创建一个矩阵，该矩阵在任意偏移对角线上 [英] Create a matrix with ones on any offset diagonal

问题描述

我的目标是创建一个在偏移对角线上带有矩阵的函数:

I'm aiming for a function that creates matrices with ones on offset diagonals:

与eye(5)函数类似，但现在在对角线偏移上.最好不要使用双for循环.我不需要完整的矩阵，而是必须将它们插入现有的矩阵中.我该怎么做?

So similar to the eye(5) function, but now on offset diagonals. Preferably not using double for loops. I do not want the full matrix, rather I have to insert them into an existing matrix. How can I accomplish this?

推荐答案

diag 内置了以下功能:

diag has this functionality built in:

diag(ones(4,1),1)
ans =

     0     1     0     0     0
     0     0     1     0     0
     0     0     0     1     0
     0     0     0     0     1
     0     0     0     0     0

diag(ones(4,1),-1)

ans =

     0     0     0     0     0
     1     0     0     0     0
     0     1     0     0     0
     0     0     1     0     0
     0     0     0     1     0

diag(V,k)的语法是:V是要放在对角线上的向量(可以是1或任何奇数向量)，而k是对角线的标签. 0是主对角线，正整数逐渐远离上对角线，负整数与下对角线相同；即k=1给出了第一个上对角线，k=-4给出了左下角.

Where the syntax of diag(V,k) is: V is the vector to be put on the diagonal (be it ones, or any odd vector), and k is the label of the diagonal. 0 is the main diagonal, positive integers are increasingly further away upper diagonals and negative integers the same for the lower diagonals; i.e. k=1 gives the first upper diagonal, k=-4 gives the lower left corner in this example.

出于完整性考虑，如果您只希望索引而不是完整矩阵(因为您建议将向量插入当前矩阵)，则可以使用以下函数:

For completeness, if you just want the indices instead of a full matrix (since you suggested you wanted to insert a vector into a present matrix) you can use the following function:

function [idx] = diagidx(n,k)
% n size of square matrix
% k number of diagonal
if k==0 % identity
    idx = [(1:n).' (1:n).']; % [row col]
elseif k>0 % Upper diagonal
    idx = [(1:n-k).' (1+k:n).'];
elseif k<0 % lower diagonal
    idx = [(1+abs(k):n).' (1:n-abs(k)).'];
end
end

其中idx的每一行都包含矩阵的索引.

where each row of idx contains the indices for the matrix.

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