Question

我有以下熊猫DataFrame df：

SIGN        TYPE    TIME                         ADDITIONAL
ABC5245     10      2017-01-01 01:52:25.000      2017-01-01 01:39:04.000
ABC5245     20      2017-01-01 01:53:22.000      2017-01-01 02:39:04.000
DEF1111     20      2017-01-01 01:57:00.000      2017-01-01 03:39:04.000
DEF1111     10      2017-01-01 01:55:15.000      2017-01-01 01:39:04.000
AAA2222     10      2017-01-01 01:57:00.000      2017-01-01 01:39:04.000

我需要按SIGN对数据进行分组，并根据TYPE创建四个新列：TIME_10，TIME_20，ADDITIONAL_10和ADDITIONAL_20

这是预期的结果：

SIGN        TIME_10                  TIME_20                  ADDITIONAL_10             ADDITIONAL_20
ABC5245     2017-01-01 01:52:25.000  2017-01-01 01:53:22.000  2017-01-01 01:39:04.000   2017-01-01 02:39:04.000         
DEF1111     2017-01-01 01:55:15.000  2017-01-01 01:57:00.000  2017-01-01 01:39:04.000   2017-01-01 03:39:04.000
AAA2222     2017-01-01 01:57:00.000  NaN                      2017-01-01 01:39:04.000   NaN

Answer 1

使用重塑和展平列标题多索引

df_out = df.set_index(['SIGN','TYPE']).unstack('TYPE')
df_out.columns = [f'{i}_{j}' for i, j in df_out.columns]
print(df_out)

输出：

                         TIME_10                  TIME_20  \
SIGN                                                        
AAA2222  2017-01-01 01:57:00.000                      NaN   
ABC5245  2017-01-01 01:52:25.000  2017-01-01 01:53:22.000   
DEF1111  2017-01-01 01:55:15.000  2017-01-01 01:57:00.000   

                   ADDITIONAL_10            ADDITIONAL_20  
SIGN                                                       
AAA2222  2017-01-01 01:39:04.000                      NaN  
ABC5245  2017-01-01 01:39:04.000  2017-01-01 02:39:04.000  
DEF1111  2017-01-01 01:39:04.000  2017-01-01 03:39:04.000

Answer 2

您可以使用透视图获得结果。如果您可以将列设为MultiIndex，则不需要第二行。

感谢@ScottBoston提供有关列格式的提示。

df = df.pivot('SIGN', 'TYPE', ['TIME', 'ADDITIONAL'])                          
df.columns = df.columns.map('{0[0]}_{0[1]}'.format)

编辑

在上下文中：

import pandas as pd
data = [
    ['ABC5245', 10, '2017-01-01 01:52:25.000', '2017-01-01 01:39:04.000'],
    ['ABC5245', 20, '2017-01-01 01:53:22.000', '2017-01-01 02:39:04.000'],
    ['DEF1111', 20, '2017-01-01 01:57:00.000', '2017-01-01 03:39:04.000'],
    ['DEF1111', 10, '2017-01-01 01:55:15.000', '2017-01-01 01:39:04.000'],
    ['AAA2222', 10, '2017-01-01 01:57:00.000', '2017-01-01 01:39:04.000'],
]
columns = ['SIGN', 'TYPE', 'TIME', 'ADDITIONAL']

df = pd.DataFrame(data=data, columns=columns)
print(df)

df = df.pivot('SIGN', 'TYPE', ['TIME', 'ADDITIONAL'])
df.columns = df.columns.map('{0[0]}_{0[1]}'.format)
print(df)

输出：

      SIGN  TYPE                     TIME               ADDITIONAL
0  ABC5245    10  2017-01-01 01:52:25.000  2017-01-01 01:39:04.000
1  ABC5245    20  2017-01-01 01:53:22.000  2017-01-01 02:39:04.000
2  DEF1111    20  2017-01-01 01:57:00.000  2017-01-01 03:39:04.000
3  DEF1111    10  2017-01-01 01:55:15.000  2017-01-01 01:39:04.000
4  AAA2222    10  2017-01-01 01:57:00.000  2017-01-01 01:39:04.000
                         TIME_10                  TIME_20            ADDITIONAL_10            ADDITIONAL_20
SIGN
AAA2222  2017-01-01 01:57:00.000                      NaN  2017-01-01 01:39:04.000                      NaN
ABC5245  2017-01-01 01:52:25.000  2017-01-01 01:53:22.000  2017-01-01 01:39:04.000  2017-01-01 02:39:04.000
DEF1111  2017-01-01 01:55:15.000  2017-01-01 01:57:00.000  2017-01-01 01:39:04.000  2017-01-01 03:39:04.000

如何针对我的情况将行转换为列？

2 个答案: