numpy逐元素外积 [英] numpy elementwise outer product

问题描述

我想在numpy中做两个2d数组的元素外部乘积.

I want to do the element-wise outer product of two 2d arrays in numpy.

A.shape = (100, 3) # A numpy ndarray
B.shape = (100, 5) # A numpy ndarray

C = element_wise_outer_product(A, B) # A function that does the trick
C.shape = (100, 3, 5) # This should be the result
C[i] = np.outer(A[i], B[i]) # This should be the result

天真的实现可以做到以下几点.

A naive implementation can the following.

tmp = []
for i in range(len(A):
    outer_product = np.outer(A[i], B[i])
    tmp.append(outer_product)
C = np.array(tmp)

从堆栈溢出中得到启发的更好的解决方案.

A better solution inspired from stack overflow.

big_outer = np.multiply.outer(A, B)
tmp = np.swapaxes(tmp, 1, 2)
C_tmp = [tmp[i][i] for i in range(len(A)]
C = np.array(C_tmp)

我正在寻找摆脱了for循环的矢量化实现. 有人有主意吗? 谢谢！

I'm looking for a vectorized implementation that gets rid the for loop. Does anyone have an idea? Thank you!

推荐答案

将A和B扩展为3D，以保持其第一轴对齐，并分别使用 None/np.newaxis ，然后彼此相乘.这将允许 broadcasting 参与矢量化解决方案.

Extend A and B to 3D keeping their first axis aligned and introducing new axes along the third and second ones respectively with None/np.newaxis and then multiply with each other. This would allow broadcasting to come into play for a vectorized solution.

因此，一个实现将是-

A[:,:,None]*B[:,None,:]

我们可以使用 ellipsis将其缩短一点表示A::,:，并跳过用B列出剩余的最后一个轴，就像这样-

We could shorten it a bit by using ellipsis for A's : :,: and skip listing the leftover last axis with B, like so -

A[...,None]*B[:,None]

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作为另一种矢量化方法，我们还可以使用 np.einsum ，一旦我们跳过了字符串表示法的语法，并认为这些表示法是天真的循环实现中涉及的迭代器的代表，它可能会更直观，

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As another vectorized approach we could also use np.einsum, which might be more intuitive once we get past the string notation syntax and consider those notations being representatives of the iterators involved in a naive loopy implementation, like so -

np.einsum('ij,ik->ijk',A,B)

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