查找矩阵中Matlab中的行是否“下降".在另一个矩阵的行内

问题描述

我在Matlab中有一个大小为Gx(l*N)的矩阵Ksets和一个大小为MxN的矩阵A.

I have a matrix Ksets in Matlab with size Gx(l*N) and a matrix A with size MxN.

Ksets的每一行都可以分解为大小为1xN的l子行.

Each row of Ksets can be decomposed into l sub-rows with size 1xN.

让我用一个例子更好地解释.

Let me explain better with an example.

clear
N=3;
l=4;
G=2;
M=5;

Ksets=[1 2 3      5 6 7      9 10 11   0 0 0;
       13 14 15   1 2 3      21 22 23  1 1 1]; %Gx(l*N)

A=[1 2 3;
   5 6 7;
   21 22 23;
   1 2 3;
   0 0 0]; %MxN

在示例中:

Ksets的

• 行1由大小为1xN的l个子行组成:[1 2 3]，[5 6 7]，[9 10 11]，[0 0 0]；

• row 1 of Ksets is composed of l sub-rows with size 1xN: [1 2 3], [5 6 7], [9 10 11], [0 0 0];

行2由大小为1xN的l个子行组成:[13 14 15]，[1 2 3]，[21 22 23]，[1 1 1].

row 2 of Ksets is composed of l sub-rows with size 1xN: [13 14 15], [1 2 3], [21 22 23], [1 1 1].

我假设Ksets的每一行都不包含相等的子行.

I assume that each row of Ksets does not contain equal sub-rows.

如果行A(m,:)等于Ksets(g,:)的l子行之一，否则，我希望您的帮助以Response(g,m)=1构造大小为GxM的矩阵Response.

I would like your help to construct a matrix Response with size GxM with Response(g,m)=1 if the row A(m,:) is equal to one of the l sub-rows of Ksets(g,:) and zero otherwise.

继续上面的示例

Response= [1 1 0 1 1;
           1 0 1 1 0]; %GxM

这段代码可以满足我的要求:

This code does what I want:

Responsecorrectmy=zeros(G, M);
for x=1:G
    for c=1:l
        Responsecorrectmy(x,:)=Responsecorrectmy(x,:)+...
                               ismember(A,Ksets(x,(c-1)*N+1:c*N), 'rows').';
    end
end

我的代码包含2个循环，这是不希望的，因为在我的实际算法中，G,l很大.您有没有循环的建议吗?

My code consists of 2 loops which is undesirable because in my real algorithm G,l are big. Do you have suggestions without loops?

推荐答案

只需稍作调整即可进行尺寸调整:

It can be done with a little reshaping and dimension-permuting:

Response = permute(any(all(bsxfun(@eq, reshape(Ksets.', N, [], G), permute(A, [2 3 4 1])), 1), 2), [3 4 1 2]);

其工作原理如下:

1. reshape(Ksets.', N, [], G)将Ksets整形为N×l×G 3D阵列，以便每个子行现在都彼此分开.
2. bsxfun(@eq, ..., permute(A, [2 3 4 1]))创建一个N×l×G×M 4D数组，其结果是将步骤1中的3D数组的每个元素与A中的每个值进行比较. /li>
3. all(..., 1)测试每个子行(即第一维)的元素 all 是否匹配. any(...,2)测试是否发生在原始行之一(即第二维)的任何子行中.
4. permute(..., [3 4 1 2])删除前两个维度(在步骤3中变为单例)，从而提供所需的G×M结果. (为此使用squeeze是不安全的，因为如果G=1，它将错误地删除三维尺寸.)

1. reshape(Ksets.', N, [], G) reshapes Ksets into an N×l×G 3D-array so that each subrow is now separated from the others.
2. bsxfun(@eq, ..., permute(A, [2 3 4 1])) creates an N×l×G×M 4D-array with the result of comparing each element of the 3D-array from step 1 with each value in A.
3. all(..., 1) tests if all the elements of each subrow (i.e. first dimension) match. any(...,2) tests if this happens for any subrow of one of the original rows (i.e. second dimension).
4. permute(..., [3 4 1 2]) removes the first two dimensions (which became singleton in step 3), giving the desired G×M result. (It would not be safe to use squeeze for this because it would incorrectly remove the third dimension if G=1).

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