【问题标题】:How to get last wednesday for current month in python?如何在python中获取当前月份的最后一个星期三?
【发布时间】:2022-01-19 06:23:55
【问题描述】:

我的代码如下

today = todayte
print('today1 =', today)
offset = (today.weekday() - 2) % 7
print('offset1=', offset)
last_wednesday = today - timedelta(days=offset)            
print('last_wednesday1 = ', last_wednesday)

我目前的输出如下

today1 = 2018-03-05
offset1 = 5
last_wednesday1 =  2018-02-28

在上述情况下,我得到 上个月的上周三
但我需要上周三的当前月份。

我的预期输出如下

last_wednesday = 2018-03-28

【问题讨论】:

  • 通用算法:获取当月的第一个星期三+1,然后减去7天。

标签: python-3.x time-series python-datetime timedelta


【解决方案1】:

这是一种方法:

from datetime import datetime , timedelta

todayDT = datetime.today()
currentMonth = todayDT.month

nWed = todayDT
while todayDT.month == currentMonth:
    todayDT += timedelta(days=1)
    if todayDT.weekday()==2: #this is Wednesday 
        nWed = todayDT
print (nWed)

【讨论】:

  • 我尝试了上面的代码,但出现以下错误Traceback (most recent call last): File "E:\nifty_banknifty\banknifty_nifty_2018\bnf_nf_2018.py", line 141, in <module> todayDT += timedelta(days=1) OverflowError: date value out of range@Adedoyin Akande
  • 刚刚编辑,有错别字
  • 现在上面的代码正在工作我得到nWed= 2018-03-28 00:00:00这个输出。我怎样才能将2018-03-28 从上面的日期时间对象中分离出来? @Adedoyin Akande @AdedoyinAkande
  • 我明白了!当我做nWed = nWed.date() 时,我得到nWed= 2018-03-28 这个日期是分开的。谢谢!!!
  • 太棒了!您也可以为答案投票:winks
【解决方案2】:

您可以结合使用日期时间和日历模块:

from datetime import datetime, timedelta
import calendar

today = datetime.now()
# find first day of the month and then, find first wednesday of the month and replace it
# weekday of wednesday == 2
first_day = datetime.today().replace(day=1)
while first_day.weekday() != 2:
    first_day += timedelta(days=1)

number_of_days_in_month = calendar.monthrange(today.year, today.month)[1]
last_wend = first_day + timedelta(days=(((number_of_days_in_month - first_day.day) // 7) * 7))

print(last_wend)

或如@Mark Ransom 建议的那样:

from datetime import datetime, timedelta

day_ = (datetime.now().replace(day=1) + timedelta(days=32)).replace(day=1)
while True:
    day_ -= timedelta(days=1)
    if day_.weekday() == 2:
        break
print(day_)

【讨论】:

  • 感谢您接受我的建议。虽然这不是我应该做的那样,但当一个简单的数学可以做同样的事情时,我讨厌迭代。
【解决方案3】:

这个怎么样,我们先跳到下个月,重用你现有的代码:

import datetime as dt

todayte = dt.date(2018, 3, 5)
today = todayte
d = dt.date(today.year, today.month, 28) # the 28th day of a month must exist
d = d + dt.timedelta(days=7) # so we are sure d is in the next month
# then we apply your original logic
offset = (d.weekday() - 2) % 7
last_wednesday = d - dt.timedelta(days=offset)
print(last_wednesday)

结果:

2018-04-04

【讨论】:

  • 我试过上面的代码我得到以下错误Traceback (most recent call last): File "E:\nifty_banknifty\banknifty_nifty_2018\bnf_nf_2018.py", line 136, in <module> d = d + dt.timedelta(days=7) # so we are sure d is in the next month AttributeError: type object 'datetime.datetime' has no attribute 'timedelta'@AlexKong @Alex Kong
  • 您需要注意导入,如果您将我的代码复制并粘贴到新文件中,它应该可以工作——因为我可以获得正确的结果。
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