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
您可以简单而可靠地使用pandas库在两个日期之间生成一系列日期
import pandas as pd
print pd.date_range(start='1/1/2010', end='1/08/2018', freq='M')
您可以通过设置“freq”为D, M, Q, Y来改变生成日期的频率
(每天，每月，每季，每年
）
我也有类似的问题，但我需要每月而不是每天迭代一次。
这就是我的解
import calendar
from datetime import datetime, timedelta
def days_in_month(dt):
return calendar.monthrange(dt.year, dt.month)[1]
def monthly_range(dt_start, dt_end):
forward = dt_end >= dt_start
finish = False
dt = dt_start
while not finish:
yield dt.date()
if forward:
days = days_in_month(dt)
dt = dt + timedelta(days=days)
finish = dt > dt_end
else:
_tmp_dt = dt.replace(day=1) - timedelta(days=1)
dt = (_tmp_dt.replace(day=dt.day))
finish = dt < dt_end
示例# 1
date_start = datetime(2016, 6, 1)
date_end = datetime(2017, 1, 1)
for p in monthly_range(date_start, date_end):
print(p)
输出
2016-06-01
2016-07-01
2016-08-01
2016-09-01
2016-10-01
2016-11-01
2016-12-01
2017-01-01
例# 2
date_start = datetime(2017, 1, 1)
date_end = datetime(2016, 6, 1)
for p in monthly_range(date_start, date_end):
print(p)
输出
2017-01-01
2016-12-01
2016-11-01
2016-10-01
2016-09-01
2016-08-01
2016-07-01
2016-06-01
使用dateutil库:
from datetime import date
from dateutil.rrule import rrule, DAILY
a = date(2009, 5, 30)
b = date(2009, 6, 9)
for dt in rrule(DAILY, dtstart=a, until=b):
print dt.strftime("%Y-%m-%d")
这个python库有许多更高级的特性，其中一些非常有用，比如相对增量，并且被实现为单个文件(模块)，很容易包含到项目中。
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'.")
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))
