迭代Python中的一系列日期

时间:2009-06-29 20:16:02

标签: python datetime iteration

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())

备注

  • 我实际上并没有使用它来打印。这只是为了演示目的。
  • start_dateend_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)。

样本输出

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

25 个答案:

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。

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

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

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:

  • 可以传递与DATE_FORMAT匹配的字符串,用于开头或结尾,并将其转换为日期对象
  • 可以传递开始或结束的日期对象
  • 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
  • startDate和stopDate是datetime.date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.date或datetime.datetime对象 (两者应该是相同的类型)
  • stepTime是timedelta对象

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