循环并创建考虑闰年的日期列表

Question

我想创建一个日期列表，在某个年份范围内每年选择相同的日期。

start_date = dt.date(2009, 2, 10)
end_date = dt.date(2019, 5, 1)
new_date = []

current_date = start_date
while current_date < end_date:
    if (current_date.year % 4) == 0:
        new_date.append(current_date + dt.timedelta(days=366))
    else: 
        new_date.append(current_date + dt.timedelta(days=365))
    if (current_date.year % 4) == 0:
        current_date += dt.timedelta(days=366)     
    else:
        current_date += dt.timedelta(days=365) 
new_date

这个输出：

[datetime.date(2010, 2, 10),
 datetime.date(2011, 2, 10),
 datetime.date(2012, 2, 10),
 datetime.date(2013, 2, 10),
 datetime.date(2014, 2, 10),
 datetime.date(2015, 2, 10),
 datetime.date(2016, 2, 10),
 datetime.date(2017, 2, 10),
 datetime.date(2018, 2, 10),
 datetime.date(2019, 2, 10),
 datetime.date(2020, 2, 10)]

但当我将日期更改为 2 月 29 日之后，闰年的日期会偏移一天。

start_date = dt.date(2009, 3, 10)
end_date = dt.date(2019, 5, 1)
new_date = []

current_date = start_date
while current_date < end_date:
    if (current_date.year % 4) == 0:
        new_date.append(current_date + dt.timedelta(days=366))
    else: 
        new_date.append(current_date + dt.timedelta(days=365))
    if (current_date.year % 4) == 0:
        current_date += dt.timedelta(days=366)     
    else:
        current_date += dt.timedelta(days=365) 
new_date

[datetime.date(2010, 3, 10),
 datetime.date(2011, 3, 10),
 datetime.date(2012, 3, 9),
 datetime.date(2013, 3, 10),
 datetime.date(2014, 3, 10),
 datetime.date(2015, 3, 10),
 datetime.date(2016, 3, 9),
 datetime.date(2017, 3, 10),
 datetime.date(2018, 3, 10),
 datetime.date(2019, 3, 10),
 datetime.date(2020, 3, 9)]

这是什么原因，我该如何解决？

Answer 1

如果我理解正确，您可以像这样实现您所需要的：

from datetime import date

start_date, end_date = date(2009, 3, 10), date(2019, 5, 1)

new_date = []
for y in range(start_date.year+1, end_date.year+2): # +1 and +2 is arbitrary, adjust as needed
    new_date.append(date(y, start_date.month, start_date.day))

new_date
# [datetime.date(2010, 3, 10),
#  datetime.date(2011, 3, 10),
#  datetime.date(2012, 3, 10),
#  datetime.date(2013, 3, 10),
#  datetime.date(2014, 3, 10),
#  datetime.date(2015, 3, 10),
#  datetime.date(2016, 3, 10),
#  datetime.date(2017, 3, 10),
#  datetime.date(2018, 3, 10),
#  datetime.date(2019, 3, 10),
#  datetime.date(2020, 3, 10)]

或作为列表理解

d0, d1 = date(2009, 3, 10), date(2019, 5, 1)
new_date = [date(y, d0.month, d0.day) for y in range(d0.year+1, d1.year+2)]

Answer 2

如果您选择闰日，这将没有意义，所以我假设您没有。

import datetime

def date_range(start, end):
    ret = []
    while start < end:
        ret.append(start)
        start = start.replace(year=start.year + 1)
    return ret

那么你的例子给出了

>>> start_date = datetime.date(2009, 2, 10)
>>> end_date = datetime.date(2019, 5, 1)
>>> date_range(start_date, end_date)
[datetime.date(2009, 2, 10),
 datetime.date(2010, 2, 10),
 datetime.date(2011, 2, 10),
 datetime.date(2012, 2, 10),
 datetime.date(2013, 2, 10),
 datetime.date(2014, 2, 10),
 datetime.date(2015, 2, 10),
 datetime.date(2016, 2, 10),
 datetime.date(2017, 2, 10),
 datetime.date(2018, 2, 10),
 datetime.date(2019, 2, 10)]

