numpy数组上的算术比较 [英] arithmetic comparisons on numpy arrays

问题描述

>>> import numpy as np
>>> x = np.eye(3)
>>> x[1, 2] = .5
>>> x
array([[ 1. ,  0. ,  0. ],
       [ 0. ,  1. ,  0.5],
       [ 0. ,  0. ,  1. ]])
>>> 0 < x.any() < 1
False
>>> 

我想检查numpy数组是否包含0到1之间的任何值. 我将0 < x.any() < 1读为如果有任何大小大于0且小于1的元素，则返回true"，但是显然不是这种情况.

I would like to check if numpy array contains any value between 0 and 1.
I read 0 < x.any() < 1 as 'if there is any element with size greater then 0 and less then 1, return true', but that's obviously not the case.

如何在numpy数组上进行算术比较?

How can I do arithmetic comparison on numpy array?

推荐答案

>>> np.any((0 < x) & (x < 1))
True

x.any()实际执行的操作:与np.any(x)相同，这意味着如果x中的任何元素都不为零，则返回True.因此，您的比较结果是0 < True < 1，这是错误的，因为在python 2中，0 < True是true，但由于True == 1，True < 1不是.

What x.any() actually does: it's the same as np.any(x), meaning it returns True if any elements in x are nonzero. So your comparison is 0 < True < 1, which is false because in python 2 0 < True is true, but True < 1 is not, since True == 1.

相反，在这种方法中，我们对每个元素进行比较是否为真的布尔数组，然后检查该数组中是否有任何元素为真:

In this approach, by contrast, we make boolean arrays of whether the comparison is true for each element, and then check if any element of that array is true:

>>> 0 < x
array([[ True, False, False],
       [False,  True,  True],
       [False, False,  True]], dtype=bool)
>>> x < 1
array([[False,  True,  True],
       [ True, False,  True],
       [ True,  True, False]], dtype=bool)
>>> (0 < x) & (x < 1)
array([[False, False, False],
       [False, False,  True],
       [False, False, False]], dtype=bool)

您必须执行显式的&，因为不幸的是numpy不能(而且我认为不能)与python的比较运算符的内置链接一起工作.

You have to do the explicit &, because unfortunately numpy doesn't (and I think can't) work with python's built-in chaining of comparison operators.

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