Converting a Pandas GroupBy object to DataFrame

问题:

I\'m starting with input data like this

df1 = pandas.DataFrame( { 
    \"Name\" : [\"Alice\", \"Bob\", \"Mallory\", \"Mallory\", \"Bob\" , \"Mallory\"] , 
    \"City\" : [\"Seattle\", \"Seattle\", \"Portland\", \"Seattle\", \"Seattle\", \"Portland\"] } )

Which when printed appears as this:

   City     Name
0   Seattle    Alice
1   Seattle      Bob
2  Portland  Mallory
3   Seattle  Mallory
4   Seattle      Bob
5  Portland  Mallory

Grouping is simple enough:

g1 = df1.groupby( [ \"Name\", \"City\"] ).count()

and printing yields a GroupBy object:

                  City  Name
Name    City
Alice   Seattle      1     1
Bob     Seattle      2     2
Mallory Portland     2     2
        Seattle      1     1

But what I want eventually is another DataFrame object that contains all the rows in the GroupBy object. In other words I want to get the following result:

                  City  Name
Name    City
Alice   Seattle      1     1
Bob     Seattle      2     2
Mallory Portland     2     2
Mallory Seattle      1     1

I can\'t quite see how to accomplish this in the pandas documentation. Any hints would be welcome.

回答1:

g1 here is a DataFrame. It has a hierarchical index, though:

In [19]: type(g1)
Out[19]: pandas.core.frame.DataFrame

In [20]: g1.index
Out[20]: 
MultiIndex([(\'Alice\', \'Seattle\'), (\'Bob\', \'Seattle\'), (\'Mallory\', \'Portland\'),
       (\'Mallory\', \'Seattle\')], dtype=object)

Perhaps you want something like this?

In [21]: g1.add_suffix(\'_Count\').reset_index()
Out[21]: 
      Name      City  City_Count  Name_Count
0    Alice   Seattle           1           1
1      Bob   Seattle           2           2
2  Mallory  Portland           2           2
3  Mallory   Seattle           1           1

Or something like:

In [36]: DataFrame({\'count\' : df1.groupby( [ \"Name\", \"City\"] ).size()}).reset_index()
Out[36]: 
      Name      City  count
0    Alice   Seattle      1
1      Bob   Seattle      2
2  Mallory  Portland      2
3  Mallory   Seattle      1

回答2:

I want to slightly change the answer given by Wes, because version 0.16.2 requires as_index=False. If you don\'t set it, you get an empty dataframe.

Source:

Aggregation functions will not return the groups that you are aggregating over if they are named columns, when as_index=True, the default. The grouped columns will be the indices of the returned object.

Passing as_index=False will return the groups that you are aggregating over, if they are named columns.

Aggregating functions are ones that reduce the dimension of the returned objects, for example: mean, sum, size, count, std, var, sem, describe, first, last, nth, min, max. This is what happens when you do for example DataFrame.sum() and get back a Series.

nth can act as a reducer or a filter, see here.

import pandas as pd

df1 = pd.DataFrame({\"Name\":[\"Alice\", \"Bob\", \"Mallory\", \"Mallory\", \"Bob\" , \"Mallory\"],
                    \"City\":[\"Seattle\",\"Seattle\",\"Portland\",\"Seattle\",\"Seattle\",\"Portland\"]})
print df1
#
#       City     Name
#0   Seattle    Alice
#1   Seattle      Bob
#2  Portland  Mallory
#3   Seattle  Mallory
#4   Seattle      Bob
#5  Portland  Mallory
#
g1 = df1.groupby([\"Name\", \"City\"], as_index=False).count()
print g1
#
#                  City  Name
#Name    City
#Alice   Seattle      1     1
#Bob     Seattle      2     2
#Mallory Portland     2     2
#        Seattle      1     1
#

EDIT:

In version 0.17.1 and later you can use subset in count and reset_index with parameter name in size:

print df1.groupby([\"Name\", \"City\"], as_index=False ).count()
#IndexError: list index out of range

print df1.groupby([\"Name\", \"City\"]).count()
#Empty DataFrame
#Columns: []
#Index: [(Alice, Seattle), (Bob, Seattle), (Mallory, Portland), (Mallory, Seattle)]

print df1.groupby([\"Name\", \"City\"])[[\'Name\',\'City\']].count()
#                  Name  City
#Name    City                
#Alice   Seattle      1     1
#Bob     Seattle      2     2
#Mallory Portland     2     2
#        Seattle      1     1

print df1.groupby([\"Name\", \"City\"]).size().reset_index(name=\'count\')
#      Name      City  count
#0    Alice   Seattle      1
#1      Bob   Seattle      2
#2  Mallory  Portland      2
#3  Mallory   Seattle      1

The difference between count and size is that size counts NaN values while count does not.

回答3:

Simply, this should do the task:

import pandas as pd

grouped_df = df1.groupby( [ \"Name\", \"City\"] )

pd.DataFrame(grouped_df.size().reset_index(name = \"Group_Count\"))

Here, grouped_df.size() pulls up the unique groupby count, and reset_index() method resets the name of the column you want it to be. Finally, the pandas Dataframe() function is called upon to create DataFrame object.

回答4:

I found this worked for me.

import numpy as np
import pandas as pd

df1 = pd.DataFrame({ 
    \"Name\" : [\"Alice\", \"Bob\", \"Mallory\", \"Mallory\", \"Bob\" , \"Mallory\"] , 
    \"City\" : [\"Seattle\", \"Seattle\", \"Portland\", \"Seattle\", \"Seattle\", \"Portland\"]})

df1[\'City_count\'] = 1
df1[\'Name_count\'] = 1

df1.groupby([\'Name\', \'City\'], as_index=False).count()

回答5:

Maybe I misunderstand the question but if you want to convert the groupby back to a dataframe you can use .to_frame(). I wanted to reset the index when I did this so I included that part as well.

example code unrelated to question

df = df[\'TIME\'].groupby(df[\'Name\']).min()
df = df.to_frame()
df = df.reset_index(level=[\'Name\',\"TIME\"])

回答6:

I have aggregated with Qty wise data and store to dataframe

almo_grp_data = pd.DataFrame({\'Qty_cnt\' :
almo_slt_models_data.groupby( [\'orderDate\',\'Item\',\'State Abv\']
          )[\'Qty\'].sum()}).reset_index()

回答7:

Below solution may be simpler:

df1.reset_index().groupby( [ \"Name\", \"City\"],as_index=False ).count()

回答8:

These solutions only partially worked for me because I was doing multiple aggregations. Here is a sample output of my grouped by that I wanted to convert to a dataframe:

Because I wanted more than the count provided by reset_index(), I wrote a manual method for converting the image above into a dataframe. I understand this is not the most pythonic/pandas way of doing this as it is quite verbose and explicit, but it was all I needed. Basically, use the reset_index() method explained above to start a \"scaffolding\" dataframe, then loop through the group pairings in the grouped dataframe, retrieve the indices, perform your calculations against the ungrouped dataframe, and set the value in your new aggregated dataframe.

df_grouped = df[[\'Salary Basis\', \'Job Title\', \'Hourly Rate\', \'Male Count\', \'Female Count\']]
df_grouped = df_grouped.groupby([\'Salary Basis\', \'Job Title\'], as_index=False)

# Grouped gives us the indices we want for each grouping
# We cannot convert a groupedby object back to a dataframe, so we need to do it manually
# Create a new dataframe to work against
df_aggregated = df_grouped.size().to_frame(\'Total Count\').reset_index()
df_aggregated[\'Male Count\'] = 0
df_aggregated[\'Female Count\'] = 0
df_aggregated[\'Job Rate\'] = 0

def manualAggregations(indices_array):
    temp_df = df.iloc[indices_array]
    return {
        \'Male Count\': temp_df[\'Male Count\'].sum(),
        \'Female Count\': temp_df[\'Female Count\'].sum(),
        \'Job Rate\': temp_df[\'Hourly Rate\'].max()
    }

for name, group in df_grouped:
    ix = df_grouped.indices[name]
    calcDict = manualAggregations(ix)

    for key in calcDict:
        #Salary Basis, Job Title
        columns = list(name)
        df_aggregated.loc[(df_aggregated[\'Salary Basis\'] == columns[0]) & 
                          (df_aggregated[\'Job Title\'] == columns[1]), key] = calcDict[key]

If a dictionary isn\'t your thing, the calculations could be applied inline in the for loop:

    df_aggregated[\'Male Count\'].loc[(df_aggregated[\'Salary Basis\'] == columns[0]) & 
                                (df_aggregated[\'Job Title\'] == columns[1])] = df[\'Male Count\'].iloc[ix].sum()