R:逐元素矩阵除法 [英] R: element-wise matrix division

问题描述

给出相同维的两个数值矩阵A和B.按元素划分的最佳方法是什么:A[i,j] / B[i,j]?我知道可以使用双for循环.但是我想要最有效的方法.

Given two numeric matrices A and B of the same dimension. What's the best way for element-wise division: A[i,j] / B[i,j]? I know it is possible to do using double for loops. But I want the most efficient way.

编辑:当存在B[i,j] == 0时，必须为A[i,j] <- 0.

When there is a B[i,j] == 0 it will have to be A[i,j] <- 0.

推荐答案

如果矩阵是A和B，则只需使用A / B.

If your matrices are A and B, you can just use A / B.

A <- matrix(1:4, 2, 2)
#     [,1] [,2]
#[1,]    1    3
#[2,]    2    4

B <- matrix((1:4) * 2, 2, 2)
#     [,1] [,2]
#[1,]    2    6
#[2,]    4    8

C <- A / B
#     [,1] [,2]
#[1,]  0.5  0.5
#[2,]  0.5  0.5

---

如果有B[i,j] == 0，则必须为A[i,j] <- 0.

如果B中有0个元素，则可能会得到NaN，Inf或-Inf，具体取决于A中的对应内容.

In case you have 0 elements in B, you may get NaN, Inf or -Inf, depending on its counterpart in A.

0 / 0
# NA

1 / 0
# Inf

-1 / 0
# -Inf

所有这些都不是有限的.如果要将它们替换为0，只需执行以下操作:

All these are not finite. If you want to replace them with 0, simply do:

C <- A / B
C[!is.finite(C)] <- 0

---

很难记住R如何对待NA，NaN，Inf和-Inf.您可以阅读?is.finite和?NA以获得常规信息.在这里，我将做一个简单的测试.

---

It is difficult to remember how R treats NA, NaN, Inf and -Inf. You can read ?is.finite and ?NA for general info. Here I will give a simple test.

x <- c(NA, NaN, Inf, -Inf)

is.finite(x)
# [1] FALSE FALSE FALSE FALSE

is.infinite(x)
# [1] FALSE FALSE  TRUE  TRUE

is.na(x)
# [1]  TRUE  TRUE FALSE FALSE

is.nan(x)
# [1] FALSE  TRUE FALSE FALSE

请注意，is.infinite不是is.finite的倒数，而是is.na的倒数.这就是为什么我使用!is.finite.

Note, is.infinite is not the inverse of is.finite, but the inverse of is.na. That is why I have used !is.finite.

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