numpy.where返回元组的目的是什么? [英] What is the purpose of numpy.where returning a tuple?

问题描述

当我运行此代码时:

import numpy as np
a = np.array([1, 2, 3, 4, 5, 6])
print(np.where(a > 2))

自然会得到一个索引数组，其中 a>2 ，即 [2，3，4，5] ，但是我们得到:

it would be natural to get an array of indices where a > 2, i.e. [2, 3, 4, 5], but instead we get:

(array([2, 3, 4, 5], dtype=int64),)

即具有空第二个成员的元组.

i.e. a tuple with empty second member.

然后，要获得 numpy.where 的自然"答案，我们必须做:

Then, to get the the "natural" answer of numpy.where, we have to do:

np.where(a > 2)[0]

该元组有什么意义?在哪种情况下有用?

注意:我在这里只谈论用例 numpy.where(cond)，而不是同样存在的 numpy.where(cond，x，y)(请参阅文档).

Note: I'm speaking here only about the use case numpy.where(cond) and not numpy.where(cond, x, y) that also exists (see documentation).

推荐答案

numpy.where 返回一个元组，因为该元组的每个元素都引用一个维度.

numpy.where returns a tuple because each element of the tuple refers to a dimension.

请从2维角度考虑此示例:

Consider this example in 2 dimensions:

a = np.array([[1, 2, 3, 4, 5, 6],
              [-2, 1, 2, 3, 4, 5]])

print(np.where(a > 2))

(array([0, 0, 0, 0, 1, 1, 1], dtype=int64),
 array([2, 3, 4, 5, 3, 4, 5], dtype=int64))

如您所见，元组的第一个元素指的是相关元素的第一个维度；第二个元素是第二个维度.

As you can see, the first element of the tuple refers to the first dimension of relevant elements; the second element refers to the second dimension.

这是 numpy 经常使用的约定.当您要求数组的形状时，您也会看到它，即一维数组的形状将返回包含1个元素的元组:

This is a convention numpy often uses. You will see it also when you ask for the shape of an array, i.e. the shape of a 1-dimensional array will return a tuple with 1 element:

a = np.array([[1, 2, 3, 4, 5, 6],
              [-2, 1, 2, 3, 4, 5]])

print(a.shape, a.ndim)  # (2, 6) 2

b = np.array([1, 2, 3, 4, 5, 6])

print(b.shape, b.ndim)  # (6,) 1

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