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()

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

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:

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)

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"])

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()