我有以下代码来做到这一点,但我如何能做得更好?现在我认为它比嵌套循环更好,但是当您在列表理解中使用生成器时,它开始变得像perl一行程序。

day_count = (end_date - start_date).days + 1
for single_date in [d for d in (start_date + timedelta(n) for n in range(day_count)) if d <= end_date]:
    print strftime("%Y-%m-%d", single_date.timetuple())

笔记

我不是用这个来打印的。这只是为了演示。 start_date和end_date变量是datetime。date对象,因为我不需要时间戳。(它们将用于生成报告)。

样例输出

开始日期为2009-05-30,结束日期为2009-06-09:

2009-05-30
2009-05-31
2009-06-01
2009-06-02
2009-06-03
2009-06-04
2009-06-05
2009-06-06
2009-06-07
2009-06-08
2009-06-09

当前回答

for i in range(16):
    print datetime.date.today() + datetime.timedelta(days=i)

其他回答

> pip install DateTimeRange

from datetimerange import DateTimeRange

def dateRange(start, end, step):
        rangeList = []
        time_range = DateTimeRange(start, end)
        for value in time_range.range(datetime.timedelta(days=step)):
            rangeList.append(value.strftime('%m/%d/%Y'))
        return rangeList

    dateRange("2018-09-07", "2018-12-25", 7)  

    Out[92]: 
    ['09/07/2018',
     '09/14/2018',
     '09/21/2018',
     '09/28/2018',
     '10/05/2018',
     '10/12/2018',
     '10/19/2018',
     '10/26/2018',
     '11/02/2018',
     '11/09/2018',
     '11/16/2018',
     '11/23/2018',
     '11/30/2018',
     '12/07/2018',
     '12/14/2018',
     '12/21/2018']

Numpy的arange函数可以应用于日期:

import numpy as np
from datetime import datetime, timedelta
d0 = datetime(2009, 1,1)
d1 = datetime(2010, 1,1)
dt = timedelta(days = 1)
dates = np.arange(d0, d1, dt).astype(datetime)

astype的用途是从numpy转换。Datetime64到datetime数组。datetime对象。

import datetime

def daterange(start, stop, step=datetime.timedelta(days=1), inclusive=False):
  # inclusive=False to behave like range by default
  if step.days > 0:
    while start < stop:
      yield start
      start = start + step
      # not +=! don't modify object passed in if it's mutable
      # since this function is not restricted to
      # only types from datetime module
  elif step.days < 0:
    while start > stop:
      yield start
      start = start + step
  if inclusive and start == stop:
    yield start

# ...

for date in daterange(start_date, end_date, inclusive=True):
  print strftime("%Y-%m-%d", date.timetuple())

此函数通过支持负步长等功能,可以实现超出严格要求的功能。只要分解了范围逻辑,就不需要单独的day_count,最重要的是,当从多个地方调用函数时,代码变得更容易阅读。

使用pendulum.period:

import pendulum

start = pendulum.from_format('2020-05-01', 'YYYY-MM-DD', formatter='alternative')
end = pendulum.from_format('2020-05-02', 'YYYY-MM-DD', formatter='alternative')

period = pendulum.period(start, end)

for dt in period:
    print(dt.to_date_string())

如果你打算使用动态timedelta,那么你可以使用:

1. 使用while循环

def datetime_range(start: datetime, end: datetime, delta: timedelta) -> Generator[datetime, None, None]:
    while start <= end:
        yield start
        start += delta

2. 使用for循环

from datetime import datetime, timedelta
from typing import Generator


def datetime_range(start: datetime, end: datetime, delta: timedelta) -> Generator[datetime, None, None]:
    delta_units = int((end - start) / delta)

    for _ in range(delta_units + 1):
        yield start
        start += delta

3.如果你正在使用async/await

async def datetime_range(start: datetime, end: datetime, delta: timedelta) -> AsyncGenerator[datetime, None]:
    delta_units = int((end - start) / delta)

    for _ in range(delta_units + 1):
        yield start
        start += delta

4. 列表理解

def datetime_range(start: datetime, end: datetime, delta: timedelta) -> List[datetime]:
    delta_units = int((end - start) / delta)
    return [start + (delta * index) for index in range(delta_units + 1)]

那么1和2解可以简单地像这样使用

start = datetime(2020, 10, 10, 10, 00)
end = datetime(2022, 10, 10, 18, 00)
delta = timedelta(minutes=30)

result = [time_part for time_part in datetime_range(start, end, delta)]
# or 
for time_part in datetime_range(start, end, delta):
    print(time_part)

3- 3 / 3解决方案可以在异步上下文中使用。因为它运行一个异步生成器对象,该对象只能在异步上下文中使用

start = datetime(2020, 10, 10, 10, 00)
end = datetime(2022, 10, 10, 18, 00)
delta = timedelta(minutes=30)

result = [time_part async for time_part in datetime_range(start, end, delta)]

async for time_part in datetime_range(start, end, delta):
    print(time_part)

这些解决方案的优点是它们都使用了动态的timedelta。这在你不知道你将得到哪个时间增量的情况下非常有用。