Python numpy.nan 和逻辑函数:错误的结果
问题描述
我在尝试评估时得到了一些令人惊讶的结果可能包含 nan 值的数据的逻辑表达式(在 numpy 中定义).
I get some surprising results when trying to evaluate
logical expressions on data that might contain nan values (as defined in numpy).
我想了解为什么会出现这种结果以及如何正确实施.
I would like to understand why this results arise and how to implement the correct way.
我不明白为什么这些表达式的计算结果是它们所做的值:
What I don't understand is why these expressions evaluate to the value they do:
from numpy import nan
nan and True
>>> True
# this is wrong.. I would expect to evaluate to nan
True and nan
>>> nan
# OK
nan and False
>>> False
# OK regardless the value of the first element
# the expression should evaluate to False
False and nan
>>> False
#ok
或也类似:
True or nan
>>> True #OK
nan or True
>>> nan #wrong the expression is True
False or nan
>>> nan #OK
nan or False
>>> nan #OK
如何(以有效的方式)实现正确的布尔函数，同时处理 nan 值?
How can I implement (in an efficient way) the correct boolean functions, handling also nan values?
解决方案
您可以使用 numpy 命名空间中的谓词:
You can use predicates from the numpy namespace:
>>> np.logical_and(True, np.nan), np.logical_and(False, np.nan)
(True, False)
>>> np.logical_and(np.nan, True), np.logical_and(np.nan, False)
(True, False)
>>>
>>> np.logical_or(True, np.nan), np.logical_or(False, np.nan)
(True, True)
>>> np.logical_or(np.nan, True), np.logical_or(np.nan, False)
(True, True)
内置布尔运算符略有不同.来自文档:x and y 等价于 if x is false, then x, else y.因此，如果第一个参数的计算结果为 False，它们将返回它(不是它的布尔等价物，因为它是).因此:
The built-in boolean operators are slightly different. From the docs :
x and y is equivalent to if x is false, then x, else y. So, if the first argument evaluates to False, they return it (not its boolean equivalent, as it were). Therefore:
>>> (None and True) is None
True
>>> [] and True
[]
>>> [] and False
[]
>>>
等
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