Best way to apply diffrent aggregations on grouped DataFrame

Question

Suppose, i group my DataFrame A by key

 A = pd.DataFrame({    'key':['II','I','I','III','II'],
                       'Z':['a', 'b', 'c', 'd', 'e'],
                       'd':[1,2,0,2,0],
                       'e':[0,2,0,3,0],
                       'f':[0,3,0,4,0],})

And i want diffrent aggregation for each column. E.g.:

• sum() for f
• max() for e
• mean() for d
• concatenation for Z (ae, bc, d)

As i'm not able to extract columns separatly from DataFrameGroupedBy, i have to split A in 4 diffrent DataFrames with columns [key, c],[key, d],[key, e],[key, f] before the groupby, apply diffrent agregations to each, then merge by key.

This seems a little rediculous and needs a lot of code. Are there more elegant ways?

Answer 1

You can use agg by dict of columns and aggregate functions:

df = A.groupby('key').agg({'f':'sum','e':'max','d':'mean', 'Z': ''.join})
print (df)
       d   Z  f  e
key               
I    1.0  bc  3  2
II   0.5  ae  0  0
III  2.0   d  4  3

Answer 2

You can also use agg and the passed dictionary to name the columns.

f = dict(
    f={'Sum of f': 'sum'},
    e={'Max of e': 'max'},
    d={'Mean of d': 'mean'},
    Z={'Concat of Z': 'sum'},
)

A.groupby('key').agg(f)

           f        e         d           Z
    Sum of f Max of e Mean of d Concat of Z
key                                        
I          3        2       1.0          bc
II         0        0       0.5          ae
III        4        3       2.0           d

However, if you don't want the pd.MultiIndex it's probably easier to use rename

f = dict(
    f='sum',
    e='max',
    d='mean',
    Z='sum',
)
m = dict(
    f='Sum of f',
    e='Max of e',
    d='Mean of d',
    Z='Concat of Z'
)

A.groupby('key').agg(f).rename(columns=m)

     Sum of f  Max of e  Mean of d Concat of Z
key                                           
I           3         2        1.0          bc
II          0         0        0.5          ae
III         4         3        2.0           d