在python中查找日期范围重叠

我试图找到一种更有效的方法来根据特定列（id）在数据框中查找重叠的数据范围（每行提供的开始/结束日期）。数据框按“来自”列排序。我认为有一种方法可以避免双重apply像我一样运行：

import pandas as pd
from datetime import datetime

df = pd.DataFrame(columns=['id','from','to'], index=range(5), \
                  data=[[878,'2006-01-01','2007-10-01'],
                        [878,'2007-10-02','2008-12-01'],
                        [878,'2008-12-02','2010-04-03'],
                        [879,'2010-04-04','2199-05-11'],
                        [879,'2016-05-12','2199-12-31']])

df['from'] = pd.to_datetime(df['from'])
df['to'] = pd.to_datetime(df['to'])


    id  from        to
0   878 2006-01-01  2007-10-01
1   878 2007-10-02  2008-12-01
2   878 2008-12-02  2010-04-03
3   879 2010-04-04  2199-05-11
4   879 2016-05-12  2199-12-31

我使用“apply”函数在所有组上循环，并且在每个组内，我每行使用“apply”：

def check_date_by_id(df):
    
    df['prevFrom'] = df['from'].shift()
    df['prevTo'] = df['to'].shift()
    
    def check_date_by_row(x):
        
        if pd.isnull(x.prevFrom) or pd.isnull(x.prevTo):
            x['overlap'] = False
            return x
        
        latest_start = max(x['from'], x.prevFrom)
        earliest_end = min(x['to'], x.prevTo)
        x['overlap'] = int((earliest_end - latest_start).days) + 1 > 0
        return x
    
    return df.apply(check_date_by_row, axis=1).drop(['prevFrom','prevTo'], axis=1)

df.groupby('id').apply(check_date_by_id)

    id  from        to          overlap
0   878 2006-01-01  2007-10-01  False
1   878 2007-10-02  2008-12-01  False
2   878 2008-12-02  2010-04-03  False
3   879 2010-04-04  2199-05-11  False
4   879 2016-05-12  2199-12-31  True

我的代码的灵感来自以下链接：

• python中有效的日期范围重叠计算？ https://stackoverflow.com/questions/9044084/efficient-date-range-overlap-calculation-in-python

• 如何在Python中找到范围重叠？ https://stackoverflow.com/questions/6821156/how-to-find-range-overlap-in-python

你可以直接改变to列并执行日期时间的直接减法。

df['overlap'] = (df['to'].shift()-df['from']) > timedelta(0)

在分组时应用此id可能看起来像

df['overlap'] = (df.groupby('id')
                   .apply(lambda x: (x['to'].shift() - x['from']) > timedelta(0))
                   .reset_index(level=0, drop=True))

Demo

>>> df
    id       from         to
0  878 2006-01-01 2007-10-01
1  878 2007-10-02 2008-12-01
2  878 2008-12-02 2010-04-03
3  879 2010-04-04 2199-05-11
4  879 2016-05-12 2199-12-31

>>> df['overlap'] = (df.groupby('id')
                       .apply(lambda x: (x['to'].shift() - x['from']) > timedelta(0))
                       .reset_index(level=0, drop=True))

>>> df
    id       from         to overlap
0  878 2006-01-01 2007-10-01   False
1  878 2007-10-02 2008-12-01   False
2  878 2008-12-02 2010-04-03   False
3  879 2010-04-04 2199-05-11   False
4  879 2016-05-12 2199-12-31    True