API: update nth to use the _set_selection_from_grouper makes first==nth(0) and last==nth(-1)

[jreback]

closes #6732

nth sets the index appropriately and the same as first/last.

this becomes less like head/tail in that as_index determines when you have an index

here's the revised behavior:

In [1]: df = DataFrame([[1, np.nan], [1, 4], [5, 6]], columns=['A', 'B'])

In [3]: df
Out[3]: 
   A   B
0  1 NaN
1  1   4
2  5   6

[3 rows x 2 columns]

In [2]: g = df.groupby('A')

In [9]: gni = df.groupby('A',as_index=False)

In [5]: g.first()
Out[5]: 
   B
A   
1  4
5  6

[2 rows x 1 columns]

# this was a regression from 0.13.1 (in that before this PR, this was returning like ``as_index=False``)
In [6]: g.nth(0)
Out[6]: 
    B
A    
1 NaN
5   6

[2 rows x 1 columns]

In [7]: g.nth(0,dropna='all')
Out[7]: 
   B
A   
1  4
5  6

[2 rows x 1 columns]

In [10]: gni.nth(0)
Out[10]: 
   A   B
0  1 NaN
2  5   6

[2 rows x 2 columns]

In [11]: gni.nth(0,dropna='all')
Out[11]: 
   A  B
0  1  4
1  5  6

[2 rows x 2 columns]

[jreback]

@hayd @jorisvandenbossche pls review for saneness

[jorisvandenbossche]

The question is a little bit if we see nth as a filtering function, or a reducing function? It looks like #6569 changed it be filtering, while this does revert it again to reducing?

[jreback]

I get it, the reason I 'changed' it was two-fold:

• nth is a reducer, much like first and last, in fact first == nth(0)
• you can now pass as_index=False to get a filter, so this preserves the behavior while making the default a more natural, show me the nth element.

so to answer your question, this is really a reducer (as it can return at most 1 per group!)

head/tail can return multiple per group (so they are filters)

[jreback]

@hayd ping!

[jreback]

@hayd any comments?

[hayd]

Looks good, thanks for putting this together, apologies for being AWOL of late!

[jreback]

glad 2 have u back!