替换由另一个矩阵索引的矩阵元素 [英] Replacing matrix elements indexed by another matrix

问题描述

经过几个小时的搜索，我开始求助于您的专业知识.R 初学者，我尝试加速我的代码.我的目标是替换矩阵 A 中的值.但是，我想替换基于另一个矩阵 B 的两个向量的值.B[, 1] 是矩阵A 的行i 的名称.第二列B[, 2]对应矩阵A的列名.

After several hours of searching, I am turning to your expertise. Beginner in R, I try to speed up my code. My goal is to replace the values in a matrix A. However, I want to replace values based on two vectors of another matrix B. B[, 1] is the name of row i of the matrix A. The second column, B[, 2] corresponds to the name of column of the matrix A.

我的代码的第一个版本是在循环中使用 match 函数.

The first version of my code was to use the match function in a loop.

for(k in 1:L){
  i <- B[k,1]
  j <- B[k,2]
  d <- match(i,rownames(A))
  e <- match(j,colnames(A))
  A[d, e] <- 0
  }

第二个版本让我加快了一点速度:

The second version allowed me to speed a little bit:

for( k in 1:L) {
  A[match(B[k,1],rownames(A)), match(B[k,2],colnames(A))] <- 0
  }

但是，处理时间很长，太长了.所以我想使用 apply 函数.为此，我必须在 B 的每个行向量中使用 apply.

However, the processing time is long, too long. So I thought to use the apply function. For this, I have to use apply in each row vectors of B.

使用 apply 函数是个好方法吗?还是我走错路了?

Is Using apply function a great way? Or I am going in the wrong way?

推荐答案

在我看来，您可以简单地执行 A[B[, 1:2]] <- 0，通过使用矩阵索引的威力.

It appears to me that you can simply do A[B[, 1:2]] <- 0, by using the power of matrix indexing.

例如，A[cbind(1:4, 1:4)] <- 0 将替换 A[1,1], A[2,2]、A[3,3] 和 A[4,4] 为 0.实际上，如果 Acode> 有dimnames"属性(你所指的rownames"和colnames")，我们也可以使用字符串作为索引.

For example, A[cbind(1:4, 1:4)] <- 0 will replace A[1,1], A[2,2], A[3,3] and A[4,4] to 0. In fact, if A has "dimnames" attributes (the "rownames" and "colnames" you refer to), we can also use the character strings as index.

可重现的示例

A <- matrix(1:16, 4, 4, dimnames = list(letters[1:4], LETTERS[1:4]))
#  A B  C  D
#a 1 5  9 13
#b 2 6 10 14
#c 3 7 11 15
#d 4 8 12 16

set.seed(0); B <- cbind(sample(letters[1:4])), sample(LETTERS[1:4]))
#     [,1] [,2]
#[1,] "d"  "D" 
#[2,] "a"  "A" 
#[3,] "c"  "B" 
#[4,] "b"  "C" 

## since `B` has just 2 columns, we can use `B` rather than `B[, 1:2]`
A[B] <- 0

#  A B  C  D
#a 0 5  9 13
#b 2 6  0 14
#c 3 0 11 15
#d 4 8 12  0

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