在给定日期后找到第一个星期一的日期

给定一个特定的日期（例如2011-07-02），我如何找到该日期之后的下一个星期一（或该日的任何工作日）的日期？

import datetime
def next_weekday(d, weekday):
    days_ahead = weekday - d.weekday()
    if days_ahead <= 0: # Target day already happened this week
        days_ahead += 7
    return d + datetime.timedelta(days_ahead)

d = datetime.date(2011, 7, 2)
next_monday = next_weekday(d, 0) # 0 = Monday, 1=Tuesday, 2=Wednesday...
print(next_monday)

这是上面略显笨拙的答案的简洁概括。

# Returns the date of the next given weekday after
# the given date. For example, the date of next Monday.
# NB: if it IS the day we're looking for, this returns 0.
# consider then doing onDay(foo, day + 1).
onDay = lambda date, day: date + datetime.timedelta(days=(day-date.weekday()+7)%7)

尝试

>>> dt = datetime(2011, 7, 2)
>>> dt + timedelta(days=(7 - dt.weekday()))
datetime.datetime(2011, 7, 4, 0, 0)

使用，下一个星期一是星期一之后的7天，星期二之后是6天，依此类推，并且还使用Python的datetime类型将星期一报告为0，...，星期日报告为6。

这是ring中的计算示例mod 7。

import datetime


def next_day(given_date, weekday):
    day_shift = (weekday - given_date.weekday()) % 7
    return given_date + datetime.timedelta(days=day_shift)

now = datetime.date(2018, 4, 15) # sunday
names = ['monday', 'tuesday', 'wednesday', 'thursday', 'friday',    
         'saturday', 'sunday']
for weekday in range(7):
    print(names[weekday], next_day(now, weekday))

将打印：

monday 2018-04-16
tuesday 2018-04-17
wednesday 2018-04-18
thursday 2018-04-19
friday 2018-04-20
saturday 2018-04-21
sunday 2018-04-15

如您所见，它可以正确地在下周一，周二，周三，周四，周五和周六给您。而且它也理解2018-04-15为是星期日，并且返回了当前的星期日，而不是下一个星期日。

我相信7年后您会发现这个答案非常有帮助;-)

您可以开始添加一天的日期对象，并在星期一停止。

>>> d = datetime.date(2011, 7, 2)
>>> while d.weekday() != 0: #0 for monday
...     d += datetime.timedelta(days=1)
... 
>>> d
datetime.date(2011, 7, 4)

另一个简单而优雅的解决方案是使用pandas偏移量。
我发现在玩日期时它非常有用且功能强大。
-如果您想要第一个星期日，只需将频率更改为freq ='W-SUN'。
-如果您要在下几个星期日进行更改，请更改offsets.Day（天）。
-使用大熊猫偏移量可以忽略假期，仅在工作日等时使用。
您还可以使用apply方法轻松地将此方法应用于整个DataFrame。

# Getting the closest monday from a given date
closest_monday = pd.date_range(start=date, end=date + offsets.Day(6), freq='W-MON')[0]

# Adding a 'ClosestMonday' column with the closest monday for each row in a pandas df using apply
# Require you to have a 'Date' column in your df
def get_closest_monday(row):
    return pd.date_range(start=row.Date, end=row.Date + offsets.Day(6), freq='W-MON')[0]

df['ClosestMonday'] = df.apply(lambda row: get_closest_monday(row), axis=1)

另一种选择使用规则

from dateutil.rrule import rrule, WEEKLY, MO
from datetime import date

next_monday = rrule(freq=WEEKLY, dtstart=date.today(), byweekday=MO, count=1)[0]

规则文档：https：//dateutil.readthedocs.io/en/stable/rrule.html

import datetime

d = datetime.date(2011, 7, 2)
while d.weekday() != 0:
    d += datetime.timedelta(1)

weekday = 0 ## Monday
dt = datetime.datetime.now().replace(hour=0, minute=0, second=0) ## or any specific date
days_remaining = (weekday - dt.weekday() - 1) % 7 + 1
next_dt = dt + datetime.timedelta(days_remaining)

给定日期后的下一个星期一：

import datetime

def get_next_monday(year, month, day):
    date0 = datetime.date(year, month, day)
    next_monday = date0 + datetime.timedelta(7 - date0.weekday() or 7)
    return next_monday

print get_next_monday(2011, 7, 2)
print get_next_monday(2015, 8, 31)
print get_next_monday(2015, 9, 1)

2011-07-04
2015-09-07
2015-09-07

通过列表理解？

from datetime import *
[datetime.today()+timedelta(days=x) for x in range(0,7) if (datetime.today()+timedelta(days=x)).weekday() % 7 == 0]

(0 at the end is for next monday, returns current date when run on monday)