The second half of the currently accepted answer is outdated and has two deprecations. First and most important, you can no longer pass a dictionary of dictionaries to the agg
groupby method. Second, never use .ix
.
If you desire to work with two separate columns at the same time I would suggest using the apply
method which implicitly passes a DataFrame to the applied function. Let's use a similar dataframe as the one from above
df = pd.DataFrame(np.random.rand(4,4), columns=list('abcd'))
df['group'] = [0, 0, 1, 1]
df
a b c d group
0 0.418500 0.030955 0.874869 0.145641 0
1 0.446069 0.901153 0.095052 0.487040 0
2 0.843026 0.936169 0.926090 0.041722 1
3 0.635846 0.439175 0.828787 0.714123 1
A dictionary mapped from column names to aggregation functions is still a perfectly good way to perform an aggregation.
df.groupby('group').agg({'a':['sum', 'max'],
'b':'mean',
'c':'sum',
'd': lambda x: x.max() - x.min()})
a b c d
sum max mean sum <lambda>
group
0 0.864569 0.446069 0.466054 0.969921 0.341399
1 1.478872 0.843026 0.687672 1.754877 0.672401
If you don't like that ugly lambda column name, you can use a normal function and supply a custom name to the special __name__
attribute like this:
def max_min(x):
return x.max() - x.min()
max_min.__name__ = 'Max minus Min'
df.groupby('group').agg({'a':['sum', 'max'],
'b':'mean',
'c':'sum',
'd': max_min})
a b c d
sum max mean sum Max minus Min
group
0 0.864569 0.446069 0.466054 0.969921 0.341399
1 1.478872 0.843026 0.687672 1.754877 0.672401
Using apply
and returning a Series
Now, if you had multiple columns that needed to interact together then you cannot use agg
, which implicitly passes a Series to the aggregating function. When using apply
the entire group as a DataFrame gets passed into the function.
I recommend making a single custom function that returns a Series of all the aggregations. Use the Series index as labels for the new columns:
def f(x):
d = {}
d['a_sum'] = x['a'].sum()
d['a_max'] = x['a'].max()
d['b_mean'] = x['b'].mean()
d['c_d_prodsum'] = (x['c'] * x['d']).sum()
return pd.Series(d, index=['a_sum', 'a_max', 'b_mean', 'c_d_prodsum'])
df.groupby('group').apply(f)
a_sum a_max b_mean c_d_prodsum
group
0 0.864569 0.446069 0.466054 0.173711
1 1.478872 0.843026 0.687672 0.630494
If you are in love with MultiIndexes, you can still return a Series with one like this:
def f_mi(x):
d = []
d.append(x['a'].sum())
d.append(x['a'].max())
d.append(x['b'].mean())
d.append((x['c'] * x['d']).sum())
return pd.Series(d, index=[['a', 'a', 'b', 'c_d'],
['sum', 'max', 'mean', 'prodsum']])
df.groupby('group').apply(f_mi)
a b c_d
sum max mean prodsum
group
0 0.864569 0.446069 0.466054 0.173711
1 1.478872 0.843026 0.687672 0.630494
You can pass a dictionary to agg
with column names as keys and the functions you want as values.
import pandas as pd
import numpy as np
# Create some randomised data
N = 20
date_range = pd.date_range('01/01/2015', periods=N, freq='W')
df = pd.DataFrame({'ages':np.arange(N), 'payments':np.arange(N)*10}, index=date_range)
print(df.head())
# ages payments
# 2015-01-04 0 0
# 2015-01-11 1 10
# 2015-01-18 2 20
# 2015-01-25 3 30
# 2015-02-01 4 40
# Apply np.mean to the ages column and np.sum to the payments.
agg_funcs = {'ages':np.mean, 'payments':np.sum}
# Groupby each individual month and then apply the funcs in agg_funcs
grouped = df.groupby(df.index.to_period('M')).agg(agg_funcs)
print(grouped)
# ages payments
# 2015-01 1.5 60
# 2015-02 5.5 220
# 2015-03 10.0 500
# 2015-04 14.5 580
# 2015-05 18.0 540
Best Answer
Since the two other inputs are constants, you can simply use a lambda expression: