在 NumPy 数组操作中避免双循环 [英] Avoiding double for-loops in NumPy array operations

问题描述

假设我有两个二维 NumPy 数组 A 和 B，我想计算矩阵 C，其条目为 C[i, j] = f(A[i, :], B[:, j])，其中 f 是一些函数，它接受两个一维数组并返回一个数字.

Suppose I have two 2D NumPy arrays A and B, I would like to compute the matrix C whose entries are C[i, j] = f(A[i, :], B[:, j]), where f is some function that takes two 1D arrays and returns a number.

例如，如果 def f(x, y): return np.sum(x * y) 那么我只会有 C = np.dot(A, B).但是，对于一般函数 f，是否有我可以利用的 NumPy/SciPy 实用程序比执行双 for 循环更有效?

For instance, if def f(x, y): return np.sum(x * y) then I would simply have C = np.dot(A, B). However, for a general function f, are there NumPy/SciPy utilities I could exploit that are more efficient than doing a double for-loop?

例如取def f(x, y): return np.sum(x != y)/len(x)，其中x和y 不是简单的 0/1 位向量.

For example, take def f(x, y): return np.sum(x != y) / len(x), where x and y are not simply 0/1-bit vectors.

推荐答案

以下是使用广播的合理通用方法.

Here is a reasonably general approach using broadcasting.

首先，将两个矩阵重塑为四阶张量.

First, reshape your two matrices to be rank-four tensors.

A = A.reshape(A.shape + (1, 1))
B = B.reshape((1, 1) + B.shape)

其次，一个元素一个元素地应用你的函数，而不执行任何归约.

Second, apply your function element by element without performing any reduction.

C = f(A, B)  # e.g. A != B

重构矩阵后，numpy 可以广播.结果张量 C 具有形状 A.shape + B.shape.

Having reshaped your matrices allows numpy to broadcast. The resulting tensor C has shape A.shape + B.shape.

第三，应用任何所需的减少，例如，对要丢弃的索引求和:

Third, apply any desired reduction by, for example, summing over the indices you want to discard:

C = C.sum(axis=(1, 3)) / C.shape[0]

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