Slicing a DataGrameGroupBy object

Question

Is there a way to slice a DataFrameGroupBy object?

For example, if I have:

df = pd.DataFrame({'A': [2, 1, 1, 3, 3], 'B': ['x', 'y', 'z', 'r', 'p']})

   A  B
0  2  x
1  1  y
2  1  z
3  3  r
4  3  p

dfg = df.groupby('A')

Now, the returned GroupBy object is indexed by values from A, and I would like to select a subset of it, e.g. to perform aggregation. It could be something like

dfg.loc[1:2].agg(...)

or, for a specific column,

dfg['B'].loc[1:2].agg(...)

EDIT. To make it more clear: by slicing the GroupBy object I mean accessing only a subset of groups. In the above example, the GroupBy object will contain 3 groups, for A = 1, A = 2, and A = 3. For some reasons, I may only be interested in groups for A = 1 and A = 2.

Answer 1

It seesm you need custom function with iloc - but if use agg is necessary return aggregate value:

df = df.groupby('A')['B'].agg(lambda x: ','.join(x.iloc[0:3]))
print (df)
A
1    y,z
2      x
3    r,p
Name: B, dtype: object

---

df = df.groupby('A')['B'].agg(lambda x: ','.join(x.iloc[1:3]))
print (df)
A
1    z
2     
3    p
Name: B, dtype: object

For multiple columns:

df = pd.DataFrame({'A': [2, 1, 1, 3, 3], 
                   'B': ['x', 'y', 'z', 'r', 'p'], 
                   'C': ['g', 'y', 'y', 'u', 'k']})
print (df)
   A  B  C
0  2  x  g
1  1  y  y
2  1  z  y
3  3  r  u
4  3  p  k

df = df.groupby('A').agg(lambda x: ','.join(x.iloc[1:3]))
print (df)
   B  C
A      
1  z  y
2      
3  p  k

Answer 2

You can slice with apply like this:

if you want to slice [1:3] from each group

n [53]: df
Out[53]: 
   A  B
0  2  x
1  1  y
2  1  z
3  3  r
4  3  p

In [54]: dfg = df.groupby('A')

In [56]: dfg.apply(lambda x: x.loc[1:3])
Out[56]: 
     A  B
A        
1 1  1  y
  2  1  z
3 3  3  r

---

if you want to slice only a column (B for example)

In [55]: dfg.apply(lambda x: x['B'].loc[1:3])
Out[55]: 
A   
1  1    y
   2    z
3  3    r
Name: B, dtype: object

---

Then, to aggregate, you just chain the call like this:

dfg.apply(lambda x: x['B'].loc[1:3]).agg(...)