获取跨两个二维 numpy 数组的相交行 [英] Get intersecting rows across two 2D numpy arrays

问题描述

我想在两个二维 numpy 数组中获取相交(公共)行.例如，如果以下数组作为输入传递:

I want to get the intersecting (common) rows across two 2D numpy arrays. E.g., if the following arrays are passed as inputs:

array([[1, 4],
       [2, 5],
       [3, 6]])

array([[1, 4],
       [3, 6],
       [7, 8]])

输出应该是:

array([[1, 4],
       [3, 6])

我知道如何用循环来做到这一点.我正在寻找一种 Pythonic/Numpy 方法来做到这一点.

I know how to do this with loops. I'm looking at a Pythonic/Numpy way to do this.

推荐答案

对于短数组，使用集合可能是最清晰、最易读的方法.

For short arrays, using sets is probably the clearest and most readable way to do it.

另一种方法是使用 numpy.intersect1d.您必须欺骗它以将行视为单个值，但是...这使事情变得不那么可读...

Another way is to use numpy.intersect1d. You'll have to trick it into treating the rows as a single value, though... This makes things a bit less readable...

import numpy as np

A = np.array([[1,4],[2,5],[3,6]])
B = np.array([[1,4],[3,6],[7,8]])

nrows, ncols = A.shape
dtype={'names':['f{}'.format(i) for i in range(ncols)],
       'formats':ncols * [A.dtype]}

C = np.intersect1d(A.view(dtype), B.view(dtype))

# This last bit is optional if you're okay with "C" being a structured array...
C = C.view(A.dtype).reshape(-1, ncols)

对于大型数组，这应该比使用集合快得多.

For large arrays, this should be considerably faster than using sets.

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