如果使用生成器表达式，为什么Numpy.all()和any()给出错误的结果? [英] Why do Numpy.all() and any() give wrong results if you use generator expressions?

问题描述

使用其他人的代码时，我偶然发现了这个陷阱.那么，对numpy行为的解释是什么?

Working with somebody else's code I stumbled across this gotcha. So what is the explanation for numpy's behavior?

In [1]: import numpy as np

In [2]: foo = [False, False]

In [3]: print np.any(x == True for x in foo)
True  # <- bad numpy!

In [4]: print np.all(x == True for x in foo)
True  # <- bad numpy!

In [5]: print np.all(foo)
False  # <- correct result

p.s.我从这里获得了列表理解代码:检查列表是否仅包含项x

p.s. I got the list comprehension code from here: Check if list contains only item x

推荐答案

np.any和np.all在生成器上不起作用.他们需要序列.当给定非序列时，他们将其视为任何其他对象，并在其上调用bool(或执行等效操作)，这将返回True:

np.any and np.all don't work on generators. They need sequences. When given a non-sequence, they treat this as any other object and call bool on it (or do something equivalent), which will return True:

>>> false = [False]
>>> np.array(x for x in false)
array(<generator object <genexpr> at 0x31193c0>, dtype=object)
>>> bool(x for x in false)
True

列表理解有效，但是:

>>> np.all([x for x in false])
False
>>> np.any([x for x in false])
False

我建议在需要生成器时使用Python内置的any和all，因为它们通常比使用NumPy和列表理解要快(由于两次转换，首先是转换为list，然后转换为).

I advise using Python's built-in any and all when generators are expected, since they are typically faster than using NumPy and list comprehensions (because of a double conversion, first to list, then to array).

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