numpy:具有多个元素的数组的真值不明确 [英] Numpy : The truth value of an array with more than one element is ambiguous

问题描述

我真的很困惑为什么出现此错误.这是我的代码:

I am really confused on why this error is showing up. Here is my code:

import numpy as np

x = np.array([0, 0])
y = np.array([10, 10])
a = np.array([1, 6])
b = np.array([3, 7])
points = [x, y, a, b]
max_pair = [x, y]
other_pairs = [p for p in points if p not in max_pair]
>>>ValueError: The truth value of an array with more than one element is ambiguous. 
Use a.any() or a.all()
(a not in max_paix)
>>>ValueError: The truth ...

让我感到困惑的是，以下方法工作正常:

What confuses me is that the following works fine:

points = [[1, 2], [3, 4], [5, 7]]
max_pair = [[1, 2], [5, 6]]
other_pairs = [p for p in points if p not in max_pair]
>>>[[3, 4], [5, 7]]
([5, 6] not in max_pair)
>>>False

为什么在使用numpy数组时会发生这种情况? not in/in是否存在歧义?
any()\all()的正确语法是什么?

Why is this happening when using numpy arrays? Is not in/in ambiguous for existance?
What is the correct syntax using any()\all()?

推荐答案

Numpy数组定义了一个自定义的相等运算符，即它们是实现__eq__魔术函数的对象.因此，==运算符和所有依赖于这种相等性的其他函数/运算符都称为此自定义相等性函数.

Numpy arrays define a custom equality operator, i.e. they are objects that implement the __eq__ magic function. Accordingly, the == operator and all other functions/operators that rely on such an equality call this custom equality function.

Numpy的相等性基于数组的逐元素比较.因此，作为回报，您将获得另一个具有布尔值的numpy数组.例如:

Numpy's equality is based on element-wise comparison of arrays. Thus, in return you get another numpy array with boolean values. For instance:

x = np.array([1,2,3])
y = np.array([1,4,5])
x == y

返回

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

但是，将in运算符与 lists 结合使用时，要求相等比较仅返回单个布尔值.这就是错误要求all或any的原因.例如:

However, the in operator in combination with lists requires equality comparisons that only return a single boolean value. This is the reason why the error asks for all or any. For instance:

any(x==y)

返回True，因为结果数组的至少一个值为True. 相反

returns True because at least one value of the resulting array is True. In contrast

all(x==y) 

返回False，因为不是结果数组的所有值都是True.

returns False because not all values of the resulting array are True.

因此，在您的情况下，解决该问题的方法如下:

So in your case, a way around the problem would be the following:

other_pairs = [p for p in points if all(any(p!=q) for q in max_pair)]

和print other_pairs打印预期结果

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

为什么呢?好吧，我们从 points 中寻找一个 p 项，其中任何项与 all 项不相等 max_pair 中的项目 q .

Why so? Well, we look for an item p from points where any of its entries are unequal to the entries of all items q from max_pair.

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