下面做一个按天递增的范围怎么样:

for d in map( lambda x: startDate+datetime.timedelta(days=x), xrange( (stopDate-startDate).days ) ):
  # Do stuff here

startDate和stopDate是datetime。日期对象

对于通用版本:

for d in map( lambda x: startTime+x*stepTime, xrange( (stopTime-startTime).total_seconds() / stepTime.total_seconds() ) ):
  # Do stuff here

startTime和stopTime是datetime。日期或datetime。datetime对象 (两者应是同一类型) stepTime是一个timedelta对象

注意.total_seconds()只在python 2.7之后才被支持。如果你被早期版本困住了，你可以写自己的函数:

def total_seconds( td ):
  return float(td.microseconds + (td.seconds + td.days * 24 * 3600) * 10**6) / 10**6

显示从今天开始的最后n天:

import datetime
for i in range(0, 100):
    print((datetime.date.today() + datetime.timedelta(i)).isoformat())

输出:

2016-06-29
2016-06-30
2016-07-01
2016-07-02
2016-07-03
2016-07-04

下面是一个通用日期范围函数的代码，类似于Ber的答案，但更灵活:

def count_timedelta(delta, step, seconds_in_interval):
    """Helper function for iterate.  Finds the number of intervals in the timedelta."""
    return int(delta.total_seconds() / (seconds_in_interval * step))


def range_dt(start, end, step=1, interval='day'):
    """Iterate over datetimes or dates, similar to builtin range."""
    intervals = functools.partial(count_timedelta, (end - start), step)

    if interval == 'week':
        for i in range(intervals(3600 * 24 * 7)):
            yield start + datetime.timedelta(weeks=i) * step

    elif interval == 'day':
        for i in range(intervals(3600 * 24)):
            yield start + datetime.timedelta(days=i) * step

    elif interval == 'hour':
        for i in range(intervals(3600)):
            yield start + datetime.timedelta(hours=i) * step

    elif interval == 'minute':
        for i in range(intervals(60)):
            yield start + datetime.timedelta(minutes=i) * step

    elif interval == 'second':
        for i in range(intervals(1)):
            yield start + datetime.timedelta(seconds=i) * step

    elif interval == 'millisecond':
        for i in range(intervals(1 / 1000)):
            yield start + datetime.timedelta(milliseconds=i) * step

    elif interval == 'microsecond':
        for i in range(intervals(1e-6)):
            yield start + datetime.timedelta(microseconds=i) * step

    else:
        raise AttributeError("Interval must be 'week', 'day', 'hour' 'second', \
            'microsecond' or 'millisecond'.")

为什么有两个嵌套迭代?对我来说，它只用一次迭代就产生了相同的数据列表:

for single_date in (start_date + timedelta(n) for n in range(day_count)):
    print ...

没有列表被存储，只有一个生成器被迭代。此外，生成器中的“if”似乎是不必要的。

毕竟，线性序列应该只需要一个迭代器，而不是两个。

与John Machin讨论后更新:

也许最优雅的解决方案是使用生成器函数来完全隐藏/抽象日期范围内的迭代:

from datetime import date, timedelta

def daterange(start_date, end_date):
    for n in range(int((end_date - start_date).days)):
        yield start_date + timedelta(n)

start_date = date(2013, 1, 1)
end_date = date(2015, 6, 2)
for single_date in daterange(start_date, end_date):
    print(single_date.strftime("%Y-%m-%d"))

注意:为了与内置的range()函数保持一致，此迭代在到达end_date之前停止。因此，对于包容性迭代使用第二天，就像使用range()一样。

下面做一个按天递增的范围怎么样:

for d in map( lambda x: startDate+datetime.timedelta(days=x), xrange( (stopDate-startDate).days ) ):
  # Do stuff here

startDate和stopDate是datetime。日期对象

对于通用版本:

for d in map( lambda x: startTime+x*stepTime, xrange( (stopTime-startTime).total_seconds() / stepTime.total_seconds() ) ):
  # Do stuff here

startTime和stopTime是datetime。日期或datetime。datetime对象 (两者应是同一类型) stepTime是一个timedelta对象

注意.total_seconds()只在python 2.7之后才被支持。如果你被早期版本困住了，你可以写自己的函数:

def total_seconds( td ):
  return float(td.microseconds + (td.seconds + td.days * 24 * 3600) * 10**6) / 10**6

这可能更清楚:

from datetime import date, timedelta

start_date = date(2019, 1, 1)
end_date = date(2020, 1, 1)
delta = timedelta(days=1)
while start_date <= end_date:
    print(start_date.strftime("%Y-%m-%d"))
    start_date += delta