提取大矩阵的非对角线切片

问题描述

我有一个很大的nxn矩阵，想取不同大小的非对角切片.例如:

I've got a large nxn matrix and would like to take off-diagonal slices of varying sizes. For example:

1 2 3 4 5 6
1 2 3 4 5 6
1 2 3 4 5 6
1 2 3 4 5 6
1 2 3 4 5 6
1 2 3 4 5 6

我想要一个R函数，当给定矩阵和对角切片的宽度"时，它将返回仅包含这些值的nxn矩阵.因此，对于上面的矩阵，比如说3，我会得到:

I'd like an R function which, when given the matrix and "width of diagonal slice" would return an nxn matrix of just those values. So for the matrix above and, say, 3, I'd get:

1 x x x x x
1 2 x x x x
1 2 3 x x x
x 2 3 4 x x
x x 3 4 5 x
x x x 4 5 6

此刻，我正在使用(原谅我)一个for循环，该循环非常慢:

At the moment I'm using (forgive me) a for loop which is incredibly slow:

getDiags<-function(ndiags, cormat){
  resmat=matrix(ncol=ncol(cormat),nrow=nrow(cormat))
  dimnames(resmat)<-dimnames(cormat)
  for(j in 1:ndiags){
    resmat[row(resmat) == col(resmat) + j] <-
      cormat[row(cormat) == col(cormat) + j]
  }
  return(resmat)
}

我意识到这是解决该问题的一种非常"un-R"的方式.有没有更好的方法，可能使用diag或lower.tri?

I realise that this is a very "un-R" way to go about solving this problem. Is there a better way to do it, probably using diag or lower.tri?

推荐答案

size <- 6
mat <- matrix(seq_len(size ^ 2), ncol = size)


low <- 0
high <- 3

delta <- rep(seq_len(ncol(mat)), nrow(mat)) -
    rep(seq_len(nrow(mat)), each = ncol(mat))
#or Ben Bolker's better alternative
delta <- row(mat) - col(mat)
mat[delta < low | delta > high] <- NA
mat

这在我的机器上可以处理5000 x 5000矩阵

this works with 5000 x 5000 matrices on my machine

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