pandas.Series.__eq__ is broken for series with different index

[lesteve]

Something seems to be wrong with s1 == s2 when s1 and s2 don't have the same index. Here is a snippet example:

import operator
import pandas
s1 = pandas.Series([1,2], ['a','b'])
s2 = pandas.Series([2,3], ['b','c'])
s1 == s2
s2 == s1

with the output:

InIn [5]: s1 == s2
Out[5]: 
a    False
b    False

In [6]: s2 == s1
Out[6]: 
b    False
c    False

On the other hand using combine works fine:

In [7]: s1.combine(s2, operator.eq)
Out[7]: 
a    0
b    1
c    0

In [8]: s2.combine(s1, operator.eq)
Out[8]: 
a    0
b    1
c    0

I guess you can first align s1 and s2 and then compare them, but is there a good reason why this couldn't work out of the box?

There doesn't seem to be any tests for pandas.Series. eq for two series with a different index in pandas/pandas/tests/test_series.py. I have a patch lying around to add such a test and I could commit it if that's useful.

[wesm]

This is actually a feature / deliberate choice and not a bug-- it's related to #652. Back in January I changed the comparison methods to do auto-alignment, but found that it led to a large amount of bugs / breakage for users and, in particular, many NumPy functions (which regularly do things like arr[1:] == arr[:-1]; example: np.unique) stopped working.

This gets back to the issue that Series isn't quite ndarray-like enough and should probably not be a subclass of ndarray.

So, I haven't got a good answer for you except for that; auto-alignment would be ideal but I don't think I can do it unless I make Series not a subclass of ndarray. I think this is probably a good idea but not likely to happen until 0.9 or 0.10 (several months down the road).

[lesteve]

Interesting, thanks for the answer. Is s[1:] == s[:-1] the main use case though, where you need to have this not completely intuitive == operator?

Out of interest, is there a way to figure out whether s1 and s2 are a view on the same underlying series and in this case have s1 == s2 do the current comparison. When s1 and s2 don't have anything to do with each other you would do the equivalent of aligning + comparison.

Not sure whether you would want to do that even if it was possible, e.g. s1 == s2.copy() would potentially return something different thans1 == s2.

[wesm]

Interesting, that would be a hack around the np.unique problem. I'll take a look into it sooner rather than later to see

[jreback]

closing as not a bug

[jorisvandenbossche]

@jreback I was answering this SO question: http://stackoverflow.com/questions/22983523/comparing-pandas-series-for-equality-when-they-are-in-different-orders/22983621#22983621. And I was wondering:

I understand that s1 == s2 is not flexible, just like df1 == df2 gives the error message ValueError: Can only compare identically-labeled DataFrame objects.
But for dataframes, you can overcome this with the flexible DataFrame.eq method. There is also a Series.eq method, but this is a not flexible method (not aligning). Is there a reason that Series.eq is not flexible?

In [154]: x = pd.Series(index=["A", "B", "C"], data=[1,2,3])
In [155]: y = pd.Series(index=["C", "B", "A"], data=[3,2,1])
In [156]: x == y
Out[156]: 
A    False
B     True
C    False
dtype: bool

In [157]: x.eq(y)
Out[157]: 
A    False
B     True
C    False
dtype: bool

In [158]: x.to_frame() == y.to_frame()
Traceback (most recent call last):
...
ValueError: Can only compare identically-labeled DataFrame objects

In [159]: x.to_frame().eq(y.to_frame())
Out[159]: 
      0
A  True
B  True
C  True

[jreback]

@jorisvandenbossche interesting, let's reopen and i'll take a look