比较包含NaN的列表 [英] Comparing lists containing NaNs
问题描述
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,只是发现尽管NaN == NaN -> False
,我的列表比较仍然有效.
I am trying to compare two different lists to see if they are equal, and was going to remove NaNs, only to discover that my list comparisons still work, despite NaN == NaN -> False
.
有人可以解释为什么以下内容评估True
或False
,因为我发现此行为是意外的.谢谢,
Could someone explain why the following evaluate True
or False
, as I am finding this behavior unexpected. Thanks,
我已阅读以下内容,似乎无法解决该问题:
I have read the following which don't seem to resolve the issue:
- Why in numpy
nan == nan
is False while nan in [nan] is True? - Why is NaN not equal to NaN? [duplicate]
(Python 2.7.3,numpy-1.9.2)
(Python 2.7.3, numpy-1.9.2)
我的评价令人吃惊,结尾处是*
I have marked surprising evaluations with a *
at the end
>>> 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 ***
推荐答案
要了解此处发生的情况,只需将nan = np.nan
替换为foo = float('nan')
,您将得到完全相同的结果,为什么?
To understand what happens here, simply replace nan = np.nan
by foo = float('nan')
, you will get exactly the same result, why?
>>> 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numpy.nan
只是一个包含float('nan')
的变量名.
Now think that numpy.nan
is just a variable name that holds a float('nan')
.
现在为什么[nan] == [nan]
仅仅是因为list
比较首先要测试项目之间的身份相等性,然后才是价值相等,所以将其视为:
Now why [nan] == [nan]
is simply because list
comparison first test identity equality between items before equality for value, think of it as:
def equals(l1, l2):
for u, v in zip(l1, l2):
if u is not v and u != v:
return False
return True
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