我有一个数据框，主要是日期。这就是我想要做的事情
从旧日期变量（DTDate），我想创建一个新的日期变量，如果旧日期是星期一，新日期将相同，但如果旧日期是除星期一以外的任何日期，则新日期我会告诉我下周一的日期。所以最后新日期的所有项目都只在星期一。
我一直在尝试使用功能并申请。这是我的数据集和代码
Date call DTDate weekday weekdayNo
0 31/12/2014 2014-12-31 Wednesday 3
1 29/10/2014 2014-10-29 Wednesday 3
2 28/10/2014 2014-10-28 Tuesday 2
3 27/3/2015 2015-03-27 Friday 5
4 27/2/2015 2015-02-27 Friday 5
5 27/11/2014 2014-11-27 Thursday 4
6 27/10/2014 2014-10-27 Monday 1
7 26/3/2015 2015-03-26 Thursday 4
8 26/2/2015 2015-02-26 Thursday 4
9 26/12/2014 2014-12-26 Friday 5
10 26/11/2014 2014-11-26 Wednesday 3
11 26/10/2014 2014-10-26 Sunday 0
12 25/3/2015 2015-03-25 Wednesday 3
13 25/12/2014 2014-12-25 Thursday 4
14 24/3/2015 2015-03-24 Tuesday 2
15 24/2/2015 2015-02-24 Tuesday 2
16 24/12/2014 2014-12-24 Wednesday 3
17 24/11/2014 2014-11-24 Monday 1
18 23/3/2015 2015-03-23 Monday 1
代码是
from datetime import date, timedelta
def AddDate(row):
if row['weekdayNo']==0:
return row['DTDate'] + timedelta(days=1)
elif row['weekdayNo'] ==2:
return row['DTDate'] + timedelta(days=6)
elif row['weekdayNo'] ==3:
return row['DTDate'] + timedelta(days=5)
elif row['weekdayNo'] ==4:
return row['DTDate'] + timedelta(days=4)
elif row['weekdayNo'] ==5:
return row['DTDate'] + timedelta(days=3)
elif row['weekdayNo'] ==6:
return row['DTDate'] + timedelta(days=2)
else:
return row['DTDate']
DF['newDate'] = DF.apply(AddDate, axis=1)
我得到以下内容，它完全相同，没有任何改变
Date call DTDate weekday weekdayNo newDate
0 31/12/2014 2014-12-31 Wednesday 3 2014-12-31
1 29/10/2014 2014-10-29 Wednesday 3 2014-10-29
2 28/10/2014 2014-10-28 Tuesday 2 2014-10-28
3 27/3/2015 2015-03-27 Friday 5 2015-03-27
4 27/2/2015 2015-02-27 Friday 5 2015-02-27
5 27/11/2014 2014-11-27 Thursday 4 2014-11-27
6 27/10/2014 2014-10-27 Monday 1 2014-10-27
7 26/3/2015 2015-03-26 Thursday 4 2015-03-26
8 26/2/2015 2015-02-26 Thursday 4 2015-02-26
9 26/12/2014 2014-12-26 Friday 5 2014-12-26
10 26/11/2014 2014-11-26 Wednesday 3 2014-11-26
11 26/10/2014 2014-10-26 Sunday 0 2014-10-26
12 25/3/2015 2015-03-25 Wednesday 3 2015-03-25
13 25/12/2014 2014-12-25 Thursday 4 2014-12-25
14 24/3/2015 2015-03-24 Tuesday 2 2015-03-24
15 24/2/2015 2015-02-24 Tuesday 2 2015-02-24
16 24/12/2014 2014-12-24 Wednesday 3 2014-12-24
17 24/11/2014 2014-11-24 Monday 1 2014-11-24
18 23/3/2015 2015-03-23 Monday 1 2015-03-23
我也认为，这个想法并不好，如果有更好的东西，请有人愿意建议，那可能是什么？提前致谢
答案 0 :(得分：2)
您不需要import datetime或timedelta来执行此操作。
df['DTDate'] = pd.to_datetime(df['DTDate']) # can skip this if column 'DTDate' is already of the right type
x.weekday()以星期一= 0和星期日= 6提取星期几。
df['newDate'] = df.DTDate.apply(lambda x: x + pd.DateOffset(days=7-x.weekday()) if x.weekday() else x)
的产率：
Date_call DTDate weekday weekdayNo newDate
