使用R将矩阵的对角线相加 [英] Adding Diagonals of a matrix using R

问题描述

我想将矩阵的上部对角线从中间开始相加，在列中递增直到(1，n)，n是最后一列，并保存每个对角线的每个和.我的代码只添加了中间的对角线，如何遍历矩阵以获得对角线的总和

I would like to add up the upper part diagonals of a matrix starting from the middle, with increment in column until (1,n), n being the last column and save each sum of every diagonal. My code only add the middle diagonal, how can I loop through the matrix to get the sum of the diagonals

A <- matrix(c(2, 4, 3, 1,
             5, 7, 1, 2,
             3, 2, 3, 4, 
             1, 5, 6, 0), # the data elements 
    nrow = 4, # number of rows 
    ncol = 4, # number of columns 
    byrow = TRUE) # fill matrix by rows

sum <- 0
print(A)
for (a in 1){
  for (b in 1:ncol){
    if (a<-b){
      sum = sum + A[a,b]
      print (sum) 
    }
  }
}

这是我的结果

> print(A)
     [,1] [,2] [,3] [,4]
[1,]    2    4    3    1
[2,]    5    7    1    2
[3,]    3    2    3    4
[4,]    1    5    6    0

for (a in 1){
  for (b in 1:ncol){ 
    if (a<-b){
      sum = sum + A[a,b]
      tail(sum, n=1)
    }
  }
}


12

推荐答案

您需要diag提取所有主要对角元素，并需要sum以获得它们的总和

You need diag to extract all main diagonal elements and sum to get the sum of them

sum(diag(A))

我不确定您要的内容是什么，但是如果您还想提取上三角矩阵，则可以使用A[upper.tri(A)]排除主要的对角线元素，也可以设置diag=TRUE包括它们A[upper.tri(A, diag = TRUE)]

I am not sure about what you're asking for, but if you also want to extract the upper triangular matrix, you can use A[upper.tri(A)] which excludes the main diagonal elements, you can also set diag=TRUE to include them A[upper.tri(A, diag = TRUE)]

@shegzter根据您的评论，可以将col和row与逻辑比较==结合使用以获得所需的数字.

@shegzter based on your comment, you can use col and row combined with logical comparison == to get the numbers you want.

> A[row(A)==col(A)] # this gives the same out put as `diag(A)`
[1] 2 7 3 0
> A[row(A)+1==col(A)]
[1] 4 1 4
> A[row(A)+2==col(A)]
[1] 3 2
> A[row(A)+3==col(A)]
[1] 1

如果您希望每个元素的总和，请在这些元素上使用sum:

If you want the sum of each of them, so use sum over those elements:

> sum(A[row(A)==col(A)])
[1] 12
> sum(A[row(A)+1==col(A)])
[1] 9
> sum(A[row(A)+2==col(A)])
[1] 5
> sum(A[row(A)+3==col(A)])
[1] 1

如果您的目标是获得12 + 9 + 5 + 1的总和，则可以使用upper.tri和sum

If your objective is getting the following sum 12+9+5+1, then you can do it all at once using upper.tri and sum

> sum(A[upper.tri(A, diag = TRUE)])
[1] 27

或者没有对角线元素:

> sum(A[upper.tri(A)])
[1] 15

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