如何在Matlab中计算两个矩阵之间的快速外积？

发布于 2024-12-26 03:22:11 字数 306 浏览 2 评论 0原文

我有两个 n×m 矩阵，A 和 B。我想创建一个新的矩阵 C ，它类似于：

for i = 1:n
    C = C + outerProduct(A(i,:), B(i,:));
end

即 C 是一个大小为 mxm 的矩阵，是 A 的行的所有外积的总和 和B。

有没有一种无需 for 循环的快速方法（考虑到 for 循环在 Matlab 中非常慢）？

原文

I have two n-by-m matrices, A and B. I want to create a new matrix C which is something like:

for i = 1:n
    C = C + outerProduct(A(i,:), B(i,:));
end

i.e. C is a matrix of size m x m, the sum of all outer products of the rows of A and B.

Is there a fast way to do it without a for loop (given that for loops are notoriously slow in Matlab)?

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╰つ倒转 2025-01-02 03:22:11

您正在执行的操作（行外积之和）相当于 A 的转置版本与 B 的乘法：

C = A.'*B;

您可以使用以下示例查看这一点：

>> mat = magic(5);  %# A sample 5-by-5 matrix
>> A = mat(1:4,:);  %# Create a 4-by-5 matrix
>> B = mat(2:5,:);  %# Create another 4-by-5 matrix

>> C = zeros(5);  %# Initialize C to be 5-by-5
>> for i = 1:4, C = C + A(i,:).'*B(i,:); end;  %'# Calculate C as you are now

>> isequal(C, A.'*B)  %'# Test for equality with the shorter solution

ans =

     1  %# Equal!

The operation you are performing (the sum of the row outer products) is equivalent to the multiplication of a transposed version of A with B:

C = A.'*B;

You can see this using the following example:

>> mat = magic(5);  %# A sample 5-by-5 matrix
>> A = mat(1:4,:);  %# Create a 4-by-5 matrix
>> B = mat(2:5,:);  %# Create another 4-by-5 matrix

>> C = zeros(5);  %# Initialize C to be 5-by-5
>> for i = 1:4, C = C + A(i,:).'*B(i,:); end;  %'# Calculate C as you are now

>> isequal(C, A.'*B)  %'# Test for equality with the shorter solution

ans =

     1  %# Equal!

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