Naming returned columns in Pandas aggregate function? [duplicate]
Solution 1:
For pandas >= 0.25
The functionality to name returned aggregate columns has been reintroduced in the master branch and is targeted for pandas 0.25. The new syntax is .agg(new_col_name=('col_name', 'agg_func'). Detailed example from the PR linked above:
In [2]: df = pd.DataFrame({'kind': ['cat', 'dog', 'cat', 'dog'],
...: 'height': [9.1, 6.0, 9.5, 34.0],
...: 'weight': [7.9, 7.5, 9.9, 198.0]})
...:
In [3]: df
Out[3]:
kind height weight
0 cat 9.1 7.9
1 dog 6.0 7.5
2 cat 9.5 9.9
3 dog 34.0 198.0
In [4]: df.groupby('kind').agg(min_height=('height', 'min'),
max_weight=('weight', 'max'))
Out[4]:
min_height max_weight
kind
cat 9.1 9.9
dog 6.0 198.0
It will also be possible to use multiple lambda expressions with this syntax and the two-step rename syntax I suggested earlier (below) as per this PR. Again, copying from the example in the PR:
In [2]: df = pd.DataFrame({"A": ['a', 'a'], 'B': [1, 2], 'C': [3, 4]})
In [3]: df.groupby("A").agg({'B': [lambda x: 0, lambda x: 1]})
Out[3]:
B
<lambda> <lambda 1>
A
a 0 1
and then .rename(), or in one go:
In [4]: df.groupby("A").agg(b=('B', lambda x: 0), c=('B', lambda x: 1))
Out[4]:
b c
A
a 0 0
For pandas < 0.25
The currently accepted answer by unutbu describes are great way of doing this in pandas versions <= 0.20. However, as of pandas 0.20, using this method raises a warning indicating that the syntax will not be available in future versions of pandas.
Series:
FutureWarning: using a dict on a Series for aggregation is deprecated and will be removed in a future version
DataFrames:
FutureWarning: using a dict with renaming is deprecated and will be removed in a future version
According to the pandas 0.20 changelog, the recommended way of renaming columns while aggregating is as follows.
# Create a sample data frame
df = pd.DataFrame({'A': [1, 1, 1, 2, 2],
'B': range(5),
'C': range(5)})
# ==== SINGLE COLUMN (SERIES) ====
# Syntax soon to be deprecated
df.groupby('A').B.agg({'foo': 'count'})
# Recommended replacement syntax
df.groupby('A').B.agg(['count']).rename(columns={'count': 'foo'})
# ==== MULTI COLUMN ====
# Syntax soon to be deprecated
df.groupby('A').agg({'B': {'foo': 'sum'}, 'C': {'bar': 'min'}})
# Recommended replacement syntax
df.groupby('A').agg({'B': 'sum', 'C': 'min'}).rename(columns={'B': 'foo', 'C': 'bar'})
# As the recommended syntax is more verbose, parentheses can
# be used to introduce line breaks and increase readability
(df.groupby('A')
.agg({'B': 'sum', 'C': 'min'})
.rename(columns={'B': 'foo', 'C': 'bar'})
)
Please see the 0.20 changelog for additional details.
Update 2017-01-03 in response to @JunkMechanic's comment.
With the old style dictionary syntax, it was possible to pass multiple lambda functions to .agg, since these would be renamed with the key in the passed dictionary:
>>> df.groupby('A').agg({'B': {'min': lambda x: x.min(), 'max': lambda x: x.max()}})
B
max min
A
1 2 0
2 4 3
Multiple functions can also be passed to a single column as a list:
>>> df.groupby('A').agg({'B': [np.min, np.max]})
B
amin amax
A
1 0 2
2 3 4
However, this does not work with lambda functions, since they are anonymous and all return <lambda>, which causes a name collision:
>>> df.groupby('A').agg({'B': [lambda x: x.min(), lambda x: x.max]})
SpecificationError: Function names must be unique, found multiple named <lambda>
To avoid the SpecificationError, named functions can be defined a priori instead of using lambda. Suitable function names also avoid calling .rename on the data frame afterwards. These functions can be passed with the same list syntax as above:
>>> def my_min(x):
>>> return x.min()
>>> def my_max(x):
>>> return x.max()
>>> df.groupby('A').agg({'B': [my_min, my_max]})
B
my_min my_max
A
1 0 2
2 3 4
Solution 2:
This will drop the outermost level from the hierarchical column index:
df = data.groupby(...).agg(...)
df.columns = df.columns.droplevel(0)
If you'd like to keep the outermost level, you can use the ravel() function on the multi-level column to form new labels:
df.columns = ["_".join(x) for x in df.columns.ravel()]
For example:
import pandas as pd
import pandas.rpy.common as com
import numpy as np
data = com.load_data('Loblolly')
print(data.head())
# height age Seed
# 1 4.51 3 301
# 15 10.89 5 301
# 29 28.72 10 301
# 43 41.74 15 301
# 57 52.70 20 301
df = data.groupby('Seed').agg(
{'age':['sum'],
'height':['mean', 'std']})
print(df.head())
# age height
# sum std mean
# Seed
# 301 78 22.638417 33.246667
# 303 78 23.499706 34.106667
# 305 78 23.927090 35.115000
# 307 78 22.222266 31.328333
# 309 78 23.132574 33.781667
df.columns = df.columns.droplevel(0)
print(df.head())
yields
sum std mean
Seed
301 78 22.638417 33.246667
303 78 23.499706 34.106667
305 78 23.927090 35.115000
307 78 22.222266 31.328333
309 78 23.132574 33.781667
Alternatively, to keep the first level of the index:
df = data.groupby('Seed').agg(
{'age':['sum'],
'height':['mean', 'std']})
df.columns = ["_".join(x) for x in df.columns.ravel()]
yields
age_sum height_std height_mean
Seed
301 78 22.638417 33.246667
303 78 23.499706 34.106667
305 78 23.927090 35.115000
307 78 22.222266 31.328333
309 78 23.132574 33.781667
Solution 3:
I agree with the OP that it seems more natural and consistent to name and define the output columns in the same place (e.g. as is done with tidyverse's summarize in R), but a work-around in pandas for now is to create the new columns with desired names via assign before doing the aggregation:
data.assign(
f=data['column1'],
mean=data['column2'],
std=data['column2']
).groupby('Country').agg(dict(f=sum, mean=np.mean, std=np.std)).reset_index()
(Using reset_index turns 'Country', 'f', 'mean', and 'std' all into regular columns with a separate integer index.)
Solution 4:
If you want to have a behavior similar to JMP, creating column titles that keep all info from the multi index you can use:
newidx = []
for (n1,n2) in df.columns.ravel():
newidx.append("%s-%s" % (n1,n2))
df.columns=newidx
It will change your dataframe from:
I V
mean std first
V
4200.0 25.499536 31.557133 4200.0
4300.0 25.605662 31.678046 4300.0
4400.0 26.679005 32.919996 4400.0
4500.0 26.786458 32.811633 4500.0
to
I-mean I-std V-first
V
4200.0 25.499536 31.557133 4200.0
4300.0 25.605662 31.678046 4300.0
4400.0 26.679005 32.919996 4400.0
4500.0 26.786458 32.811633 4500.0