Question

假设您有两个熊猫的日期时间：from_date和end_date。我需要一个函数将其拆分为n个月的折叠数（让我们说n=3）。例如：

import pandas as pd

from_date = pd.to_datetime("2020-02-15")
to_date = pd.to_datetime("2020-05-20")

应分为2折：

{
"1": {"from_date": 2020-02-15, "to_date": 2020-05-15}, 
"2": {"from_date": 2020-05-16, "to_date": 2020-05-20}
}

每折需要满足以下条件： from_date + pd.DateOffset(months=2) >= end_date。因此，这与开始日期和结束日期之间的天数无关。

最Python的方式是什么？熊猫里有东西吗？

Answer 1

根据您希望使用两个日期的方式替换相应的打印语句！

根据How do I calculate the date six months from the current date using the datetime Python module?，dateutil.relativedelta可以帮助解决有无第31天的那几个月！

import pandas as pd
from dateutil.relativedelta import relativedelta

from_date = pd.to_datetime("2020-02-15")
to_date = pd.to_datetime("2020-05-20")
fold = 0
result = {}


while from_date+relativedelta(months=+3)<to_date:
    curfrom = from_date #retain current 'from_date'
    from_date =from_date+relativedelta(months=+3)
    result[fold] = {"from_date": curfrom, "to_date": from_date}
    fold = fold+1
    from_date = from_date + relativedelta(days=+1) #So that the next 'from_date' starts 1 day after
result[fold] = {"from_date": curfrom, "to_date": to_date}

print(result)

Answer 2

我的解决方案：

import pandas as pd

def check_and_split(from_date, to_date):
    from_date = pd.to_datetime(from_date)
    to_date = pd.to_datetime(to_date)

    done = False
    fold = 0
    result = {}

    start = from_date
    end = to_date

    while not done:
        if start + pd.DateOffset(months=2) > to_date:
            done = True
            end = to_date
        else:
            end = start + pd.DateOffset(months=3)

        result[fold] = {"from_date": start, "to_date": end}

        if not done:
            start = end + pd.DateOffset(days=1)
            fold += 1

    return result

没有更Python化的方式吗？熊猫里有东西吗？

熊猫日期时间：将开始和结束日期分成n个月的倍数

2 个答案: