Question

我通过API访问一些数据，我需要为我的请求提供日期范围，例如。 start =＆＃39; 20100101＆＃39;，end =＆＃39; 20150415＆＃39;。我想我会通过将日期范围分解为非重叠间隔并在每个间隔上使用多处理来加快速度。

我的问题是我如何分解日期范围并不能始终如一地给我预期的结果。这就是我所做的：

from datetime import date

begin = '20100101'
end = '20101231'

假设我们想把它分解成几个季度。首先，我将字符串更改为日期：

def get_yyyy_mm_dd(yyyymmdd):
    # given string 'yyyymmdd' return (yyyy, mm, dd)
    year = yyyymmdd[0:4]
    month = yyyymmdd[4:6]
    day = yyyymmdd[6:]
    return int(year), int(month), int(day)

y1, m1, d1 = get_yyyy_mm_dd(begin)
d1 = date(y1, m1, d1)
y2, m2, d2 = get_yyyy_mm_dd(end)
d2 = date(y2, m2, d2)

然后将此范围划分为子间隔：

def remove_tack(dates_list):
    # given a list of dates in form YYYY-MM-DD return a list of strings in form 'YYYYMMDD'
    tackless = []
    for d in dates_list:
        s = str(d)
        tackless.append(s[0:4]+s[5:7]+s[8:])
    return tackless

def divide_date(date1, date2, intervals):
    dates = [date1]
    for i in range(0, intervals):
        dates.append(dates[i] + (date2 - date1)/intervals)
    return remove_tack(dates)

使用上面的开头和结尾我们得到：

listdates = divide_date(d1, d2, 4)
print listdates # ['20100101', '20100402', '20100702', '20101001', '20101231'] looks correct

但如果我使用日期：

begin = '20150101'
end = '20150228'

...

listdates = divide_date(d1, d2, 4)
print listdates # ['20150101', '20150115', '20150129', '20150212', '20150226']

我在2月底错过了两天。我的申请不需要时间或时区，我也不介意安装另一个图书馆。

Answer 1

我实际上会采用不同的方法，依靠timedelta和date添加来确定非重叠范围

<强>实施

def date_range(start, end, intv):
    from datetime import datetime
    start = datetime.strptime(start,"%Y%m%d")
    end = datetime.strptime(end,"%Y%m%d")
    diff = (end  - start ) / intv
    for i in range(intv):
        yield (start + diff * i).strftime("%Y%m%d")
    yield end.strftime("%Y%m%d")

<强>执行

>>> begin = '20150101'
>>> end = '20150228'
>>> list(date_range(begin, end, 4))
['20150101', '20150115', '20150130', '20150213', '20150228']

Answer 2

您应该更改日期时间的日期

from datetime import date, datetime, timedelta

begin = '20150101'
end = '20150228'

def get_yyyy_mm_dd(yyyymmdd):
  # given string 'yyyymmdd' return (yyyy, mm, dd)
  year = yyyymmdd[0:4]
  month = yyyymmdd[4:6]
  day = yyyymmdd[6:]
  return int(year), int(month), int(day)

y1, m1, d1 = get_yyyy_mm_dd(begin)
d1 = datetime(y1, m1, d1)
y2, m2, d2 = get_yyyy_mm_dd(end)
d2 = datetime(y2, m2, d2)

def remove_tack(dates_list):
  # given a list of dates in form YYYY-MM-DD return a list of strings in form 'YYYYMMDD'
  tackless = []
  for d in dates_list:
    s = str(d)
    tackless.append(s[0:4]+s[5:7]+s[8:])
  return tackless

def divide_date(date1, date2, intervals):
  dates = [date1]
  delta = (date2-date1).total_seconds()/4
  for i in range(0, intervals):
    dates.append(dates[i] + timedelta(0,delta))
  return remove_tack(dates)

listdates = divide_date(d1, d2, 4)
print listdates

结果：

[＆＃39; 20150101 00：00：00＆＃39;，＆＃39; 20150115 12：00：00＆＃39;，＆＃39; 20150130 00：00：00＆＃39;，＆＃39; 20150213 12：00：00＆＃39;，＆＃39; 20150228 00：00：00＆＃39;]

Answer 3

您可以使用datetime.date对象吗？

如果你这样做：

import datetime
begin = datetime.date(2001, 1, 1)
end = datetime.date(2010, 12, 31)

intervals = 4

date_list = []

delta = (end - begin)/4
for i in range(1, intervals + 1):
    date_list.append((begin+i*delta).strftime('%Y%m%d'))

和date_list应该包含每个inteval的结束日期。

Answer 4

使用来自Pandas的Datetimeindex和Periods以及字典理解：

import pandas as pd

begin = '20100101'
end = '20101231'

start = dt.datetime.strptime(begin, '%Y%m%d')
finish = dt.datetime.strptime(end, '%Y%m%d')

dates = pd.DatetimeIndex(start=start, end=finish, freq='D').tolist()
quarters = [d.to_period('Q') for d in dates]
df = pd.DataFrame([quarters, dates], index=['Quarter', 'Date']).T

quarterly_dates = {str(q): [ts.strftime('%Y%m%d') 
                            for ts in df[df.Quarter == q].Date.values.tolist()]
                           for q in quarters}

>>> quarterly_dates
{'2010Q1': ['20100101',
  '20100102',
  '20100103',
  '20100104',
  '20100105',
...
  '20101227',
  '20101228',
  '20101229',
  '20101230',
  '20101231']}

>>> quarterly_dates.keys()
['2010Q1', '2010Q2', '2010Q3', '2010Q4']

Answer 5

我创建了一个函数，其中包括日期分割中的结束日期。


from dateutil import rrule
from dateutil.relativedelta import relativedelta
from dateutil.rrule import DAILY


def date_split(start_date, end_date, freq=DAILY, interval=1):
    """

    :param start_date:
    :param end_date:
    :param freq: refer rrule arguments can be SECONDLY, MINUTELY, HOURLY, DAILY, WEEKLY etc
    :param interval: The interval between each freq iteration.
    :return: iterator object
    """
    # remove microsecond from date object as minimum allowed frequency is in seconds.
    start_date = start_date.replace(microsecond=0)
    end_date = end_date.replace(microsecond=0)
    assert end_date > start_date, "end_date should be greated than start date."
    date_intervals = rrule.rrule(freq, interval=interval, dtstart=start_date, until=end_date)
    for date in date_intervals:
        yield date
    if date != end_date:
        yield end_date

Answer 6

如果要按天数划分日期。您可以使用以下代码段。

import datetime

firstDate = datetime.datetime.strptime("2019-01-01", "%Y-%m-%d")
lastDate = datetime.datetime.strptime("2019-03-30", "%Y-%m-%d")
numberOfDays = 15
startdate = firstDate
startdatelist = []
enddatelist = []

while startdate <= lastDate:
    enddate = startdate + datetime.timedelta(days=numberOfDays - 1)
    startdatelist.append(startdate.strftime("%Y-%m-%d 00:00:00"))
    if enddate > lastDate: enddatelist.append(lastDate.strftime("%Y-%m-%d 23:59:59"))
    enddatelist.append(enddate.strftime("%Y-%m-%d 23:59:59"))
    startdate = enddate + datetime.timedelta(days=1)

for a, b in zip(startdatelist, enddatelist):
    print(str(a) + "  -  " + str(b))

给定一个日期范围，我们如何将其分解为N个连续的子区间？

6 个答案: