如何找到一周的开始和开始(Python)

f45qwnt8 于 2023-03-04 发布在 Python

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我正在尝试创建一个程序，它将获取今天的日期，并查找最近的历史开始和结束的一周（不包括银行假日和周末）。想法是，它将获取今天的日期"02/26/23"，并返回02/21/23作为开始（02/20/23是银行假日）和02/24/23作为结束。
我当前的代码是

import pandas as pd
import datetime as dt
from pandas.tseries.holiday import USFederalHolidayCalendar

end = dt.datetime.now()
start = dt.datetime(end.year, end.month, end.day-5)

dr = pd.date_range(start=start, end=end)
cal = USFederalHolidayCalendar()
holidays = cal.holidays(start=dr.min(), end=dr.max())
A = dr[~dr.isin(holidays)]
B = A[A.weekday != 5]
B = B[B.weekday != 6]

for year in set(B.year):
    tmp = B[B.year == year]
    for week in set(tmp.week):
        temp = tmp[tmp.week == week]
        print(temp[temp.weekday == temp.weekday.min()])  # begining of week
        print(temp[temp.weekday == temp.weekday.max()])  # ending of week

这导致：

DatetimeIndex(['2023-02-21'], dtype='datetime64[ns]', freq=None)
DatetimeIndex(['2023-02-24'], dtype='datetime64[ns]', freq=None)

但也会给出"未来警告：weekofyear和week已被弃用..."
有没有办法把它转换成日期时间格式，这样如果我打印它，它会显示为2023 - 02 - 21。我怎样才能避免FutureWarning？
(BTW我从https://stackoverflow.com/a/43674721/21292653中获得了其中一些

python-3.x

来源：https://stackoverflow.com/questions/75574377/how-to-find-the-beginning-and-start-of-a-week-python

1条答案

按热度按时间

xxhby3vn1#

您应该尝试pd.Index(tmp.isocalendar().week)，而不是tmp.week，它是：

import pandas as pd
import datetime as dt
from pandas.tseries.holiday import USFederalHolidayCalendar

end = dt.datetime.now()
start = dt.datetime(end.year, end.month, end.day-5)

dr = pd.date_range(start=start, end=end)
cal = USFederalHolidayCalendar()
holidays = cal.holidays(start=dr.min(), end=dr.max())
A = dr[~dr.isin(holidays)]
B = A[A.weekday != 5]
B = B[B.weekday != 6]

for year in set(B.year):
    tmp = B[B.year == year]
    for week in set(pd.Index(tmp.isocalendar().week)):
        temp = tmp[pd.Index(tmp.isocalendar().week) == week]
        print(temp[temp.weekday == temp.weekday.min()])  # begining of week
        print(temp[temp.weekday == temp.weekday.max()])  # ending of week

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如何找到一周的开始和开始(Python)

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