I am trying to find a more efficient way of finding overlapping data ranges (start/end dates provided per row) in a dataframe based on a specific column (id). Dataframe is sorted on the 'from' column. I think there is a way to avoid the double apply function as I did:
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
I used the "apply" function to loop on all groups and within each group, I use "apply" per row:
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
My code was inspired from the following links :
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
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
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)]