评估所有成对的矩阵

问题描述

我有一个函数f，它使用2个具有相同行数的矩阵并求出一个标量值.现在，我正在寻找创建新函数的可能性，该函数需要两个矩阵列表并为所有对调用f.

I have a function f which takes 2 matrices with the same number of rows and procudes a scalar value. A am now looking for a possibility to create a new function which takes two lists of matrices and calls f for all pairs.

我需要一个高效的摩尔循环实现:

I need a moore efficient implementation of this loop:

% X = cell of matrices
% Y = cell of matrices
for k=1:length(X)
    for l=1:length(Y)
        M(k,l) = f(X{k},Y{l});
    end
end

(不要求X和Y是单元格)

(It is not a requirement that X and Y are cells)

例如f可以是距离平方的均值

For example f could be the mean of squared distances

f = @(X,Y) mean(mean(bsxfun(@plus,dot(X,X,1)',dot(Y,Y,1))-2*(X'*Y)));

但不要质疑f，这只是一个更为复杂的问题的示例.

but don't question f, it is just an example of a much more complicated problem.

推荐答案

首先，您应该预先分配M.然后，您可以使用meshgrid轻松生成一个循环，以生成X和Y中所有成对的元素的列表:

Firstly you should be pre-allocating M. Then you can cut out one loop quite easily by using meshgrid to generate a list of all pairs of elements in X and Y:

[K, L] = meshgrid(1:length(X), 1:length(Y));

M = zeros(size(K));  %//This preallocation alone should give you a significant speed up
for ii = 1:numel(K)
    M(ii) = f(X{K(ii)},Y{L(ii)});
end

但是，如果可以实现f以便对其进行正确的矢量化处理，则可以将其传递给两个3D矩阵，即X和Y对的整个列表，并且完全没有循环.但这完全取决于您的f是做什么的.

However, if you can implement f so that it is properly vectorized, you might be able to pass it two 3D matrices, i.e. the entire list of X and Y pairs and do it without a loop at all. But this depends entirely on what it is that your f does.

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