groupby filtering is missing some groups

[phobson]

I brought this up on the mailing list. @cpcloud modified my example into very concise sample showing expected and resulting output:

In [20]: df  = pd.DataFrame([
    ['best', 'a', 'x'],
    ['worst', 'b', 'y'],
    ['best', 'c', 'x'],
    ['best','d', 'y'],
    ['worst','d', 'y'],
    ['worst','d', 'y'],
    ['best','d', 'z'],
], columns=['a', 'b', 'c'])

In [21]: pd.concat(v[v.a == 'best'] for _, v in df.groupby('c'))
Out[21]:
      a  b  c
0  best  a  x
2  best  c  x
3  best  d  y # <--- missing from the next statement
6  best  d  z

In [22]: df.groupby('c').filter(lambda g: g.a == 'best')
Out[22]:
      a  b  c
0  best  a  x
2  best  c  x
6  best  d  z

[dsm054]

I'm not sure this is what filter is meant for. I think it acts on the groups, not on elements of the groups.

If I'm right, then the bug is that this isn't tripping raise TypeError("the filter must return a boolean result"), not that the result differs from the concat result. If not, then the docs are confusing. (Maybe they are in any case.)

[cpcloud]

I'll repost how I thought it might've worked

In [20]: df
Out[20]:
       a  b  c
0   best  a  x
1  worst  b  y
2   best  c  x
3   best  d  y
4  worst  d  y
5  worst  d  y
6   best  d  z

In [21]: pd.concat(v[v.a == 'best'] for _, v in df.groupby('c'))
Out[21]:
      a  b  c
0  best  a  x
2  best  c  x
3  best  d  y
6  best  d  z

In [22]: df.groupby('c').filter(lambda g: g.a == 'best')
Out[22]:
      a  b  c
0  best  a  x
2  best  c  x
6  best  d  z

[cpcloud]

@dsm054 by groups do you mean ['x', 'y', 'z'] or the (name, frame) pairs?

[jreback]

I agree with @dsm054 here. This should return the SAME df, (as 'best' is in EACH group). The bug is that the complete groups are not being returned (because the filter function is tricking the detector).

[dsm054]

I mean the subframes that filter is passed. I think you're given a group and you're supposed to return a yea or nay bool based on the group, not an iterable of bools about your opinions on each row in the group.

The example in the function itself is (lambda x: x['A'].sum() + x['B'].sum() > 0), which is a reduction-to-bool operation.

[cpcloud]

well then this is very weird :)

[cpcloud]

pr coming

[dsm054]

Just to be clear, I'm not sure I agree with @jreback that the right thing to do is to return the same df because "best" is in each group. That's equivalent to saying that we're going to use the convention bool(array) == any(array), which is only going to work here.

[hayd]

Posted on ML before this. I think raise is the correct course of action here. Filter is for removing/including entire groups. Picking all or any (or first!) of the boolean Series would be strange IMO...

Function to apply to each subframe. Should return True or False.

[phobson]

Very insightful comments here. Much appreciated. I'm 👍 on raising an error, which I believe version 0.12 did.

[jreback]

I concur

[cpcloud]

Okay, this mainly had to do with the slow_path/fast_path business, where the slow path calls apply on the group (usually a frame) and the fast path just calls the function directly, which of course in the case of this:

In [8]: df
Out[8]:
   a       b
0  1  0.9712
1  1  0.7578
2  1  0.6300
3  2  1.7625
4  2 -0.0895

In [9]: df[df.a == 1].apply(lambda x: len(x) > 1)
Out[9]:
a    True
b    True
dtype: bool

In [10]: len(df[df.a == 1]) > 1
Out[10]: True

gives two different unrelated results and there were some hacks around trying to figure out which one was "correct".

i gutted all that code and now just call the fast path, we'll see what happens on travis but this doesn't break anything locally.

[cpcloud]

I went with raise for the same reason that np.array([False, True]).item() raises