矩阵的对角线总和 [英] Sum of antidiagonal of a matrix

问题描述

我正在尝试将矩阵的对角线(次级对角线，次对角线)上的元素求和.

I'm trying to sum the elements along the antidiagonal (secondary diagonal, minor diagonal) of a matrix.

所以，如果我有一个矩阵m:

So, if I have a matrix m:

m <- matrix(c(2, 3, 1, 4, 2, 5, 1, 3, 7), 3)
m

     [,1] [,2] [,3]
[1,]    2    4    1
[2,]    3    2    3
[3,]    1    5    7

我正在寻找总和m[3, 1] + m[2, 2] + m[1, 3]，即1 + 2 + 1

我不知道如何设置迭代.据我了解，对此没有任何功能(例如diag()表示另一个对角线).

I can't figure out how to set up an iteration. As far as I know there is no function for this (like diag() for the other diagonal).

推荐答案

使用

m <- matrix(c(2, 3, 1, 4, 2, 5, 1, 3, 7), 3)

1)反转显示的行(或反转列-未显示)，取对角线和总和:

1) Reverse the rows as shown (or the columns - not shown), take the diagonal and sum:

sum(diag(m[nrow(m):1, ]))
## [1] 4

2)，或像这样使用row和col:

sum(m[c(row(m) + col(m) - nrow(m) == 1)])
## [1] 4

由于row(m) + col(m) - nrow(m)沿所有反对角线都是恒定的，因此可以推广到其他反对角线.对于这种概括，将c(...)中的部分写为row(m) + col(m) - nrow(m) - 1 == 0可能更方便，因为然后用-1替换0使用超对角线，用+1替换使用对角线. -2和2分别使用第二个超对角线和次对角线，依此类推.

This generalizes to other anti-diagonals since row(m) + col(m) - nrow(m) is constant along all anti-diagonals. For such a generalization it might be more convenient to write the part within c(...) as row(m) + col(m) - nrow(m) - 1 == 0 since then replacing 0 with -1 uses the superdiagonal and with +1 uses the subdiagonal. -2 and 2 use the second superdiagonal and subdiagonal respectively and so on.

3)或使用以下索引顺序:

3) or use this sequence of indexes:

n <- nrow(m)
sum(m[seq(n, by = n-1, length = n)])
## [1] 4

4)或使用outer这样:

n <- nrow(m)
sum(m[!c(outer(1:n, n:1, "-"))])
## [1] 4

由于outer(1:n, n:1, "-")沿反对角线是恒定的，因此此对其他反对角线也很好地概括.我们可以将其中的部分写为outer(1:n, n:1) == 0，如果将0替换为-1，我们将得到超级反对角线，而+1则得到子反对角线. -2和2给出super super和sub sub对角线.例如，sum(m[c(outer(1:n, n:1, "-") == 1)])是子反对角线的总和.

This one generalizes nicely to other anti-diagonals too as outer(1:n, n:1, "-") is constant along anti-diagonals. We can write the part within [...] as outer(1:n, n:1) == 0 and if we replace 0 with -1 we get the super anti-diagonal and with +1 we get the sub anti-diagonal. -2 and 2 give the super super and sub sub antidiagonals. For example sum(m[c(outer(1:n, n:1, "-") == 1)]) is the sum of the sub anti-diagonal.

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