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
> 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']
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。
import datetime
def daterange(start, stop, step_days=1):
current = start
step = datetime.timedelta(step_days)
if step_days > 0:
while current < stop:
yield current
current += step
elif step_days < 0:
while current > stop:
yield current
current += step
else:
raise ValueError("daterange() step_days argument must not be zero")
if __name__ == "__main__":
from pprint import pprint as pp
lo = datetime.date(2008, 12, 27)
hi = datetime.date(2009, 1, 5)
pp(list(daterange(lo, hi)))
pp(list(daterange(hi, lo, -1)))
pp(list(daterange(lo, hi, 7)))
pp(list(daterange(hi, lo, -7)))
assert not list(daterange(lo, hi, -1))
assert not list(daterange(hi, lo))
assert not list(daterange(lo, hi, -7))
assert not list(daterange(hi, lo, 7))
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。
对于那些对python函数方式感兴趣的人:
from datetime import date, timedelta
from itertools import count, takewhile
for d in takewhile(lambda x: x<=date(2009,6,9), map(lambda x:date(2009,5,30)+timedelta(days=x), count())):
print(d)