0 2014-12-31 2014-12-31 Wednesday 3 2015-01-05
1 2014-10-29 2014-10-29 Wednesday 3 2014-11-03
2 2014-10-28 2014-10-28 Tuesday 2 2014-11-03
3 2015-03-27 2015-03-27 Friday 5 2015-03-30
4 2015-02-27 2015-02-27 Friday 5 2015-03-02
5 2014-11-27 2014-11-27 Thursday 4 2014-12-01
6 2014-10-27 2014-10-27 Monday 1 2014-10-27
7 2015-03-26 2015-03-26 Thursday 4 2015-03-30
8 2015-02-26 2015-02-26 Thursday 4 2015-03-02
9 2014-12-26 2014-12-26 Friday 5 2014-12-29
10 2014-11-26 2014-11-26 Wednesday 3 2014-12-01
11 2014-10-26 2014-10-26 Sunday 0 2014-10-27
12 2015-03-25 2015-03-25 Wednesday 3 2015-03-30
13 2014-12-25 2014-12-25 Thursday 4 2014-12-29
14 2015-03-24 2015-03-24 Tuesday 2 2015-03-30
15 2015-02-24 2015-02-24 Tuesday 2 2015-03-02
16 2014-12-24 2014-12-24 Wednesday 3 2014-12-29
17 2014-11-24 2014-11-24 Monday 1 2014-11-24
18 2015-03-23 2015-03-23 Monday 1 2015-03-23
答案 1 :(得分：1)
AddDate函数可以更简单，实际上是单个衬里
In [34]: df['newDate'] = df['DTDate'].apply(lambda x: x + timedelta(days=7-x.dayofweek)
if x.dayofweek else x)
此处，lambda函数lambda x: x + timedelta(days=7-x.dayofweek) if x.dayofweek else x如果不是星期一，则添加delta = 7-x.dayofweek天。
要验证新weekday，请创建新列newdayofweek
In [35]: df['newdayofweek'] = df['newDate'].apply(lambda x: x.dayofweek)
In [36]: df
Out[36]:
Date call DTDate weekday weekdayNo newDate newdayofweek
0 0 31/12/2014 2014-12-31 Wednesday 3 2015-01-05 0
1 1 29/10/2014 2014-10-29 Wednesday 3 2014-11-03 0
2 2 28/10/2014 2014-10-28 Tuesday 2 2014-11-03 0
3 3 27/3/2015 2015-03-27 Friday 5 2015-03-30 0
4 4 27/2/2015 2015-02-27 Friday 5 2015-03-02 0
5 5 27/11/2014 2014-11-27 Thursday 4 2014-12-01 0
6 6 27/10/2014 2014-10-27 Monday 1 2014-10-27 0
7 7 26/3/2015 2015-03-26 Thursday 4 2015-03-30 0
8 8 26/2/2015 2015-02-26 Thursday 4 2015-03-02 0
9 9 26/12/2014 2014-12-26 Friday 5 2014-12-29 0
10 10 26/11/2014 2014-11-26 Wednesday 3 2014-12-01 0
11 11 26/10/2014 2014-10-26 Sunday 0 2014-10-27 0
12 12 25/3/2015 2015-03-25 Wednesday 3 2015-03-30 0
13 13 25/12/2014 2014-12-25 Thursday 4 2014-12-29 0
14 14 24/3/2015 2015-03-24 Tuesday 2 2015-03-30 0
15 15 24/2/2015 2015-02-24 Tuesday 2 2015-03-02 0
16 16 24/12/2014 2014-12-24 Wednesday 3 2014-12-29 0
17 17 24/11/2014 2014-11-24 Monday 1 2014-11-24 0
18 18 23/3/2015 2015-03-23 Monday 1 2015-03-23 0
注意：星期一= 0，星期日= 6
的星期几答案 2 :(得分：0)
这是一种更高效的方法。
In [54]: %timeit s.apply(lambda x: x + pd.DateOffset(days=7-x.weekday()) if x.weekday() else x)
1 loops, best of 3: 244 ms per loop
这基本上是在python空间中循环。
.where这里我们要构建一个添加日期的TimedeltaIndex。 if-then与In [55]: %timeit s.where(s.dt.weekday==0,pd.TimedeltaIndex(7-s.dt.weekday,unit='d')+s)
100 loops, best of 3: 9.69 ms per loop
相同，但这是一个矢量化表达式。
.owl