Find date range overlap in python

You can compare the 'from' time with the previous 'to' time:

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

Output:

0    False
1    False
2    False
3    False
4     True

You could just shift the to column and perform a direct subtraction of the datetimes.

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

Applying this while grouping by id may look like

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

Another solution. This could be rewritten to leverage Interval.overlaps in pandas 24 and later.

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)]