我有一个熊猫df，代表一家商店的营业时间，如下所示：
Dates Open
0 2016-01-01 0
1 2016-01-02 0
2 2016-01-03 0
3 2016-01-04 1
4 2016-01-05 1
5 2016-01-06 1
6 2016-01-07 1
7 2016-01-08 1
8 2016-01-09 0
9 2016-01-10 0
10 2016-01-11 1
11 2016-01-12 1
12 2016-01-13 1
13 2016-01-14 1
14 2016-01-15 1
15 2016-01-16 0
16 2016-01-17 0
17 2016-01-18 1
18 2016-01-19 1
19 2016-01-20 1
20 2016-01-21 1
21 2016-01-22 1
22 2016-01-23 0
23 2016-01-24 0
24 2016-01-25 1
25 2016-01-26 1
26 2016-01-27 1
27 2016-01-28 1
28 2016-01-29 1
可以通过以下方式重新创建：
Dates =['2016-01-01',
'2016-01-02',
'2016-01-03',
'2016-01-04',
'2016-01-05',
'2016-01-06',
'2016-01-07',
'2016-01-08',
'2016-01-09',
'2016-01-10',
'2016-01-11',
'2016-01-12',
'2016-01-13',
'2016-01-14',
'2016-01-15',
'2016-01-16',
'2016-01-17',
'2016-01-18',
'2016-01-19',
'2016-01-20',
'2016-01-21',
'2016-01-22',
'2016-01-23',
'2016-01-24',
'2016-01-25',
'2016-01-26',
'2016-01-27',
'2016-01-28',
'2016-01-29']
Open = [0,0,0,1,1,1,1,1,0,0,1,1,1,1,1,0,0,1,1,1,1,1,0,0,1,1,1,1,1]
df = DataFrame({'Dates':Dates, 'Open':Open})
“开店”列表示商店开店的日子。我想在最左边的列中为每个日期的下一个营业日创建一个新列。我无法使用预定义的工作日功能，但必须使用“打开”列来确定商店是否营业。理想的结果将是：
Dates Open Desired
0 2016-01-01 0 2016-01-04
1 2016-01-02 0 2016-01-04
2 2016-01-03 0 2016-01-04
3 2016-01-04 1 2016-01-05
4 2016-01-05 1 2016-01-06
5 2016-01-06 1 2016-01-07
6 2016-01-07 1 2016-01-08
7 2016-01-08 1 2016-01-11
8 2016-01-09 0 2016-01-11
9 2016-01-10 0 2016-01-11
10 2016-01-11 1 2016-01-12
11 2016-01-12 1 2016-01-13
12 2016-01-13 1 2016-01-14
13 2016-01-14 1 2016-01-15
14 2016-01-15 1 2016-01-18
15 2016-01-16 0 2016-01-18
16 2016-01-17 0 2016-01-18
17 2016-01-18 1 2016-01-19
18 2016-01-19 1 2016-01-20
19 2016-01-20 1 2016-01-21
20 2016-01-21 1 2016-01-22
21 2016-01-22 1 2016-01-25
22 2016-01-23 0 2016-01-25
23 2016-01-24 0 2016-01-25
24 2016-01-25 1 2016-01-26
25 2016-01-26 1 2016-01-27
26 2016-01-27 1 2016-01-28
27 2016-01-28 1 2016-01-29
28 2016-01-29 1
最佳答案
IIUC，然后您可以使用Business Day offset将字符串转换为datetime dtype后添加to_datetime：
In [141]:
df['Dates'] = pd.to_datetime(df['Dates'])
df.info()
<class 'pandas.core.frame.DataFrame'>
Int64Index: 29 entries, 0 to 28
Data columns (total 2 columns):
Dates 29 non-null datetime64[ns]
Open 29 non-null int64
dtypes: datetime64[ns](1), int64(1)
memory usage: 696.0 bytes
In [146]:
from pandas.tseries.offsets import *
df['Desired'] = df['Dates'] + BDay()
df
Out[146]:
Dates Open Desired
0 2016-01-01 0 2016-01-04
1 2016-01-02 0 2016-01-04
2 2016-01-03 0 2016-01-04
3 2016-01-04 1 2016-01-05
4 2016-01-05 1 2016-01-06
5 2016-01-06 1 2016-01-07
6 2016-01-07 1 2016-01-08
7 2016-01-08 1 2016-01-11
8 2016-01-09 0 2016-01-11
9 2016-01-10 0 2016-01-11
10 2016-01-11 1 2016-01-12
11 2016-01-12 1 2016-01-13
12 2016-01-13 1 2016-01-14
13 2016-01-14 1 2016-01-15
14 2016-01-15 1 2016-01-18
15 2016-01-16 0 2016-01-18
16 2016-01-17 0 2016-01-18
17 2016-01-18 1 2016-01-19
18 2016-01-19 1 2016-01-20
19 2016-01-20 1 2016-01-21
20 2016-01-21 1 2016-01-22
21 2016-01-22 1 2016-01-25
22 2016-01-23 0 2016-01-25
23 2016-01-24 0 2016-01-25
24 2016-01-25 1 2016-01-26
25 2016-01-26 1 2016-01-27
26 2016-01-27 1 2016-01-28
27 2016-01-28 1 2016-01-29
28 2016-01-29 1 2016-02-01
关于python - 在 Pandas 专栏中查找下一个工作日，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/34808639/