我正在使用datetime Python模块。我希望从当前日期计算6个月的日期。有人能帮我一下吗?
我想从当前日期生成一个6个月后的日期的原因是为了生成一个回顾日期。如果用户在系统中输入数据,系统将有从输入数据之日起6个月的审查日期。
我正在使用datetime Python模块。我希望从当前日期计算6个月的日期。有人能帮我一下吗?
我想从当前日期生成一个6个月后的日期的原因是为了生成一个回顾日期。如果用户在系统中输入数据,系统将有从输入数据之日起6个月的审查日期。
当前回答
我有一个更好的办法来解决“2月31日”的问题:
def add_months(start_date, months):
import calendar
year = start_date.year + (months / 12)
month = start_date.month + (months % 12)
day = start_date.day
if month > 12:
month = month % 12
year = year + 1
days_next = calendar.monthrange(year, month)[1]
if day > days_next:
day = days_next
return start_date.replace(year, month, day)
我认为它也适用于负数(减去月份),但我还没有对此进行过多测试。
其他回答
我使用这个函数来更改年和月,但保留日:
def replace_month_year(date1, year2, month2):
try:
date2 = date1.replace(month = month2, year = year2)
except:
date2 = datetime.date(year2, month2 + 1, 1) - datetime.timedelta(days=1)
return date2
你应该这样写:
new_year = my_date.year + (my_date.month + 6) / 12
new_month = (my_date.month + 6) % 12
new_date = replace_month_year(my_date, new_year, new_month)
Python可以使用datautil包,请参阅下面的示例
它不仅限于此,你还可以同时通过日、月和年的组合。
import datetime
from dateutil.relativedelta import relativedelta
# subtract months
proc_dt = datetime.date(2021,8,31)
proc_dt_minus_3_months = proc_dt + relativedelta(months=-3)
print(proc_dt_minus_3_months)
# add months
proc_dt = datetime.date(2021,8,31)
proc_dt_plus_3_months = proc_dt + relativedelta(months=+3)
print(proc_dt_plus_3_months)
# subtract days:
proc_dt = datetime.date(2021,8,31)
proc_dt_minus_3_days = proc_dt + relativedelta(days=-3)
print(proc_dt_minus_3_days)
# add days days:
proc_dt = datetime.date(2021,8,31)
proc_dt_plus_3_days = proc_dt + relativedelta(days=+3)
print(proc_dt_plus_3_days)
# subtract years:
proc_dt = datetime.date(2021,8,31)
proc_dt_minus_3_years = proc_dt + relativedelta(years=-3)
print(proc_dt_minus_3_years)
# add years:
proc_dt = datetime.date(2021,8,31)
proc_dt_plus_3_years = proc_dt + relativedelta(years=+3)
print(proc_dt_plus_3_years)
结果:
2021-05-31
2021-11-30
2021-08-28
2021-09-03
2018-08-31
2024-08-31
这里有一个dateutil的例子。relativedelta,我发现它对于迭代过去一年很有用,每次跳过一个月到现在的日期:
>>> import datetime
>>> from dateutil.relativedelta import relativedelta
>>> today = datetime.datetime.today()
>>> month_count = 0
>>> while month_count < 12:
... day = today - relativedelta(months=month_count)
... print day
... month_count += 1
...
2010-07-07 10:51:45.187968
2010-06-07 10:51:45.187968
2010-05-07 10:51:45.187968
2010-04-07 10:51:45.187968
2010-03-07 10:51:45.187968
2010-02-07 10:51:45.187968
2010-01-07 10:51:45.187968
2009-12-07 10:51:45.187968
2009-11-07 10:51:45.187968
2009-10-07 10:51:45.187968
2009-09-07 10:51:45.187968
2009-08-07 10:51:45.187968
和其他答案一样,你必须弄清楚你说的“6个月后”到底是什么意思。如果你的意思是“六年后某个月的今天”,那么这个是:
datetime.datetime.now() + relativedelta(months=6)
另一种解决方案:计算下个月的天数总和,并将结果添加到当前日期。
import calendar
import datetime
def date_from_now(months):
today = datetime.datetime.today()
month = today.month
year = today.year
sum_days = 0
for i in range(int(months)):
month += 1
if month == 13:
month = 1
year += 1
sum_days += calendar.monthrange(year, month)[1]
return datetime.date.today() + datetime.timedelta(sum_days)
print(date_from_now(12)) # if to day is 2017-01-01, output: 2019-01-01
从这个答案中,请参见parsedatetime。下面是代码示例。更多细节:使用许多自然语言-> YYYY-MM-DD转换示例进行单元测试,以及明显的parsedatetime转换挑战/bug。
#!/usr/bin/env python
# -*- coding: utf-8 -*-
import time, calendar
from datetime import date
# from https://github.com/bear/parsedatetime
import parsedatetime as pdt
def print_todays_date():
todays_day_of_week = calendar.day_name[date.today().weekday()]
print "today's date = " + todays_day_of_week + ', ' + \
time.strftime('%Y-%m-%d')
def convert_date(natural_language_date):
cal = pdt.Calendar()
(struct_time_date, success) = cal.parse(natural_language_date)
if success:
formal_date = time.strftime('%Y-%m-%d', struct_time_date)
else:
formal_date = '(conversion failed)'
print '{0:12s} -> {1:10s}'.format(natural_language_date, formal_date)
print_todays_date()
convert_date('6 months')
上面的代码从MacOSX机器生成以下内容:
$ ./parsedatetime_simple.py
today's date = Wednesday, 2015-05-13
6 months -> 2015-11-13
$