如何在Matlab上展开矩阵? [英] How to unfold a Matrix on Matlab?

问题描述

我有一个给定的矩阵 H ，我想通过以下方法展开(展开)以找到矩阵 B :

I have a given matrix H and I would like to unfold (expand) it to find a matrix B by following the method below :

让 H 为尺寸为 m×n 的矩阵.让 x = gcd(m，n)

Let H be a matrix of dimension m × n. Let x = gcd (m,n)

1. 矩阵 H 分为两部分.
2. 切割模式如下:

• 对角切割"指的是是通过将 c = n/x 个单位向右交替移动(我们将 c 个单位向右移动几次)来实现的.
• 我们交替向下移动 cb = m/x 个单位(即 b =(nm)/x )(向下移动 b 个单位几次).
• The "diagonal cut" is made by alternately moving c = n/x units to the right (we move c units to the right several times).
• We alternately move c-b = m/x units down (i.e. b = (n-m)/x) (we move b units down several times).
在应用此对角切口"之后，矩阵，我们重复复制和粘贴两个部分以获得矩阵B.
1. After applying this "diagonal cut" of the matrix, we copy and paste the two parts repeatedly to obtain the matrix B.

示例:让尺寸为 m×n = 5×10 的矩阵 H 由:

Exemple : Let the matrix H of dimension m × n = 5 × 10 defined by :

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

• 让我们计算 x = gcd(m，n)= gcd(5,10)= 5 ，
• 或者向右移动: c = n/x = 10/5 = 2 ，
• 或者向下移动: b =(n-m)/x =(10-5)/5 = 1 .
1. 对角切割图:矩阵 H 分为两部分.切割方式如下:
1. Diagonal cutting diagram : The matrix H is cut in two parts. The cutting pattern is such that :
• 我们将 c = 2 个单位向右移动几次 c = 2 个单位向右
• 我们反复向下移动 c-b = 1 个单位.
• We move c = 2 units to the right several times c = 2 units to the right,
• We repeatedly move c - b = 1 unit downwards.

我们得到:

在应用此对角切口"之后，对于矩阵，我们重复复制并粘贴两个部分以获得矩阵:
1. After applying this "diagonal cut" of the matrix, we copy and paste the two parts repeatedly to obtain the matrix :

备注:在矩阵 X ， X1 和 X2 中，破折号为零.

Remark : In the matrices X, X1 and X2 the dashes are zeros.

1. 结果矩阵 B 为( L 为因数):
1. The resulting matrix B is (L is factor) :

有什么建议吗?

推荐答案

这可以通过创建带有切割图案的逻辑蒙版，然后逐个将输入乘以蒙版及其取反来实现.可以使用 blkdiag 重复 L .

This can be done by creating a logical mask with the cutting pattern, and then element-wise multiplying the input by the mask and by its negation. Repeating by L can be done with blkdiag.

H = [1 0 1 1 1 0 1 1 0 0
     0 1 1 0 0 1 1 0 1 1
     1 1 0 1 1 1 0 1 0 0
     0 1 1 0 1 0 1 0 1 1
     1 0 0 1 0 1 0 1 1 1];
L = 2;
[m, n] = size(H);
x = gcd(m, n);
c = n / x;
b = (n-m)/x;
mask = repelem(tril(true(m/b)), b, c);
A = [H.*mask; H.*~mask];
A = repmat({A}, L, 1);
B = blkdiag(A{:});

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