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())
start_date
和end_date
变量是datetime.date
个对象,因为我不需要时间戳。 (它们将用于生成报告)。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
答案 0 :(得分:456)
为什么有两个嵌套迭代?对我来说,它只产生一次迭代产生相同的数据列表:
for single_date in (start_date + timedelta(n) for n in range(day_count)):
print ...
没有列表存储,只迭代一个生成器。此外,发电机中的“if”似乎是不必要的。
毕竟,线性序列应该只需要一个迭代器,而不是两个迭代器。
也许最优雅的解决方案是使用生成器函数在日期范围内完全隐藏/抽象迭代:
from datetime import timedelta, date
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,请使用第二天,就像使用range()
一样。
答案 1 :(得分:161)
这可能更清楚:
d = start_date
delta = datetime.timedelta(days=1)
while d <= end_date:
print d.strftime("%Y-%m-%d")
d += delta
答案 2 :(得分:144)
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库有许多更高级的功能,一些非常有用,比如relative delta
s - 并且被实现为一个文件(模块),可以很容易地包含在项目中。
答案 3 :(得分:48)
Pandas一般适用于时间序列,并且直接支持日期范围。
import pandas as pd
daterange = pd.date_range(start_date, end_date)
然后,您可以遍历日期范围以打印日期:
for single_date in daterange:
print (single_date.strftime("%Y-%m-%d"))
它还有很多选项可以让生活更轻松。例如,如果您只想要工作日,则只需交换bdate_range。见http://pandas.pydata.org/pandas-docs/stable/timeseries.html#generating-ranges-of-timestamps
Pandas的强大功能实际上是它的数据帧,它支持矢量化操作(非常像numpy),可以非常快速,轻松地对大量数据进行操作。
编辑: 您也可以完全跳过for循环并直接打印,这样更容易,更有效:
print(daterange)
答案 4 :(得分:13)
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())
此功能的功能超出了您的严格要求,支持负步骤等。只要您将范围逻辑分解出来,那么您就不需要单独的day_count
,最重要的是代码变得更容易从多个地方调用函数时读取。
答案 5 :(得分:11)
为什么不尝试:
import datetime as dt
start_date = dt.datetime(2012, 12,1)
end_date = dt.datetime(2012, 12,5)
total_days = (end_date - start_date).days + 1 #inclusive 5 days
for day_number in range(total_days):
current_date = (start_date + dt.timedelta(days = day_number)).date()
print current_date
答案 6 :(得分:11)
这是我能想到的最易读的解决方案。
import datetime
def daterange(start, end, step=datetime.timedelta(1)):
curr = start
while curr < end:
yield curr
curr += step
答案 7 :(得分:7)
显示今天的最后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
答案 8 :(得分:5)
Numpy的arange
功能可以应用于日期:
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)
使用astype
是将numpy.datetime64
转换为datetime.datetime
个对象的数组。
答案 9 :(得分:4)
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))
答案 10 :(得分:3)
我有类似的问题,但我需要每月迭代而不是每天。
这是我的解决方案
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)
<强>输出强>
2016-06-01
2016-07-01
2016-08-01
2016-09-01
2016-10-01
2016-11-01
2016-12-01
2017-01-01
示例#2
date_start = datetime(2017, 1, 1)
date_end = datetime(2016, 6, 1)
for p in monthly_range(date_start, date_end):
print(p)
<强>输出强>
2017-01-01
2016-12-01
2016-11-01
2016-10-01
2016-09-01
2016-08-01
2016-07-01
2016-06-01
答案 11 :(得分:3)
可以't *相信这个问题已经存在了9年而没有人提出简单的递归函数:
from datetime import datetime, timedelta
def walk_days(start_date, end_date):
if start_date <= end_date:
print(start_date.strftime("%Y-%m-%d"))
next_date = start_date + timedelta(days=1)
walk_days(next_date, end_date)
#demo
start_date = datetime(2009, 5, 30)
end_date = datetime(2009, 6, 9)
walk_days(start_date, end_date)
输出:
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
修改 *现在我可以相信 - 请参阅Does Python optimize tail recursion?。谢谢Tim。
答案 12 :(得分:2)
for i in range(16):
print datetime.date.today() + datetime.timedelta(days=i)
答案 13 :(得分:2)
对于那些对 Pythonic 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)
答案 14 :(得分:2)
> 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']
答案 15 :(得分:1)
您可以使用Arrow
:
这是文档中的示例,迭代数小时:
from arrow import Arrow
>>> start = datetime(2013, 5, 5, 12, 30)
>>> end = datetime(2013, 5, 5, 17, 15)
>>> for r in Arrow.range('hour', start, end):
... print repr(r)
...
