Question

我正在尝试找到一种更有效的方法，根据特定列（id）查找数据框中的重叠数据范围（每行提供的开始/结束日期）。

数据框在'from'列

上排序

我认为有一种方法可以像我一样避免“双重”应用功能...

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”函数循环所有组并在每个组中，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：

Answer 1

您可以移动df['overlap'] = (df['to'].shift()-df['from']) > timedelta(0)列并直接减去日期时间。

id

在df['overlap'] = (df.groupby('id') .apply(lambda x: (x['to'].shift() - x['from']) > timedelta(0)) .reset_index(level=0, drop=True))分组时应用此功能可能看起来像

>>> 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

<强>演示

{{1}}

Answer 2

另一种解决方案。可以将其重写为利用熊猫24及更高版本中的Interval.overlaps。

def overlapping_groups(group):
    if len(group) > 1:
      for index, row in group.iterrows():
        for index2, row2 in group.drop(index).iterrows():
          int1 = pd.Interval(row2['start_date'],row2['end_date'], closed = 'both')
          if row['start_date'] in int1:
            return row['id']
          if row['end_date'] in int1:
            return row['id']

gcols = ['id']
group_output = df.groupby(gcols,group_keys=False).apply(overlapping_groups)
ids_with_overlap = set(group_output[~group_output.isnull()].reset_index(drop = True))
df[df['id'].isin(ids_with_overlap)]

Answer 3

您可以将“从”时间与前一个“到”时间进行比较：

df['to'].shift() > df['from']

输出：

0    False
1    False
2    False
3    False
4     True

Answer 4

您可以对from列进行排序，然后使用非常有效的滚动应用功能检查它是否与之前的to列重叠。

df['from'] = pd.DatetimeIndex(df['from']).astype(np.int64)
df['to'] = pd.DatetimeIndex(df['to']).astype(np.int64)

sdf = df.sort_values(by='from')
sdf[["from", "to"]].stack().rolling(window=2).apply(lambda r: 1 if r[1] >= r[0] else 0).unstack()

现在，重叠时段是from=0.0

的时段

   from   to
0   NaN  1.0
1   1.0  1.0
2   1.0  1.0
3   1.0  1.0
4   0.0  1.0

在python中查找日期范围重叠

4 个答案: