如何在 Python 中将日期范围划分为月份？

Question

我有以下日期范围：

begin: 2018-02-15
end: 2018-04-23

我想实现以下目标：

["2018-02-15 - 2018-02-28", "2018-03-01 - 2018-03-31", "2018-04-01 - 2018-04-23"]

本质上，我想将给定的日期范围划分为月份。我想不出在 Python 中实现这一点的方法。

我已经考虑了解决方案here，但是，这会根据指定的时间间隔拆分日期范围。我希望能够动态拆分日期范围。

因此，给定从 2018 年 2 月 15 日到 2018 年 4 月 23 日的日期范围，我希望能够获得该范围内的各个月份，如下所示：

• 2018 年 2 月 15 日至 2018 年 2 月 28 日
• 2018 年 3 月 1 日至 2018 年 3 月 31 日
• 2018 年 4 月 1 日至 2018 年 4 月 23 日

Answer 1

在一个循环中；从第一天开始，不断增加一天，直到您到达结束日期；每当月份更改时保存日期。

import datetime
begin = '2018-02-15'
end = '2018-04-23'

dt_start = datetime.datetime.strptime(begin, '%Y-%m-%d')
dt_end = datetime.datetime.strptime(end, '%Y-%m-%d')
one_day = datetime.timedelta(1)
start_dates = [dt_start]
end_dates = []
today = dt_start
while today <= dt_end:
    #print(today)
    tomorrow = today + one_day
    if tomorrow.month != today.month:
        start_dates.append(tomorrow)
        end_dates.append(today)
    today = tomorrow

end_dates.append(dt_end)


out_fmt = '%d %B %Y'
for start, end in zip(start_dates,end_dates):
    print('{} to {}'.format(start.strftime(out_fmt), end.strftime(out_fmt)))

结果：

>>>
15 February 2018 to 28 February 2018
01 March 2018 to 31 March 2018
01 April 2018 to 23 April 2018
>>>

---

您可能会想出一种方法来获取开始日期和结束日期之间的月份范围；为每个月份的第一天创建一个日期时间对象，存储它们以及它们之前的几天。不过，跨越年份的日期可能会有问题。

Answer 2

要使用方便的日期对象，请始终使用标准模块datetime。这会包装您的字符串格式日期，并允许更轻松的计算以及定制的输出格式。

不幸的是，它似乎遗漏了一个重要信息：每个月的最后天，给定一年（这对于二月是必要的）。还有一个额外的模块calendar，它返回the last day for a month，但是因为这就是你所需要的，并且有一个简单的基于datetime的函数可以做同样的事情，所以我选择了后者。

这样，您可以设置任何begin 日期并将其连同该月的最后一天一起附加到您的列表中，然后将begin 设置为下一个 月的第一天并继续直到你通过end。

警告/微调：我意识到如果 begin 和 end 都在相同月份内，它将不起作用。这需要临时检查，所以我将最初的 while begin < end 更改为 while True 并将结束日期的检查移到单独的行中。

另外，跨越一个年需要再次单独测试，因为否则语句month+1将在12月失败。

import datetime

# borrowed from https://stackoverflow.com/a/13565185
# as noted there, the calendar module has a function of its own
def last_day_of_month(any_day):
    next_month = any_day.replace(day=28) + datetime.timedelta(days=4)  # this will never fail
    return next_month - datetime.timedelta(days=next_month.day)

begin = "2018-02-15"
end = "2018-04-23"

def monthlist(begin,end):
    begin = datetime.datetime.strptime(begin, "%Y-%m-%d")
    end = datetime.datetime.strptime(end, "%Y-%m-%d")

    result = []
    while True:
        if begin.month == 12:
            next_month = begin.replace(year=begin.year+1,month=1, day=1)
        else:
            next_month = begin.replace(month=begin.month+1, day=1)
        if next_month > end:
            break
        result.append ([begin.strftime("%Y-%m-%d"),last_day_of_month(begin).strftime("%Y-%m-%d")])
        begin = next_month
    result.append ([begin.strftime("%Y-%m-%d"),end.strftime("%Y-%m-%d")])
    return result


date_list = monthlist(begin,end)
print (date_list)

结果

[ ['2018-02-15', '2018-02-28'],
  ['2018-03-01', '2018-03-31'],
  ['2018-04-01', '2018-04-23'] ]

（为了便于阅读，略作格式化）

Answer 3

如果你不介意使用pandas，有一个很好的助手date_range 可以实现你想要的：

import pandas as pd
start = pd.Timestamp('20180215')
end = pd.Timestamp('20180423')

parts = list(pd.date_range(start, end, freq='M')) 
# parts = [Timestamp('2018-02-28 00:00:00', freq='M'), Timestamp('2018-03-31 00:00:00', freq='M')]

if start != parts[0]:
  parts.insert(0, start)
if end != parts[-1]:
  parts.append(end)
parts[0] -= pd.Timedelta('1d')  # we add back one day later

pairs = zip(map(lambda d: d + pd.Timedelta('1d'), parts[:-1]), parts[1:])

pairs_str = list(map(lambda t: t[0].strftime('%Y-%m-%d') + ' - ' + t[1].strftime('%Y-%m-%d'), pairs))

