我有以下代码来执行此操作,但我怎样才能做得更好?现在我认为它比嵌套循环更好,但是当你在列表理解中有一个生成器时,它开始得到Perl-one-liner.
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())
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start_date和end_date变量是datetime.date因为我不需要时间戳对象.(它们将用于生成报告).对于开始日期2009-05-30和结束日期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
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