在Numpy数组中查找子列表的索引 [英] Finding the index of a sublist in a Numpy Array

问题描述

我正在努力寻找Numpy数组中子列表的索引.

I am struggling with finding the index of a sublist in a Numpy Array.

a = [[False,  True,  True,  True],
     [ True,  True,  True,  True],
     [ True,  True,  True,  True]]
sub = [True, True, True, True]
index = np.where(a.tolist() == sub)[0]
print(index)

这段代码给了我

array([0 0 0 1 1 1 1 2 2 2 2])

我无法向我解释.输出不应该是 array([1，2])，为什么不是呢?另外我该如何实现此输出?

which I cannot explain to me. Shouldn't the output be array([1, 2]) and why is it not? Also how can I achieve this output?

推荐答案

如果我正确理解，这是我的主意:

If I understand correctly, here's my idea:

>>> a
array([[False,  True,  True,  True],
       [ True,  True,  True,  True],
       [ True,  True,  True,  True]])
>>> sub
>>> array([ True,  True,  True,  True])
>>> 
>>> result, = np.where(np.all(a == sub, axis=1))
>>> result
array([1, 2])

有关此解决方案的详细信息:

a == sub 给您

>>> a == sub
array([[False,  True,  True,  True],
       [ True,  True,  True,  True],
       [ True,  True,  True,  True]])

一个布尔数组，其中每行的 True / False 值指示 a 中的值是否等于<代码>子.( sub 正在此处的行中广播.)

a boolean array where for each row the True/False value indicates if the value in a is equal to the corresponding value in sub. (sub is being broadcasted along the rows here.)

np.all(a == sub，axis = 1)给您

>>> np.all(a == sub, axis=1)
array([False,  True,  True])

一个布尔数组，对应于等于 sub 的 a 行.

a boolean array corresponding to the rows of a that are equal to sub.

在此子结果上使用 np.where 可以为您提供此布尔数组为 True 的索引.

Using np.where on this sub-result gives you the indices where this boolean array is True.

有关您的尝试的详细信息:

np.where(a == sub)(不需要 tolist )为您提供了两个数组，它们共同指示数组 a ==处的索引sub 是 True .

np.where(a == sub) (the tolist is unnecessary) gives you two arrays which together indicate the indices where the array a == sub is True.

>>> np.where(a == sub)
(array([0, 0, 0, 1, 1, 1, 1, 2, 2, 2, 2]),
 array([1, 2, 3, 0, 1, 2, 3, 0, 1, 2, 3]))

如果将这两个数组压缩在一起，则会得到行/列索引，其中 a == sub 是 True ，即

If you would zip these two arrays together you would get the row/column indices where a == sub is True, i.e.

>>> for row, col in zip(*np.where(a==sub)):
...:    print('a == sub is True at ({}, {})'.format(row, col))
a == sub is True at (0, 1)
a == sub is True at (0, 2)
a == sub is True at (0, 3)
a == sub is True at (1, 0)
a == sub is True at (1, 1)
a == sub is True at (1, 2)
a == sub is True at (1, 3)
a == sub is True at (2, 0)
a == sub is True at (2, 1)
a == sub is True at (2, 2)
a == sub is True at (2, 3)

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