Pandas:当其中一列中有列表时,如何按列中的值进行分组

Question

我正在尝试按"value_1"列中的值进行分组.但我的最后一栏是由列表组成的.当我尝试使用"value_1"列进行分组时,由列表组成的列将消失.

数据帧:

 value_1:        value_2:           value_3:               list: 
 american     california, nyc      walmart, kmart      [supermarket, connivence] 
 canadian         toronto            dunkinDonuts      [coffee]
 american          texas                               [state]
 canadian                             walmart          [supermarket] 
   ...              ...                 ...              ....

我的预期输出是:

value_1:        value_2:              value_3:             list: 
american   california, nyc, texas   walmart, kmart      [supermarket, connivence, state] 
canadian         toronto         dunkinDonuts, walmart  [coffee, supermarket]

谢谢!

Answer 1

您可以使用以下函数聚合包含字符串的列：groupby value_1

def str_cat(x):
    return x.str.cat(sep=', ')

并用于GroupBy.sum附加列中的列表list：

df.replace('',None).groupby('value_1').agg({'list':'sum', 'value_2': str_cat,
                                            'value_3': str_cat})

                        list                       value_2  \
value_1                                                              
american  [supermarket, connivence, state]  california, nyc, texas   
canadian             [coffee, sipermarket]          toronto, texas   

                    value_3  
value_1                                 
american  walmart, kmart, dunkinDonuts  
canadian         dunkinDonuts, walmart  

Answer 2

使用no list和value_1for list使用lambda函数创建动态字典,并使用带有flatenning的列表理解:

f1 = lambda x: ', '.join(x.dropna())
#alternative for join only strings
#f1 = lambda x: ', '.join([y for y in x if isinstance(y, str)])
f2 = lambda x: [z for y in x for z in y]
d = dict.fromkeys(df.columns.difference(['value_1','list']), f1)
d['list'] = f2 

df = df.groupby('value_1', as_index=False).agg(d)
print (df)
     value_1                 value_2                value_3  \
0   american  california, nyc, texas         walmart, kmart   
1   canadian                 toronto  dunkinDonuts, walmart   

                               list  
0  [supermarket, connivence, state]  
1             [coffee, supermarket]  

说明:

f1并且f2是lambda函数.

首先删除缺失值(如果存在)和join带分隔符的字符串:

f1 = lambda x: ', '.join(x.dropna())

首先只获取字符串值(省略缺失值,因为NaNs)和join带分隔符的字符串:

f1 = lambda x: ', '.join([y for y in x if isinstance(y, str)])

首先获取所有字符串值,过滤空字符串和join带分隔符的字符串:

f1 = lambda x: ', '.join([y for y in x if y != '']) 

函数f2用于展平列表,因为聚合后得到嵌套列表[['a','b'], ['c']]

f2 = lambda x: [z for y in x for z in y]