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

问题描述

我就像R中的矩阵，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

使用对角线时，我使用了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，我尝试了

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

但是我无法获得dataD b/c，它表示未找到对象'V1'

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

谢谢！

推荐答案

row()和col()辅助函数在这里很有用.使用@James A，我们可以使用以下小技巧获得对角线上方的位置:

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

和下面的对角线:

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

这些可以概括为任意所需的对角线:

These can be generalised to give whatever diagonals you want:

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

并且不需要任何子设置.

and don't require any subsetting.

然后，只需对这些值调用所需的任何函数，就很简单了.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

如果根据我的评论，您说的是除对角线以外的所有内容，请使用

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..

因此，这不会丢失，Ben Bolker建议(在评论中)可以使用我上面提到的row()和col()函数更巧妙地完成上述代码块:

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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