您如何使用我的数据将行转换为列？我当前的数据集看起来像下面显示的“原始df”，我希望它看起来像“新df2”。要明确一点，约会1与SDAP1和RDAP1匹配，因此约会2对应于SDAP2 RDAP2。
原始df：
Name Appoint1 Appoint2 Appoint1t Appoint2t SDAP1 RDAP1 SDAP2 RDAP2
Sam 23.09.2017 24.09.2017 11:00:00 11:00:00 3 -9 6 8
Sarah 24.09.2017 27.09.2017 12:00:00 12:00:00 2 Nan 7 8
Steve 23.10.2017 31.10.2017 11:00:00 12:00:00 5 9 7 9
Mark 23.09.2017 11:00:00 0 3
James 23.09.2017 26.09.2017 11:00:00 4 7 1 4
新df：
Name Appointments Appointmenttime SDAP RDAP
Sam 23.09.2017 11:00:00 3 -9
Sam 24.09.2017 11:00:00 6 8
Sarah 24.09.2017 12:00:00 2 NaN
Sarah 27.09.2017 12:00:00 7 8
Steve 23.10.2017 11:00:00 5 9
Steve 31.10.2017 12:00:00 7 9
Mark 23.09.2017 11:00:00 0 3
James 23.09.2017 4 7
James 26.09.2017 11:00:00 1 4
答案 0 :(得分：2)
是否有必要使用wide_to_long？使用concat似乎要容易得多。
df1 = df[["Name","Appoint1","Appoint1t"]]
df2 = df[["Name","Appoint2","Appoint2t"]].rename(columns={"Appoint2": "Appoint1", "Appoint2t": "Appoint1t"})
print (pd.concat([df1,df2]).dropna().sort_index())
#
Name Appoint1 Appoint1t
0 Sam 23.09.2017 11:00:00
0 Sam 24.09.2017 11:00:00
1 Sarah 24.09.2017 12:00:00
1 Sarah 27.09.2017 12:00:00
2 Steve 23.10.2017 11:00:00
2 Steve 31.10.2017 12:00:00
3 Mark 23.09.2017 11:00:00
4 James 23.09.2017 11:00:00
4 James 26.09.2017 11:00:00
首先重命名各列，以使用wide_to_long：
df.columns = ['Name', 'Appoint_1', 'Appoint_2', 'Time_1', 'Time_2']
print (pd.wide_to_long(df,stubnames=["Appoint","Time"],i="Name",j="count",sep='_')
.dropna().reset_index().drop("count",axis=1))
#
Name Appoint Time
0 Sam 23.09.2017 11:00:00
1 Sarah 24.09.2017 12:00:00
2 Steve 23.10.2017 11:00:00
3 Mark 23.09.2017 11:00:00
4 James 23.09.2017 11:00:00
5 Sam 24.09.2017 11:00:00
6 Sarah 27.09.2017 12:00:00
7 Steve 31.10.2017 12:00:00
8 James 26.09.2017 11:00:00