如何测试变量是否为pd.NaT?

问题描述

我正在尝试测试我的变量之一是否是pd.NaT.我知道它是NaT，但仍然无法通过测试.例如，以下代码不显示任何内容:

I'm trying to test if one of my variables is pd.NaT. I know it is NaT, and still it won't pass the test. As an example, the following code prints nothing :

a=pd.NaT

if a == pd.NaT:
    print("a not NaT")

有人知道吗?有没有一种方法可以有效地测试a是否为NaT?

Does anyone have a clue ? Is there a way to effectively test if a is NaT?

推荐答案

熊猫NaT的行为类似于浮点NaN，因为它不等于自身.相反，您可以使用pandas.isnull:

Pandas NaT behaves like a floating-point NaN, in that it's not equal to itself. Instead, you can use pandas.isnull:

In [21]: pandas.isnull(pandas.NaT)
Out[21]: True

这还会为None和NaN返回True.

This also returns True for None and NaN.

从技术上讲，您还可以按照用于浮点NaN的通用模式，用x != x检查熊猫NaT.但是，这可能会导致NumPy NaT出现问题，它们看起来非常相似并且代表相同的概念，但实际上是具有不同行为的不同类型:

Technically, you could also check for Pandas NaT with x != x, following a common pattern used for floating-point NaN. However, this is likely to cause issues with NumPy NaTs, which look very similar and represent the same concept, but are actually a different type with different behavior:

In [29]: x = pandas.NaT

In [30]: y = numpy.datetime64('NaT')

In [31]: x != x
Out[31]: True

In [32]: y != y
/home/i850228/.local/lib/python3.6/site-packages/IPython/__main__.py:1: FutureWarning: In the future, NAT != NAT will be True rather than False.
  # encoding: utf-8
Out[32]: False

numpy.isnat(用于检查NumPy NaT的功能)也因熊猫NaT而失败:

numpy.isnat, the function to check for NumPy NaT, also fails with a Pandas NaT:

In [33]: numpy.isnat(pandas.NaT)
---------------------------------------------------------------------------
TypeError                                 Traceback (most recent call last)
<ipython-input-33-39a66bbf6513> in <module>()
----> 1 numpy.isnat(pandas.NaT)

TypeError: ufunc 'isnat' is only defined for datetime and timedelta.

pandas.isnull适用于Pandas和NumPy NaT，所以这可能是解决方法:

pandas.isnull works for both Pandas and NumPy NaTs, so it's probably the way to go:

In [34]: pandas.isnull(pandas.NaT)
Out[34]: True

In [35]: pandas.isnull(numpy.datetime64('NaT'))
Out[35]: True

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