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"])

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()

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"])

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。

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

下面的解决方案可能更简单:

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