Comparing pandas Series for equality when they contain nan?

Question

My application needs to compare Series instances that sometimes contain nans. That causes ordinary comparison using == to fail, since nan != nan:

import numpy as np
from pandas import Series
s1 = Series([1,np.nan])
s2 = Series([1,np.nan])

>>> (Series([1, nan]) == Series([1, nan])).all()
False

What's the proper way to compare such Series?

Answer 1

How about this. First check the NaNs are in the same place (using isnull):

In [11]: s1.isnull()
Out[11]: 
0    False
1     True
dtype: bool

In [12]: s1.isnull() == s2.isnull()
Out[12]: 
0    True
1    True
dtype: bool

Then check the values which aren't NaN are equal (using notnull):

In [13]: s1[s1.notnull()]
Out[13]: 
0    1
dtype: float64

In [14]: s1[s1.notnull()] == s2[s2.notnull()]
Out[14]: 
0    True
dtype: bool

In order to be equal we need both to be True:

In [15]: (s1.isnull() == s2.isnull()).all() and (s1[s1.notnull()] == s2[s2.notnull()]).all()
Out[15]: True

You could also check name etc. if this wasn't sufficient.

If you want to raise if they are different, use assert_series_equal from pandas.util.testing:

In [21]: from pandas.util.testing import assert_series_equal

In [22]: assert_series_equal(s1, s2)

Answer 2

Currently one should just use series1.equals(series2) see docs. This also checks if nans are in the same positions.

Answer 3

In [16]: s1 = Series([1,np.nan])

In [17]: s2 = Series([1,np.nan])

In [18]: (s1.dropna()==s2.dropna()).all()
Out[18]: True