Pandas 数据透视表格式化列名

Question

我pandas.pivot_table在 Pandas 数据帧上使用了该函数，我的输出看起来与此类似：

                    Winners                 Runnerup            
         year       2016    2015    2014    2016    2015    2014
Country  Sport                              
india    badminton                              
india    wrestling  

我真正需要的是像下面这样的东西

Country Sport   Winners_2016    Winners_2015    Winners_2014    Runnerup_2016   Runnerup_2015   Runnerup_2014
india   badminton   1   1   1   1   1   1
india   wrestling   1   0   1   0   1   0

我有很多列和年份，所以我无法手动编辑它们，所以有人可以告诉我如何做到这一点吗？

Answer 1

您还可以使用列表理解：

df.columns = ['_'.join(col) for col in df.columns]
print (df)
                   Winners_2016  Winners_2015  Winners_2014  Runnerup_2016  \
Country Sport                                                                
india   badminton             1             1             1              1   
        wrestling             1             1             1              1   

                   Runnerup_2015  Runnerup_2014  
Country Sport                                    
india   badminton              1              1  
        wrestling              1              1  

另一种使用 convertcolumns to_series然后调用的解决方案join：

df.columns = df.columns.to_series().str.join('_')
print (df)
                   Winners_2016  Winners_2015  Winners_2014  Runnerup_2016  \
Country Sport                                                                
india   badminton             1             1             1              1   
        wrestling             1             1             1              1   

                   Runnerup_2015  Runnerup_2014  
Country Sport                                    
india   badminton              1              1  
        wrestling              1              1  

我对时间非常感兴趣：

In [45]: %timeit ['_'.join(col) for col in df.columns]
The slowest run took 7.82 times longer than the fastest. This could mean that an intermediate result is being cached.
100000 loops, best of 3: 4.05 µs per loop

In [44]: %timeit ['{}_{}'.format(x,y) for x,y in zip(df.columns.get_level_values(0),df.columns.get_level_values(1))]
The slowest run took 4.56 times longer than the fastest. This could mean that an intermediate result is being cached.
10000 loops, best of 3: 131 µs per loop

In [46]: %timeit df.columns.to_series().str.join('_')
The slowest run took 4.31 times longer than the fastest. This could mean that an intermediate result is being cached.
1000 loops, best of 3: 452 µs per loop