如何找到 Python DataFrame 中存在重叠的所有时间段？答案

【问题标题】：How to find all time periods in which there is overlap in a Python DataFrame?如何找到 Python DataFrame 中存在重叠的所有时间段？
【发布时间】：2023-01-30 21:57:48
【问题描述】：

我的 df 如下：

df = pd.DataFrame({'Name':['Anne','Anne','Anne','Anne','Anne','Anne','Anne','Anne','Anne','Anne','Anne','Anne',
                           'Bob','Bob','Bob','Bob','Bob','Bob','Bob','Bob','Bob','Bob','Bob','Bob'],
               
               "start":["2019-01-01", "2019-02-01", "2019-03-01", "2019-04-01", "2019-05-01", "2019-06-01", "2019-07-01", "2019-08-01", "2019-09-01", "2019-10-01", "2019-11-01", "2019-12-01",  
                        "2019-01-01", "2019-02-01", "2019-03-01", "2019-04-01", "2019-05-01", "2019-06-01", "2019-07-01", "2019-08-01", "2019-09-01", "2019-10-01", "2019-11-01", "2019-12-01"],
               
                 "end":["2019-01-31", "2019-02-28", "2019-03-31", "2019-04-30", "2019-05-31", "2019-06-30", "2019-07-31", "2019-08-31", "2019-09-30", "2019-10-31", "2019-11-30", "2019-12-31",
                        "2019-01-31", "2019-02-28", "2019-03-31", "2019-04-30", "2019-05-31", "2019-06-30", "2019-07-31", "2019-08-31", "2019-09-30", "2019-10-31", "2019-11-30", "2019-12-31"],
                 
                "percentage":[1/12, 1/12, 1/12, 1/12, 1/12, 1/12, 1/12, 1/12, 1/12, 1/12, 1/12, 1/12,
                              1/12, 1/12, 1/12, 1/12, 1/12, 1/12, 1/12, 1/12, 1/12, 1/12, 1/12, 1/12]})

# insert "wrong" row
df.loc[len(df.index)] = ['Anne', "2019-01-15", "2019-02-15", 1/12] 

df.start = df.start.apply(pd.to_datetime, format="%Y-%m-%d")
df.end   = df.end.apply(pd.to_datetime, format="%Y-%m-%d")

我现在想找到同一用户的所有行，其中有一个重叠期。在我上面的代码示例中，只有一处重叠。重叠部分是针对安妮的：

2019-01-01 至 2019-01-31
2019-02-01 至 2019-02-31
2019-01-15 至 2019-02-15

如何返回每个用户重叠的行的索引？

【问题讨论】：

标签： python pandas dataframe datetime

【解决方案1】：

利用：

df1 = df.loc[df.index.repeat(df.end.sub(df.start).dt.days + 1)].copy()
df1['start'] += pd.to_timedelta(df1.groupby(level=0).cumcount(), 'd')

df1 = df[df1.duplicated(['Name','start'], keep=False).groupby(level=0).any()]
print (df1)
    Name      start        end percentage
0   Anne 2019-01-01 2019-01-31       1/12
1   Anne 2019-02-01 2019-02-28       1/12
24  Anne 2019-01-15 2019-02-15       1/12

【讨论】：