# pairs_str = ['2018-02-15 - 2018-02-28', '2018-03-01 - 2018-03-31', '2018-04-01 - 2018-04-23']

Answer 4

使用python日历和会计年度变化

import calendar
from datetime import datetime
begin = '2018-02-15'
end= '2018-04-23'

begin_year, begin_month, begin_date = [int(i) for i in begin.split("-")]
end_year, end_month, end_date = [int(i) for i in end.split("-")]

years = end_year - begin_year
# if date range contains more than single year, we calculate total months
if years:
    months = (12 - begin_month) + end_month + (12 * (years - 1))
else:
    months = end_month - begin_month
dates = []
month = begin_month
year = begin_year

def create_datetime_object(y, m, d):
    return datetime.strptime('{}-{}-{}'.format(y, m, d), '%Y-%m-%d')
# append the first date
dates.append(create_datetime_object(begin_year, begin_month, begin_date))

for i in range(months+1):
    days_in_month = calendar.monthrange(year, month)[-1]
    if month == begin_month and year == begin_year:
        dates.append(create_datetime_object(begin_year, begin_month, days_in_month))
    elif month == end_month and year == end_year:
        dates.append(create_datetime_object(end_year, end_month, 1))
    else:
        dates.append(create_datetime_object(year, month, 1))
        dates.append(create_datetime_object(year, month, days_in_month))
    if month == 12:
        month = 0
        year += 1
    month += 1
# append the last date
dates.append(create_datetime_object(end_year, end_month, end_date))

要获得问题列表，我们可以执行以下操作 -

dates = [datetime.strftime(dt, '%Y-%m-%d') for dt in dates]

Answer 5

我不得不进行类似的操作并最终构建了这个函数。我在不同的用例（不同年份、同月...）上对其进行了测试，效果很好。

这里的灵感来自 S.Lott 的回答 Creating a range of dates in Python

import datetime

def get_segments(start_date, end_date):
    """
    Divides input date range into associated months periods

    Example:
        Input: start_date = 2018-02-15
               end_date   = 2018-04-23
        Output:
            ["2018-02-15 - 2018-02-28", 
             "2018-03-01 - 2018-03-31", 
             "2018-04-01 - 2018-04-23"]
    """
    curr_date = start_date
    curr_month = start_date.strftime("%m")
    segments = []

    loop = (curr_date!=end_date) 
    days_increment = 1

    while loop:
        # Get incremented date with 1 day
        curr_date = start_date + datetime.timedelta(days=days_increment)
        # Get associated month
        prev_month = curr_month
        curr_month = curr_date.strftime("%m")
        # Add to segments if new month
        if prev_month!=curr_month:
            # get start of segment
            if not segments:
                start_segment = start_date
            else:
                start_segment = segments[-1][1] + datetime.timedelta(days=1)
            # get end of segment
            end_segment = curr_date - datetime.timedelta(days=1)
            # define and add segment
            segment = [start_segment, end_segment]
            segments.append(segment)
        # stop if last day reached
        loop = (curr_date!=end_date) 
        # increment added days
        days_increment += 1

    if not segments or segments[-1][1]!=end_date:
        if not segments:
            start_last_segment = start_date
        else:
            start_last_segment = segments[-1][1] + datetime.timedelta(days=1)
        last_segment = [start_last_segment, end_date]
        segments.append(last_segment)

    for i in range(len(segments)):
        segments[i][0] = segments[i][0].strftime("%Y-%m-%d")
        segments[i][1] = segments[i][1].strftime("%Y-%m-%d")

    return segments

这是一个例子：

start_date = datetime.datetime(2020, 5, 27)
end_date = datetime.datetime(2021, 3, 1)

segments = get_segments(start_date, end_date)

for seg in segments:
    print(seg)

输出：

['2020-05-27', '2020-05-31']
['2020-06-01', '2020-06-30']
['2020-07-01', '2020-07-31']
['2020-08-01', '2020-08-31']
['2020-09-01', '2020-09-30']
['2020-10-01', '2020-10-31']
['2020-11-01', '2020-11-30']
['2020-12-01', '2020-12-31']
['2021-01-01', '2021-01-31']
['2021-02-01', '2021-02-28']
['2021-03-01', '2021-03-01']

Answer 6

我通过@wwii 扩展了解决方案

现在您不会有重复的开始和/或结束日期

def date_range_split_monthly(begin, end):

    dt_start = datetime.strptime(begin, '%Y-%m-%d')
    dt_end = datetime.strptime(end, '%Y-%m-%d')
    one_day = timedelta(1)
    start_dates = [dt_start]
    end_dates = []
    today = dt_start
    while today <= dt_end:
        # print(today)
        tomorrow = today + one_day
        if tomorrow.month != today.month:
            if tomorrow <= dt_end:
                start_dates.append(tomorrow)
                end_dates.append(today)
        today = tomorrow

    end_dates.append(dt_end)

    return start_dates, end_dates

Answer 7

对于使用Pendulum的人：

import pendulum
start = pendulum.now().subtract(months=6)
end = pendulum.today()
period = pendulum.period(start, end)

time_ranges = list(period.range("months"))

arr = []

for index, dt in enumerate(time_ranges):
    if index < len(time_ranges) - 1:
        start_range = time_ranges[index].format("YYYY-MM-D")
        end_range = time_ranges[index + 1].format("YYYY-MM-D")
        litt = F"{start_range} - {end_range}"
        print(litt)
        arr.append(litt)

print(arr)

More about period here