与仅基于年份的生成器不同，此答案也可以正确处理月份。

Answer 3

保留你的代码（虽然它可以更有效地完成），你只需要检查下一个年是否是闰年：

import calendar
import pprint
import datetime as dt

def increaseYear(start_date):
    end_date = dt.date(2019, 5, 1)
    new_date = []

    current_date = start_date
    while current_date < end_date:
        # using calendar.isleap here, there are more conditions
        # than year%4==0: https://stackoverflow.com/questions/11621740/how-to-determine-whether-a-year-is-a-leap-year
        _nextYearisLeap = calendar.isleap(current_date.year + 1)
        if _nextYearisLeap:
            new_date.append(current_date + dt.timedelta(days=366))
        else: 
            new_date.append(current_date + dt.timedelta(days=365))
        current_date = new_date[-1]
    return new_date


for startDate in (dt.date(2009, 2, 10), dt.date(2009, 3, 10)):
    dates = increaseYear(startDate)
    pprint.pprint(dates)

输出：

[datetime.date(2010, 2, 10),
 datetime.date(2011, 2, 10),
 datetime.date(2012, 2, 11),
 datetime.date(2013, 2, 10),
 datetime.date(2014, 2, 10),
 datetime.date(2015, 2, 10),
 datetime.date(2016, 2, 11),
 datetime.date(2017, 2, 10),
 datetime.date(2018, 2, 10),
 datetime.date(2019, 2, 10),
 datetime.date(2020, 2, 11)]
[datetime.date(2010, 3, 10),
 datetime.date(2011, 3, 10),
 datetime.date(2012, 3, 10),
 datetime.date(2013, 3, 10),
 datetime.date(2014, 3, 10),
 datetime.date(2015, 3, 10),
 datetime.date(2016, 3, 10),
 datetime.date(2017, 3, 10),
 datetime.date(2018, 3, 10),
 datetime.date(2019, 3, 10),
 datetime.date(2020, 3, 10)]

Answer 4

您可以使用dateutil 模块来计算闰年：

import datetime as dt
from dateutil.relativedelta import relativedelta

dt.date(2011, 3, 10) + relativedelta(years=1)
Output:
datetime.date(2012, 3, 10)

dt.date(2020, 2, 29) + relativedelta(years=1)
Output:
datetime.date(2021, 2, 28)

Answer 5

要了解您的解决方案出了什么问题，首先要了解，如果是闰年或非闰年，每年的计数会有所不同：

        Leap  ~Leap
Jan 31    31     31
   ...   ...    ...
Feb 28    59     59
Feb 29    60    n/a
Mar 01    61     60
Mar 02    62     61

观察 2 月 29 日之后闰年和非闰年之间的天数如何偏移 1 天。因此，您的第一个示例类似于（假设不存在 is_leap 的便利属性）：

map(lambda x: x + 366 if x.is_leap else 365, [41, 41, 41, 41, ...])

而你的第二个例子是：

map(lambda x: x + 366 if x.is_leap else 365, [69, 69, 70, 69, ...])

归结为 - 您对何时以及如何抵消闰年的处理已经错了。

您已经有了一些可以为您提供替代方案的答案，所以我将缩短这个答案，并解释为什么您的实施工作。就我个人而言，我会像 MrFuppes 那样处理它，并通过直接添加年份来创建一个新的 datetime 对象：

start_date = dt.date(2009, 3, 10)
end_date = dt.date(2019, 5, 1)
def create_new_dates(start, end):
    cur = start
    while cur < end:
        cur = dt.date(cur.year + 1, cur.month, cur.day)
        yield cur

new_dates = list(create_new_dates(start_date, end_date))

结果：

[datetime.date(2010, 3, 10),
 datetime.date(2011, 3, 10),
 datetime.date(2012, 3, 10),
 datetime.date(2013, 3, 10),
 datetime.date(2014, 3, 10),
 datetime.date(2015, 3, 10),
 datetime.date(2016, 3, 10),
 datetime.date(2017, 3, 10),
 datetime.date(2018, 3, 10),
 datetime.date(2019, 3, 10),
 datetime.date(2020, 3, 10)]