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

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()

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。

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:

下面的解决方案可能更简单:

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