.9459101
15    5    3 2.0794415

您可能并不是真的想向函数提供 row() 和 col() 函数将返回的内容:这会生成一个包含 15 个(有些冗余)3 x 5 矩阵的数组:

You probably didn't really mean to supply to the function what the row() and col() functions would have returned: This produces an array of 15 (somewhat redundant) 3 x 5 matrices:

> outer(row(mat), col(mat) , FUN=function(r,c) log(r+c) )

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