Question

假设我有两个约会。如何获得两个日期之间的所有中间日期和小时（从00到23）？非常感谢你！

from datetime import datetime,timedelta, date

date_min = (datetime.today()-timedelta(1)).strftime('%Y-%m-%d')+' 00'
date_max =  datetime.today().strftime('%Y-%m-%d')+' 00'

date_min, date_max

('2018-09-16 00', '2018-09-17 00')

how to obtain?

'2018-09-16 01', '2018-09-16 02', '2018-09-16 03', '2018-09-16 04', .... , '2018-09-16 21', '2018-09-16 22', '2018-09-16 23', '2018-09-17 00'

Answer 1

使用大熊猫，它是一种衬套：

pandas timestamp

如果您想从datetime转换回[timestamp.to_pydatetime() for timestamp in solution] [datetime.datetime(2018, 9, 16, 0, 0), datetime.datetime(2018, 9, 16, 1, 0), datetime.datetime(2018, 9, 16, 2, 0), datetime.datetime(2018, 9, 16, 3, 0), datetime.datetime(2018, 9, 16, 4, 0), datetime.datetime(2018, 9, 16, 5, 0), datetime.datetime(2018, 9, 16, 6, 0), datetime.datetime(2018, 9, 16, 7, 0), datetime.datetime(2018, 9, 16, 8, 0), datetime.datetime(2018, 9, 16, 9, 0), datetime.datetime(2018, 9, 16, 10, 0), datetime.datetime(2018, 9, 16, 11, 0), datetime.datetime(2018, 9, 16, 12, 0), datetime.datetime(2018, 9, 16, 13, 0), datetime.datetime(2018, 9, 16, 14, 0), datetime.datetime(2018, 9, 16, 15, 0), datetime.datetime(2018, 9, 16, 16, 0), datetime.datetime(2018, 9, 16, 17, 0), datetime.datetime(2018, 9, 16, 18, 0), datetime.datetime(2018, 9, 16, 19, 0), datetime.datetime(2018, 9, 16, 20, 0), datetime.datetime(2018, 9, 16, 21, 0), datetime.datetime(2018, 9, 16, 22, 0), datetime.datetime(2018, 9, 16, 23, 0), datetime.datetime(2018, 9, 17, 0, 0)]，请执行以下操作：

[timestamp.to_pydatetime().strftime('%Y-%m-%d %H') for timestamp in solution]

['2018-09-16 00', '2018-09-16 01', '2018-09-16 02', '2018-09-16 03', '2018-09-16 04', '2018-09-16 05', '2018-09-16 06', '2018-09-16 07', '2018-09-16 08', '2018-09-16 09', '2018-09-16 10', '2018-09-16 11', '2018-09-16 12', '2018-09-16 13', '2018-09-16 14', '2018-09-16 15', '2018-09-16 16', '2018-09-16 17', '2018-09-16 18', '2018-09-16 19', '2018-09-16 20', '2018-09-16 21', '2018-09-16 22', '2018-09-16 23', '2018-09-17 00']

要转换为您提到的字符串格式：

<form name="chatform">

Answer 2

这将起作用：

date = datetime.today()-timedelta(days=1)
while date < datetime.today():
    date = date + timedelta(hours=1)
    print date.strftime('%Y-%m-%d %H')

并提供以下输出：

2018-09-17 00
2018-09-17 01
2018-09-17 02
2018-09-17 03
2018-09-17 04
2018-09-17 05
2018-09-17 06
2018-09-17 07
2018-09-17 08

您还可以指定从几小时开始：

date = datetime.today().replace(hour=0, minute=0, second=0)-timedelta(days=1)

Answer 3

纯Python：

import itertools
from datetime import datetime,timedelta

date_max = datetime.today()
date_min = date_max - timedelta(hours=2)
n_hours = (date_max - date_min).total_seconds() / 3600
[date_min + timedelta(hours=x) for x in itertools.takewhile(lambda h: h <= n_hours, itertools.count())]

返回

[datetime.datetime(2018, 9, 17, 14, 49, 42, 193664),
 datetime.datetime(2018, 9, 17, 15, 49, 42, 193664),
 datetime.datetime(2018, 9, 17, 16, 49, 42, 193664)]

如何在Python中添加日期和时间

3 个答案: