Question

我正在创建一个Dataframe

import pandas as pd

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

df1.groupby( ["City"] )['Name'].transform(lambda x:     
','.join(x)).drop_duplicates()      
I want the output as      

Name    City                 
Alice,Bob,Mallory,Bob     Seattle           
Mallory,Mallory    Portland        

but i am getting only           
Name         
Alice,Bob,Mallory,Bob                
Mallory,Mallory        

This is an example with small number of columns but in my actual problem i 
have too many columns so i cannot use           
df1['Name']= df1.groupby( ['City'] )['Name'].transform(lambda x:         
','.join(x))              
df1.groupby( ['City','Name'], as_index=False )              
df1.drop_duplicates()

因为对于每一列我必须写相同的代码
没有为每列编写转换，有没有办法做到这一点个别。

Answer 1

<强> 1。列聚合

我认为您需要apply与,.join，然后对于变更单使用双[[]]：

df = df1.groupby(["City"])['Name'].apply(','.join).reset_index()
df = df[['Name','City']]
print (df)
                    Name      City
0        Mallory,Mallory  Portland
1  Alice,Bob,Mallory,Bob   Seattle

因为transform使用汇总值创建新列：

df1['new'] = df1.groupby("City")['Name'].transform(','.join)
print (df1)
       City     Name                    new
0   Seattle    Alice  Alice,Bob,Mallory,Bob
1   Seattle      Bob  Alice,Bob,Mallory,Bob
2  Portland  Mallory        Mallory,Mallory
3   Seattle  Mallory  Alice,Bob,Mallory,Bob
4   Seattle      Bob  Alice,Bob,Mallory,Bob
5  Portland  Mallory        Mallory,Mallory

<强> 2。列和更多聚合

如果更多列需要agg并在[]中指定列，或者没有指定加入所有字符串列：

df1 = pd.DataFrame( {     
"Name" : ["Alice", "Bob", "Mallory", "Mallory", "Bob" , "Mallory"] ,  
"Name2":   ["Alice1", "Bob1", "Mallory1", "Mallory1", "Bob1" , "Mallory1"],      
"City" : ["Seattle", "Seattle", "Portland", "Seattle", "Seattle",     
"Portland"]   } )   
print (df1)
       City     Name     Name2
0   Seattle    Alice    Alice1
1   Seattle      Bob      Bob1
2  Portland  Mallory  Mallory1
3   Seattle  Mallory  Mallory1
4   Seattle      Bob      Bob1
5  Portland  Mallory  Mallory1

df = df = df1.groupby('City')['Name', 'Name2'].agg(','.join).reset_index()
print (df)
       City                   Name                      Name2
0  Portland        Mallory,Mallory          Mallory1,Mallory1
1   Seattle  Alice,Bob,Mallory,Bob  Alice1,Bob1,Mallory1,Bob1

如果需要聚合所有列，请使用

。

df = df1.groupby('City').agg(','.join).reset_index()
print (df)
       City                   Name                      Name2
0  Portland        Mallory,Mallory          Mallory1,Mallory1
1   Seattle  Alice,Bob,Mallory,Bob  Alice1,Bob1,Mallory1,Bob1

df1 = pd.DataFrame( {     
"Name" : ["Alice", "Bob", "Mallory", "Mallory", "Bob" , "Mallory"] ,  
"Name2":   ["Alice1", "Bob1", "Mallory1", "Mallory1", "Bob1" , "Mallory1"],      
"City" : ["Seattle", "Seattle", "Portland", "Seattle", "Seattle", "Portland"],
'Numbers':[1,5,4,3,2,1]} )   
print (df1)
       City     Name     Name2  Numbers
0   Seattle    Alice    Alice1        1
1   Seattle      Bob      Bob1        5
2  Portland  Mallory  Mallory1        4
3   Seattle  Mallory  Mallory1        3
4   Seattle      Bob      Bob1        2
5  Portland  Mallory  Mallory1        1


df = df1.groupby('City').agg({'Name': ','.join, 
                              'Name2': ','.join, 
                              'Numbers': 'max'}).reset_index()
print (df)
       City                   Name                      Name2  Numbers
0  Portland        Mallory,Mallory          Mallory1,Mallory1        4
1   Seattle  Alice,Bob,Mallory,Bob  Alice1,Bob1,Mallory1,Bob1        5

Answer 2

你可以做到

In [42]: df1.groupby('City')['Name'].agg(','.join).reset_index(name='Name')
Out[42]:
       City                   Name
0  Portland        Mallory,Mallory
1   Seattle  Alice,Bob,Mallory,Bob

或者，

In [49]: df1.groupby('City', as_index=False).agg({'Name': ','.join})
Out[49]:
       City                   Name
0  Portland        Mallory,Mallory
1   Seattle  Alice,Bob,Mallory,Bob

对于多个聚合

df1.groupby('City', as_index=False).agg(
      {'Name': ','.join, 'Name2': ','.join, 'Number': 'max'})

在Pandas中使用group by进行转换

2 个答案: