I have a pandas data structure such as this:
>>> df
Benny Daniel Doris Eric Jack Zoe
Age 75 30 95 25 28 23
Salary 2000 9000 100000 10000 12000 20000
I would like to find the mean age and salary for several different groups, where each is a subset of the columns, and they may overlap, such as this dictionary for example:
{'Parrot lovers': ['Doris', 'Benny'], 'Tea Drinkers': ['Doris', 'Zoe'],\
'Maintainance': ['Benny', 'Jack'], 'Coffee Drinkers': ['Benny', 'Eric'],\
'Senior Management': ['Doris', 'Zoe', 'Jack']}
How can I design a groupby function that will do this?
Here is how I set up the problem...
import StringIO
import pandas as pd
df = """index Benny Daniel Doris Eric Jack Zoe
Age 75 30 95 25 28 23
Salary 2000 9000 100000 10000 12000 20000"""
df = pd.read_csv(StringIO.StringIO(df),sep="\s+").set_index('index')
d = {'Parrot lovers': ['Doris', 'Benny'], 'Tea Drinkers': ['Doris', 'Zoe'],\
'Maintainance': ['Benny', 'Jack'], 'Coffee Drinkers': ['Benny', 'Eric'],\
'Senior Management': ['Doris', 'Zoe', 'Jack']}
For the solution Just Use .loc
and iterate through the dictionary...
averages = {k:df.loc[:,v].mean(axis=1) for k,v in d.iteritems()}
print pd.DataFrame(averages).T #gives the nice printout...
index Age Salary
Coffee Drinkers 50.000000 6000
Maintainance 51.500000 7000
Parrot lovers 85.000000 51000
Senior Management 48.666667 44000
Tea Drinkers 59.000000 60000
There are probably a handful of ways to do this, here's one path.
Transpose your data, and add a True/False column for category:
In [20]: group_map = {'Parrot lovers': ['Doris', 'Benny'],
'Tea Drinkers': ['Doris', 'Zoe'],
'Maintainance': ['Benny', 'Jack'],
'Coffee Drinkers': ['Benny', 'Eric'],
'Senior Management': ['Doris', 'Zoe', 'Jack']}
In [22]: df = df.T
In [23]: for k in group_map:
...: df[k] = df.index.isin(group_map[k])
Now, you can groupby any category to get means:
In [24]: df.groupby('Parrot lovers')['Salary'].mean()
Out[24]:
Parrot lovers
False 12750
True 51000
Name: Salary, dtype: int64
Or, iterate over the columns to get the mean for each category.
In [24]: means = {}
...: for k in group_map:
...: means[k] = df.groupby(k)['Salary'].mean()[True]
...: means
...:
Out[24]:
{'Coffee Drinkers': 6000,
'Maintainance': 7000,
'Parrot lovers': 51000,
'Senior Management': 44000,
'Tea Drinkers': 60000}
The technical post webpages of this site follow the CC BY-SA 4.0 protocol. If you need to reprint, please indicate the site URL or the original address.Any question please contact:yoyou2525@163.com.