在Python中迭代一系列日期
我有下面的代码来做这件事,但是我怎样才能做得更好呢?现在我认为它比嵌套循环好,但是当您在列表中有一个生成器时,它开始得到Perl一行在Python中迭代一系列日期,python,date,datetime,iteration,date-range,Python,Date,Datetime,Iteration,Date Range,我有下面的代码来做这件事,但是我怎样才能做得更好呢?现在我认为它比嵌套循环好,但是当您在列表中有一个生成器时,它开始得到Perl一行 day\u count=(结束日期-开始日期)。days+1 如果d为什么有两个嵌套迭代?对于我来说,它只在一次迭代中生成相同的数据列表: for single_date in (start_date + timedelta(n) for n in range(day_count)): print ... 并且没有存储任何列表,只对一个生成器进行迭代。而
day\u count=(结束日期-开始日期)。days+1
如果d为什么有两个嵌套迭代?对于我来说,它只在一次迭代中生成相同的数据列表:
for single_date in (start_date + timedelta(n) for n in range(day_count)):
print ...
并且没有存储任何列表,只对一个生成器进行迭代。而且生成器中的“if”似乎没有必要
毕竟,一个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,此迭代在到达结束日期之前停止。因此,对于包含性迭代,请使用第二天,就像使用range()
一样,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
from datetime导入日期,timedelta
开始日期=日期(2019年1月1日)
结束日期=日期(2020年1月1日)
增量=时间增量(天数=1)
开始日期<代码>导入日期时间时
def daterange(开始、停止、步骤=datetime.TIMEDTA(天=1),包括=False):
#inclusive=False,默认情况下表现类似于range
如果step.days>0:
启动<停止时:
产量起点
开始=开始+步骤
#不是+=!如果传入的对象是可变的,则不要修改它
#由于此功能不限于
#仅datetime模块中的类型
elif步骤天数<0:
启动>停止时:
产量起点
开始=开始+步骤
如果包含且开始==停止:
产量起点
# ...
对于daterange中的日期(开始日期、结束日期,包括=True):
打印strftime(“%Y-%m-%d”,date.timetuple())
通过支持负阶跃等,此函数的功能超出了您严格要求的范围。只要您将范围逻辑考虑在内,就不需要单独的日计数
,最重要的是,当您从多个位置调用函数时,代码变得更容易阅读。使用库:
此python库具有许多更高级的功能,其中一些非常有用,例如相对增量
s-并作为单个文件(模块)实现,可以轻松地包含在项目中。import datetime
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))
def日期范围(开始、停止、步骤天数=1):
电流=启动
步长=datetime.timedelta(步长天数)
如果步骤天>0:
当电流<停止时:
屈服电流
电流+=阶跃
elif步骤天数<0:
当前>停止时:
屈服电流
电流+=阶跃
其他:
raise VALUERROR(“daterange()步骤\天数参数不能为零”)
如果名称=“\uuuuu main\uuuuuuuu”:
从pprint导入pprint作为pp
lo=日期时间。日期(2008年12月27日)
hi=datetime.date(2009年1月5日)
pp(列表(日期范围(lo,hi)))
pp(列表(日期范围(hi,lo,-1)))
pp(列表(日期范围(lo,hi,7)))
pp(列表(日期范围(hi,lo,-7)))
断言非列表(日期范围(lo,hi,-1))
断言非列表(日期范围(hi,lo))
断言非列表(日期范围(lo,hi,-7))
断言非列表(日期范围(hi、lo、7))
范围内的i(16):
打印datetime.date.today()+datetime.timedelta(天=i)
对于按天递增的范围,以下内容如何:
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
(两者应为同一类型)
- stepTime是一个timedelta对象
请注意,只有在python 2.7之后才支持.total_seconds(),如果您仍使用早期版本,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
为什么不试试:
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
一般来说,熊猫是时间序列的最佳选择,它直接支持日期范围
import pandas as pd
daterange = pd.date_range(start_date, end_date)
然后,您可以在日期范围内循环以打印日期:
for single_date in daterange:
print (single_date.strftime("%Y-%m-%d"))
它还提供了许多选项,使生活更轻松。例如,如果您只想要工作日,您只需在bdate_范围内进行交换。请参阅
Pandas的强大之处在于它的数据帧,它支持矢量化操作(与numpy非常相似),使得跨大量数据的操作非常快速和简单
编辑:
您也可以完全跳过for循环,直接打印,这样更容易、更高效:
print(daterange)
此功能有一些额外的功能:
- 可以为开始或结束传递与日期\格式匹配的字符串,并将其转换为日期对象
- 可以为开始或结束传递日期对象
- 如果结束时间早于开始时间,则检查错误
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))
导入日期时间
从日期时间导入时间增量
日期\格式=“%Y/%m/%d”
def日期范围(开始、结束):
def转换(日期):
尝试:
date=datetime.datetime.strtime(日期,日期格式)
返回日期。日期()
除类型错误外:
返回日期
def获取日期(n):
return datetime.datetime.strftime(转换(开始)+timedelta(天数=n),日期格式)
天=(转换(结束)-转换(开始))。天
如果天数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
Numpy的arange
功能可应用于日期:
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)
astype
的用途是将numpy.datetime64
转换为datetime.datetime
对象数组。以下是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'.")
这是我能想到的最人性化的解决方案
import datetime
def daterange(start, end, step=datetime.timedelta(1)):
curr = start
while curr < end:
yield curr
curr += step
导入日期时间
def daterange(开始、结束、步骤=datetime.timedelta(1)):
curr=开始
当前<结束时:
收益率货币
电流+=步进
我也有类似的问题,但我需要每月迭代,而不是每天迭代。
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'.")
import datetime
def daterange(start, end, step=datetime.timedelta(1)):
curr = start
while curr < end:
yield curr
curr += step
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
date_start = datetime(2016, 6, 1)
date_end = datetime(2017, 1, 1)
for p in monthly_range(date_start, date_end):
print(p)
2016-06-01
2016-07-01
2016-08-01
2016-09-01
2016-10-01
2016-11-01
2016-12-01
2017-01-01
date_start = datetime(2017, 1, 1)
date_end = datetime(2016, 6, 1)
for p in monthly_range(date_start, date_end):
print(p)
2017-01-01
2016-12-01
2016-11-01
2016-10-01
2016-09-01
2016-08-01
2016-07-01
2016-06-01
import pandas as pd
print pd.date_range(start='1/1/2010', end='1/08/2018', freq='M')
> 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']
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)
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')
2019Jan10
2019Jan11
2019Jan12
2019Jan13
2019Jan14
2019Jan15
2019Jan16
2019Jan17
2019Jan18
2019Jan19
2019Jan20
2019Jan21
2019Jan22
2019Jan23
2019Jan24
2019Jan25
2019Jan26
2019Jan27
2019Jan28
2019Jan29
2019Jan30
2019Jan31
2019Feb01
2019Feb02
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
import pendulum
start = pendulum.from_format('2020-05-01', 'YYYY-MM-DD', formatter='alternative')
end = pendulum.from_format('2020-05-02', 'YYYY-MM-DD', formatter='alternative')
period = pendulum.period(start, end)
for dt in period:
print(dt.to_date_string())
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)
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]>
>>> start = Arrow(2013, 5, 5)
>>> end = Arrow(2013, 5, 5)
>>> for r in Arrow.range('day', start, end):
... print repr(r)