R - 矩阵中非对角线元素的最小值、最大值和平均值 [英] R - min, max and mean of off-diagonal elements in a matrix

问题描述

我在 R 中有一个矩阵，我想得到:

I have like a matrix in R and I want to get:

Max off - diagonal elements
Min off – diagonal elements
Mean off –diagonal elements

我使用对角线 max(diag(A)) ， min(diag(A)) ， mean(diag(A)) 并且工作得很好

With diagonal I used max(diag(A)) , min(diag(A)) , mean(diag(A)) and worked just fine

但是对于非对角线我试过了

But for off-diagonal I tried

dataD <- subset(A, V1!=V2)

Error in subset.matrix(A, V1 != V2) : object 'V1' not found

使用:

colMeans(dataD) # get the mean for columns

但我无法获取 dataD b/c 它说找不到对象V1"

but I cannot get dataD b/c it says object 'V1' not found

谢谢！

推荐答案

这里的 row() 和 col() 辅助函数很有用.使用@James A，我们可以使用这个小技巧获得上非对角线:

Here the row() and col() helper functions are useful. Using @James A, we can get the upper off-diagonal using this little trick:

> A[row(A) == (col(A) - 1)]
[1]  5 10 15

以及通过以下对角线:

> A[row(A) == (col(A) + 1)]
[1]  2  7 12

这些可以概括为你想要的任何对角线:

These can be generalised to give whatever diagonals you want:

> A[row(A) == (col(A) - 2)]
[1]  9 14

并且不需要任何子集.

然后在这些值上调用您想要的任何函数是一件简单的事情.例如:

Then it is a simple matter of calling whatever function you want on these values. E.g.:

> mean(A[row(A) == (col(A) - 1)])
[1] 10

如果按照我的评论你的意思是除了对角线之外的所有东西，那么使用

If as per my comment you mean everything but the diagonal, then use

> diag(A) <- NA
> mean(A, na.rm = TRUE)
[1] 8.5
> max(A, na.rm = TRUE)
[1] 15
> # etc. using sum(A, na.rm = TRUE), min(A, na.rm = TRUE), etc..

所以这不会丢失，Ben Bolker 建议(在评论中)上面的代码块可以使用 row() 和 col() 上面提到的函数:

So this doesn't get lost, Ben Bolker suggests (in the comments) that the above code block can be done more neatly using the row() and col() functions I mentioned above:

mean(A[row(A)!=col(A)])
min(A[row(A)!=col(A)])
max(A[row(A)!=col(A)])
sum(A[row(A)!=col(A)])

这是一个更好的解决方案.

which is a nicer solution all round.

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