<Arrow [2013-05-05T12:30:00+00:00]>
<Arrow [2013-05-05T13:30:00+00:00]>
<Arrow [2013-05-05T14:30:00+00:00]>
<Arrow [2013-05-05T15:30:00+00:00]>
<Arrow [2013-05-05T16:30:00+00:00]>
要迭代数天,您可以这样使用:
>>> start = Arrow(2013, 5, 5)
>>> end = Arrow(2013, 5, 5)
>>> for r in Arrow.range('day', start, end):
... print repr(r)
(没有检查是否可以传递 datetime.date
个对象,但无论如何 Arrow
个对象通常更容易)
答案 16 :(得分:1)
您可以使用熊猫库简单可靠地生成两个日期之间的一系列日期
import pandas as pd
print pd.date_range(start='1/1/2010', end='1/08/2018', freq='M')
您可以通过将频率设置为D,M,Q,Y来更改生成日期的频率 (每天,每月,每季度,每年 )
答案 17 :(得分:0)
这是一般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'.")
答案 18 :(得分:0)
通过将range
args存储在元组中来实现可逆步骤的方法略有不同。
def date_range(start, stop, step=1, inclusive=False):
day_count = (stop - start).days
if inclusive:
day_count += 1
if step > 0:
range_args = (0, day_count, step)
elif step < 0:
range_args = (day_count - 1, -1, step)
else:
raise ValueError("date_range(): step arg must be non-zero")
for i in range(*range_args):
yield start + timedelta(days=i)
答案 19 :(得分:0)
为了完整起见,熊猫还为超出范围的时间戳提供了一个period_range函数:
pd.period_range(start='1/1/1626', end='1/08/1627', freq='D')
答案 20 :(得分:0)
import datetime
from dateutil.rrule import DAILY,rrule
date=datetime.datetime(2019,1,10)
date1=datetime.datetime(2019,2,2)
for i in rrule(DAILY , dtstart=date,until=date1):
print(i.strftime('%Y%b%d'),sep='\n')
输出:
2019Jan10
2019Jan11
2019Jan12
2019Jan13
2019Jan14
2019Jan15
2019Jan16
2019Jan17
2019Jan18
2019Jan19
2019Jan20
2019Jan21
2019Jan22
2019Jan23
2019Jan24
2019Jan25
2019Jan26
2019Jan27
2019Jan28
2019Jan29
2019Jan30
2019Jan31
2019Feb01
2019Feb02
答案 21 :(得分:0)
from datetime import date,timedelta
delta = timedelta(days=1)
start = date(2020,1,1)
end=date(2020,9,1)
loop_date = start
while loop_date<=end:
print(loop_date)
loop_date+=delta
答案 22 :(得分:0)
使用pendulum.period:
10
kotlin.Int
答案 23 :(得分:0)
此功能有一些额外的功能:
错误检查,以防结束时间早于开始
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))
答案 24 :(得分:0)
如果按照天增加范围,请执行以下操作:
for d in map( lambda x: startDate+datetime.timedelta(days=x), xrange( (stopDate-startDate).days ) ):
# Do stuff here
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
请注意,.total_seconds()仅在python 2.7之后支持如果您遇到早期版本,则可以编写自己的函数:
def total_seconds( td ):
return float(td.microseconds + (td.seconds + td.days * 24 * 3600) * 10**6) / 10**6