Question

我正在寻找一种优雅和pythonic的方式来获取上一季度结束的日期。

这样的事情：

def previous_quarter(reference_date):
   ...


>>> previous_quarter(datetime.date(2013, 5, 31))
datetime.date(2013, 3, 31)

>>> previous_quarter(datetime.date(2013, 2, 1))
datetime.date(2012, 12, 31)

>>> previous_quarter(datetime.date(2013, 3, 31))
datetime.date(2012, 12, 31)

>>> previous_quarter(datetime.date(2013, 11, 1))
datetime.date(2013, 9, 30)

编辑：我尝试了什么吗？

是的，这似乎有效：

def previous_quarter(ref_date):
    current_date = ref_date - timedelta(days=1)
    while current_date.month % 3:
        current_date -= timedelta(days=1)
    return current_date

但这似乎是不必要的迭代。

Answer 1

只需查看收到的月份，您就可以通过“艰难的方式”来做到这一点：

def previous_quarter(ref):
    if ref.month < 4:
        return datetime.date(ref.year - 1, 12, 31)
    elif ref.month < 7:
        return datetime.date(ref.year, 3, 31)
    elif ref.month < 10:
        return datetime.date(ref.year, 6, 30)
    return datetime.date(ref.year, 9, 30)

Answer 2

使用dateutil：

import datetime as DT
import dateutil.rrule as rrule

def previous_quarter(date):
    date = DT.datetime(date.year, date.month, date.day)
    rr = rrule.rrule(
        rrule.DAILY,
        bymonth=(3,6,9,12),    # the month must be one of these
        bymonthday=-1,         # the day has to be the last of the month
        dtstart = date-DT.timedelta(days=100))
    result = rr.before(date, inc=False)  # inc=False ensures result < date
    return result.date()

print(previous_quarter(DT.date(2013, 5, 31)))
# 2013-03-31
print(previous_quarter(DT.date(2013, 2, 1)))
# 2012-12-31
print(previous_quarter(DT.date(2013, 3, 31)))
# 2012-12-31
print(previous_quarter(DT.date(2013, 11, 1)))
# 2013-09-30

Answer 3

利用所涉及的数据模式并将问题转换为表格查找 - 您的经典space-time tradeff：

from datetime import date

PQTBL = (((12,31,-1),)*3 + ((3,31,0),)*3 + ((6,30,0),)*3 + ((9,30,0),)*3)

def previous_quarter(ref):
    entry = PQTBL[ref.month-1]
    return date(ref.year+entry[2], entry[0], entry[1])

Answer 4

几乎可以肯定你会更喜欢使用pandas（python库），它有很多“业务时间”数据的功能。

http://pandas.pydata.org/pandas-docs/dev/timeseries.html

Answer 5

为“下一个季度”版本重新编写了 Justin Ethier's 代码。还通过 pytz 和 strftime 格式添加了时区。 @justin-ethier

import pytz
from datetime import datetime, timedelta
import datetime as dt

def nextQuarter():
    ref = datetime.now(pytz.timezone('America/New_York'))
    if ref.month < 4:
        next = dt.datetime(ref.year, 3, 31, 23, 59, 59).strftime('%m-%d-%Y %H:%M:%S')
    elif ref.month < 7:
        next = dt.datetime(ref.year, 6, 30, 23, 59, 59).strftime('%m-%d-%Y %H:%M:%S')
    elif ref.month < 10:
        next = dt.datetime(ref.year, 9, 30, 23, 59, 59).strftime('%m-%d-%Y %H:%M:%S')
    else:
        next = dt.datetime(ref.year + 1, 12, 31, 23, 59, 59).strftime('%m-%d-%Y %H:%M:%S')
    return next
next = nextQuarter()

Answer 6

import datetime
def previous_quarter(ref):
    quarter = (ref.month - 1) // 3
    prev_quarter = (quarter - 1) % 4
    return datetime.datetime(ref.year if quarter>0 else ref.year-1, prev_quarter*3+1, 1)

Answer 7

仅使用python的datetime库 -

的解决方案

import datetime

def get_quarter_end(dt):
     '''
         given a datetime object, find the end of the quarter
     '''
     quarter_of_month = int((dt.month-1)/3 + 1)
     #======================================================
     # find the first day of the next quarter
     #======================================================
     # if in last quarter then go to the next year
     year = dt.year + 1 if quarter_of_month==4 else dt.year
     # if in last quarter then month is january (or 1)
     month = 1 if quarter_of_month==4 else (quarter_of_month*3) + 1

     first_of_next_quarter = datetime.datetime(year  = year, 
                                               month = month,
                                               day   = 1
                                              )
     # last day of quarter for dt will be minus 1 day of first of next quarter
     quarter_end_dt = first_of_next_quarter - datetime.timedelta(days=1)
     return quarter_end_dt

if __name__=='__main__':
    dt = datetime.datetime.strptime('2016-07-15', '%Y-%m-%d')
    target_dt = get_quarter_end(dt)

Answer 8

找到季度的第一天和月份，然后使用relativedelta减去一天。

from dateutil.relativedelta import relativedelta
def previous_quarter(ref):
    first_month_of_quarter = ((ref.month - 1) // 3) * 3 + 1
    return ref.replace(month=first_month_of_quarter, day=1) - relativedelta(days=1)

计算上一季度的结束

8 个答案: