我从这样的输入数据开始

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

印刷出来时是这样的:

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

分组非常简单:

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

打印产生一个GroupBy对象:

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

但我最终想要的是另一个DataFrame对象,它包含GroupBy对象中的所有行。换句话说,我想得到以下结果:

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

我不太清楚如何在pandas文档中实现这一点。欢迎任何提示。


当前回答

关键是使用reset_index()方法。

Use:

import pandas

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

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

现在你在g1中有了新的数据框架:

其他回答

g1这里是一个数据帧。不过,它有一个层次索引:

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)

也许你想要这样的东西?

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

或者像这样:

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

我想稍微改变一下Wes给出的答案,因为版本0.16.2要求as_index=False。如果你不设置它,你会得到一个空的数据框架。

来源:

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
#

编辑:

在0.17.1及以后版本中,你可以在count中使用子集,在reset_index中使用参数名的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

count和size的区别在于,size计算NaN值,而count不计算NaN值。

也许我误解了这个问题,但如果你想将groupby转换回数据帧,你可以使用.to_frame()。当我这样做的时候,我想重置索引,所以我也包括了这一部分。

与问题无关的示例代码

df = df['TIME'].groupby(df['Name']).min()
df = df.to_frame()
df = df.reset_index(level=['Name',"TIME"])
 grouped=df.groupby(['Team','Year'])['W'].count().reset_index()

 team_wins_df=pd.DataFrame(grouped)
 team_wins_df=team_wins_df.rename({'W':'Wins'},axis=1)
 team_wins_df['Wins']=team_wins_df['Wins'].astype(np.int32)
 team_wins_df.reset_index()
 print(team_wins_df)

我发现这对我很有用。

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()