python - 如何根据多列的值将多行合并为一行？

Question

我需要将多行合并为一行，原始数据帧如下所示：

IndividualID    DayID    TripID    JourSequence   TripPurpose
200100000001    1        1         1              3
200100000001    1        2         2              31
200100000001    1        3         3              23
200100000001    1        4         4              5
200100000009    1        55        1              3
200100000009    1        56        2              12
200100000009    1        57        3              4
200100000009    1        58        4              6
200100000009    1        59        5              19
200100000009    1        60        6              2

我试图建立某种“旅行链”，所以基本上一个人一天的所有旅行顺序和旅行目的都应该在同一行......

理想情况下，我试图将表格转换为这样的：

IndividualID    DayID     Seq1   TripPurp1     Seq2   TripPur2     Seq3   TripPurp3     Seq4   TripPur4
200100000001    1         1      3             2      31           3       23           4      5
200100000009    1         1      3             2      12           3        4           4      6

如果这是不可能的，那么以下模式也可以：

IndividualID    DayID      TripPurposes
200100000001    1          3, 31, 23, 5
200100000009    1          3, 12, 4, 6

有没有可能的解决方案？我在考虑 for 循环/ while 语句，但也许这不是一个好主意。提前致谢！

Answer 1

你可以试试：

df_out = df.set_index(['IndividualID','DayID',df.groupby(['IndividualID','DayID']).cumcount()+1]).unstack().sort_index(level=1, axis=1)
df_out.columns = df_out.columns.map('{0[0]}_{0[1]}'.format)
df_out.reset_index()

输出：

   IndividualID  DayID  JourSequence_1  TripID_1  TripPurpose_1  \
0  200100000001      1             1.0       1.0            3.0   
1  200100000009      1             1.0      55.0            3.0   

   JourSequence_2  TripID_2  TripPurpose_2  JourSequence_3  TripID_3  \
0             2.0       2.0           31.0             3.0       3.0   
1             2.0      56.0           12.0             3.0      57.0   

   TripPurpose_3  JourSequence_4  TripID_4  TripPurpose_4  JourSequence_5  \
0           23.0             4.0       4.0            5.0             NaN   
1            4.0             4.0      58.0            6.0             5.0   

   TripID_5  TripPurpose_5  JourSequence_6  TripID_6  TripPurpose_6  
0       NaN            NaN             NaN       NaN            NaN  
1      59.0           19.0             6.0      60.0            2.0  

Answer 2

要获得第二个输出，您只需要分组并应用列表：

df.groupby(['IndividualID', 'DayID'])['TripPurpose'].apply(list)

                      TripPurpose
IndividualID  DayID 
200100000001    1   [3, 31, 23, 5]
200100000009    1   [3, 12, 4, 6, 19, 2]

要获得第一个输出，您可以执行以下操作（可能不是最佳方法）：

df2 = pd.DataFrame(df.groupby(['IndividualID', 'DayID'])['TripPurpose'].apply(list))
trip = df2['TripPurpose'].apply(pd.Series).rename(columns = lambda x: 'TripPurpose'+ str(x+1))
df3 = pd.DataFrame(df.groupby(['IndividualID', 'DayID'])['JourSequence'].apply(list))
seq = df3['JourSequence'].apply(pd.Series).rename(columns = lambda x: 'seq'+ str(x+1))
pd.merge(trip,seq,on=['IndividualID','DayID'])

输出未排序