本文介绍了朱莉娅的numpy.einsum?(2)的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

来自这个问题,我想知道是否有可能更广义的einsum.让我们假设,我有问题

Coming from this question, I wonder if a more generalized einsum was possible. Let us assume, I had the problem

using PyCall
@pyimport numpy as np

a = rand(10,10,10)
b = rand(10,10)
c = rand(10,10,10)

Q = np.einsum("imk,ml,lkj->ij", a,b,c)

或者类似的东西,我如何在不循环求和的情况下解决这个问题?

Or something similar, how were I to solve this problem without looping through the sums?

致以最诚挚的问候

推荐答案

代码来执行此操作.它完全符合马特 B. 在他的评论中描述的内容.希望对你有帮助,如果有问题请告诉我.

code to do this. It follows exactly what Matt B. described in his comment. Hope it helps, let me know if there are problems with it.

https://github.com/ahwillia/Einsum.jl

这就是你将如何实现你的例子:

This is how you would implement your example:

using Einsum

a = rand(10,10,10)
b = rand(10,10)
c = rand(10,10,10)
Q = zeros(10,10)

@einsum Q[i,j] = a[i,m,k]*b[m,l]*c[l,k,j]

在底层,宏构建了以下一系列嵌套 for 循环,并在编译前将它们插入到您的代码中.(注意这不是插入的确切代码,它还会检查以确保输入的尺寸一致,使用 macroexpand 查看完整代码):

Under the hood the macro builds the following series of nested for loops and inserts them into your code before compile time. (Note this is not the exact code inserted, it also checks to make sure the dimensions of the inputs agree, using macroexpand to see the full code):

for j = 1:size(Q,2)
    for i = 1:size(Q,1)
        s = 0
        for l = 1:size(b,2)
            for k = 1:size(a,3)
                for m = 1:size(a,2)
                    s += a[i,m,k] * b[m,l] * c[l,k,j]
                end
            end
        end
        Q[i,j] = s
    end
end

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08-04 07:57
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