如何在相关矩阵中将较大的值移近矩阵对角线 [英] How to move larger values close to matrix diagonal in a correlation matrix

问题描述

我有一个包含五个元素(C1，C2，C3，C4，C5)的相关矩阵X

I have a correlation matrix X of five elements(C1,C2,C3,C4,C5)

      C1    C2    C3     C4   C5  

 C1    *     1     0     1     0
 C2    1     *     0     0     1
 C3    0     0     *     1     1
 C4    1     0     1     *     0
 C5    0     1     1     0     *

我想使用MatLab移动尽可能多的非零单元格靠近对角线，同时保持对角线单元为"*".

I want to use MatLab to move as many as non-zero cells close to diagonal, while keep the diagonal cells are "*".

例如，您可能会注意到列和行在以下矩阵中移动，而对角线单元格为"*".

For example, you may notice that the columns and rows is shifting in the following matrix, while the diagonal cells are "*".

      C1    C4    C2     C5   C3  

 C1    *     1     1     0     0
 C4    1     *     0     0     1
 C2    1     0     *     1     0
 C5    0     0     1     *     1
 C3    0     1     0     1     *

因为我想进行聚类，所以我希望有尽可能多的非零像元在平移后变得接近对角线.这是一个NP难题.

Because I want to do clustering, so I want as many as non-zero cells get close to diagonal after shifting. It's an NP-hard problem.

任何人都知道MatLab的哪些功能可以实现这一目标?

Anyone know what functions in MatLab can realize this?

推荐答案

您正在寻找的可能是反向Cuthill-McKee算法(RCM)，该算法几乎可以满足您的要求:对于给定的矩阵，它会找到一个排列，该排列倾向于使其非零元素更接近对角线. MATLAB中有一个内置函数 symrcm 可以做到这一点

What you're looking for is probably the reverse Cuthill-McKee algorithm (RCM), which pretty much does what you want: for a given matrix it finds a permutation that tends to have its non-zero elements closer to the diagonal. There's a built-in function symrcm in MATLAB that does just that.

因此，假设X是您的矩阵，则可以执行以下操作:

So assuming that X is your matrix, you can do the following:

p = symrcm(X);
Xnew = X(p, p);

Xnew是新的重新排序矩阵，p是新的行/列顺序.

Xnew is the new reordered matrix, and p is the new row/column order.

首先创建一个矩阵:

X = [10 0 0 7 0; 3 20 0 0 11; 0 0 30 0 29; 12 7 0 40 0; 0 33 0 0 50]

现在让我们重新排序:

p = symrcm(X);
Xnew = X(p, p)

结果是:

Xnew =    
    40    12     7     0     0
     7    10     0     0     0
     0     3    20    11     0
     0     0    33    50     0
     0     0     0    29    30

似乎正确.

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