比较包含NaN的列表
我试图比较两个不同的列表以查看它们是否相等,并且打算删除NaN,但是发现我的列表比较仍然有效,尽管NaN == NaN -> False
。
有人可以解释为什么下面的结果为True
或False
,因为我发现此行为是意外的。谢谢,
我已阅读以下内容,似乎无法解决该问题:
(Python 2.7.3,numpy-1.9.2)
我用*
结尾标记了令人惊讶的评估
>>> nan = np.nan
>>> [1,2,3]==[3]
False
>>> [1,2,3]==[1,2,3]
True
>>> [1,2,nan]==[1,2,nan]
True ***
>>> nan == nan
False
>>> [nan] == [nan]
True ***
>>> [nan, nan] == [nan for i in range(2)]
True ***
>>> [nan, nan] == [float(nan) for i in range(2)]
True ***
>>> float(nan) is (float(nan) + 1)
False
>>> float(nan) is float(nan)
True ***
-
要了解这里发生的情况,只需替换
nan = np.nan
为foo = float('nan')
,您将得到完全相同的结果,为什么?>>> foo = float('nan') >>> foo is foo # This is obviously True! True >>> foo == foo # This is False per the standard (nan != nan). False >>> bar = float('nan') # foo and bar are two different objects. >>> foo is bar False >>> foo is float(foo) # "Tricky", but float(x) is x if type(x) == float. True
现在认为这
numpy.nan
只是一个包含float('nan')
。的变量名。现在为什么
[nan] == [nan]
仅仅是因为list
比较首先要在项目之间进行身份平等性的检验,然后才是价值相等,所以将其视为:def equals(l1, l2): for u, v in zip(l1, l2): if u is not v and u != v: return False return True