测试2d numpy数组中的成员资格 [英] test for membership in a 2d numpy array

问题描述

我有两个大小相同的2D数组

I have two 2D arrays of the same size

a = array([[1,2],[3,4],[5,6]])
b = array([[1,2],[3,4],[7,8]])

我想知道a中b的行.

所以输出应该是:

array([ True,  True, False], dtype=bool)

不做:

array([any(i == a) for i in b])

因为a和b很大.

有一个函数可以执行此操作，但仅适用于一维数组:in1d

There is a function that does this but only for 1D arrays : in1d

推荐答案

我们真正想做的是使用np.in1d ...，除了np.in1d仅适用于一维数组.我们的数组是多维的.但是，我们可以将数组 view 视为 strings 的一维数组:

What we'd really like to do is use np.in1d... except that np.in1d only works with 1-dimensional arrays. Our arrays are multi-dimensional. However, we can view the arrays as a 1-dimensional array of strings:

arr.view(np.dtype((np.void, arr.dtype.itemsize * arr.shape[-1])))

例如，

In [15]: arr = np.array([[1, 2], [2, 3], [1, 3]])

In [16]: arr = arr.view(np.dtype((np.void, arr.dtype.itemsize * arr.shape[-1])))

In [30]: arr.dtype
Out[30]: dtype('V16')

In [31]: arr.shape
Out[31]: (3, 1)

In [37]: arr
Out[37]: 
array([[b'\x01\x00\x00\x00\x00\x00\x00\x00\x02\x00\x00\x00\x00\x00\x00\x00'],
       [b'\x02\x00\x00\x00\x00\x00\x00\x00\x03\x00\x00\x00\x00\x00\x00\x00'],
       [b'\x01\x00\x00\x00\x00\x00\x00\x00\x03\x00\x00\x00\x00\x00\x00\x00']],
      dtype='|V16')

这使arr的每一行成为一个字符串.现在，只需将其挂接即可 到np.in1d:

This makes each row of arr a string. Now it is just a matter of hooking this up to np.in1d:

import numpy as np

def asvoid(arr):
    """
    Based on http://stackoverflow.com/a/16973510/190597 (Jaime, 2013-06)
    View the array as dtype np.void (bytes). The items along the last axis are
    viewed as one value. This allows comparisons to be performed on the entire row.
    """
    arr = np.ascontiguousarray(arr)
    if np.issubdtype(arr.dtype, np.floating):
        """ Care needs to be taken here since
        np.array([-0.]).view(np.void) != np.array([0.]).view(np.void)
        Adding 0. converts -0. to 0.
        """
        arr += 0.
    return arr.view(np.dtype((np.void, arr.dtype.itemsize * arr.shape[-1])))


def inNd(a, b, assume_unique=False):
    a = asvoid(a)
    b = asvoid(b)
    return np.in1d(a, b, assume_unique)


tests = [
    (np.array([[1, 2], [2, 3], [1, 3]]),
     np.array([[2, 2], [3, 3], [4, 4]]),
     np.array([False, False, False])),
    (np.array([[1, 2], [2, 2], [1, 3]]),
     np.array([[2, 2], [3, 3], [4, 4]]),
     np.array([True, False, False])),
    (np.array([[1, 2], [3, 4], [5, 6]]),
     np.array([[1, 2], [3, 4], [7, 8]]),
     np.array([True, True, False])),
    (np.array([[1, 2], [5, 6], [3, 4]]),
     np.array([[1, 2], [5, 6], [7, 8]]),
     np.array([True, True, False])),
    (np.array([[-0.5, 2.5, -2, 100, 2], [5, 6, 7, 8, 9], [3, 4, 5, 6, 7]]),
     np.array([[1.0, 2, 3, 4, 5], [5, 6, 7, 8, 9], [-0.5, 2.5, -2, 100, 2]]),
     np.array([False, True, True]))
]

for a, b, answer in tests:
    result = inNd(b, a)
    try:
        assert np.all(answer == result)
    except AssertionError:
        print('''\
a:
{a}
b:
{b}

answer: {answer}
result: {result}'''.format(**locals()))
        raise
else:
    print('Success!')

收益

Success!

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