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
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。
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)
一般来说，Pandas非常适合时间序列，并直接支持日期范围。
import pandas as pd
daterange = pd.date_range(start_date, end_date)
然后你可以循环daterrange来打印日期:
for single_date in daterange:
print (single_date.strftime("%Y-%m-%d"))
它也有很多选择，让生活更轻松。例如，如果您只想要工作日，您只需交换bdate_range。看到http://pandas.pydata.org/pandas-docs/stable/timeseries.html generating-ranges-of-timestamps
Pandas的强大之处在于它的数据框架，它支持向量化操作(很像numpy)，使得跨大量数据的操作非常快速和简单。
编辑:
你也可以完全跳过for循环，直接打印出来，这样更简单、更高效:
print(daterange)
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))
这个函数有一些额外的特性:
can pass a string matching the DATE_FORMAT for start or end and it is converted to a date object
can pass a date object for start or end
error checking in case the end is older than the start
import datetime
from datetime import timedelta
DATE_FORMAT = '%Y/%m/%d'
def daterange(start, end):
def convert(date):
try:
date = datetime.datetime.strptime(date, DATE_FORMAT)
return date.date()
except TypeError:
return date
def get_date(n):
return datetime.datetime.strftime(convert(start) + timedelta(days=n), DATE_FORMAT)
days = (convert(end) - convert(start)).days
if days <= 0:
raise ValueError('The start date must be before the end date.')
for n in range(0, days):
yield get_date(n)
start = '2014/12/1'
end = '2014/12/31'
print list(daterange(start, end))
start_ = datetime.date.today()
end = '2015/12/1'
print list(daterange(start, end))
