为什么在numpy中，"nan == nan"是False而[nan]中的nan是True? [英] Why in numpy `nan == nan` is False while nan in [nan] is True?
问题描述
问题的第一部分(标题中)之前已经回答了几次(即如何检查包含NaN的列表)?
While the first part of the question (which is in the title) has been answered a few times before (i.e., Why is NaN not equal to NaN?), I don't see why the second piece works the way it does (inspired by this question How to Check list containing NaN)?
即:
>> nan == nan
False
>> nan in [nan]
True
考虑到@DSM的答案的问题解释性附录.那么，为什么float("nan")的行为不同于nan?它不应该再次对简单的nan求值，为什么解释器会表现为这种方式?
An explanatory addendum to the question considering the answer from @DSM. So, why float("nan") is behaving differently from nan? Shouldn't it evaluate again to simple nan and why interpreter behaves this way?
>> x = float("nan")
>> y = nan
>> x
nan
>> y
nan
>> x is nan, x is float("nan"), y is nan
(False, False, True)
基本上，在第一种情况下它引用相同的泛型nan，但是在第二种情况下创建单独的对象:
Basically, it refers to same generic nan in the first case, but creates separate object in the second:
>> nans = [nan for i in range(2)]
>> map(id, nans)
[190459300, 190459300]
>> nans = [float("nan") for i in range(2)]
>> map(id, nans)
[190459300, 190459301]
推荐答案
nan不等于nan是nan定义的一部分，因此很容易.
nan not being equal to nan is part of the definition of nan, so that part's easy.
对于nan in [nan]为True，这是因为在相等性之前对身份进行测试以包含在列表中.您正在比较相同的两个对象.
As for nan in [nan] being True, that's because identity is tested before equality for containment in lists. You're comparing the same two objects.
如果您尝试使用两个不同 nan进行相同的操作，则会得到False:
If you tried the same thing with two different nans, you'd get False:
>>> nans = [float("nan") for i in range(2)]
>>> map(id, nans)
[190459300, 190459284]
>>> nans
[nan, nan]
>>> nans[0] is nans[1]
False
>>> nans[0] in nans
True
>>> nans[0] in nans[1:]
False
您的附录与nan并没有多大关系，这只是Python的工作方式.一旦您了解到float("nan")没有义务返回某些nan单例，并且y = x不会复制x而是将名称y绑定到由x命名的对象，则一无所有.
Your addendum doesn't really have much to do with nan, that's simply how Python works. Once you understand that float("nan") is under no obligation to return some nan singleton, and that y = x doesn't make a copy of x but instead binds the name y to the object named by x, there's nothing left to get.
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