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。

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()

简单地说，这应该完成任务:

import pandas as pd

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

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

在这里，grouped_df.size()提取唯一的groupby计数，reset_index()方法重置您希望它是的列的名称。 最后，调用pandas Dataframe()函数来创建一个Dataframe对象。

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"])

我已经与Qty明智的数据聚合并存储到dataframe

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

这些解决方案只部分适用于我，因为我正在进行多个聚合。下面是我分组的一个输出示例，我想转换为一个数据框架:

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]

如果字典不是你的东西，计算可以内联应用在for循环中:

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

下面的解决方案可能更简单:

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

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:

 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)

这将以与普通groupby()方法相同的顺序返回序数级/索引。它基本上与@NehalJWani在他的评论中发布的答案相同，但存储在一个变量中，并调用了reset_index()方法。

fare_class = df.groupby(['Satisfaction Rating','Fare Class']).size().to_frame(name = 'Count')
fare_class.reset_index()

这个版本不仅返回相同的百分比数据，这是有用的统计，而且还包括一个lambda函数。

fare_class_percent = df.groupby(['Satisfaction Rating', 'Fare Class']).size().to_frame(name = 'Percentage')
fare_class_percent.transform(lambda x: 100 * x/x.sum()).reset_index()

      Satisfaction Rating      Fare Class  Percentage
0            Dissatisfied        Business   14.624269
1            Dissatisfied         Economy   36.469048
2               Satisfied        Business    5.460425
3               Satisfied         Economy   33.235294

例子:

建议在group_by方法中设置group_keys=False，避免将组键添加到索引中。

例子:

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.groupby(["Name"], group_keys=False)