'Find date range overlap in python
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 :
Solution 1:[1]
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)]
Solution 2:[2]
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
Solution 3:[3]
You can sort the from column and then simply check if it overlaps with a previous to column or not using rolling apply function which is very efficient.
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()
Now the overlapping periods are the ones with from=0.0
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
Solution 4:[4]
Since I ran into a similar issue like yours, I have been browsing quite extensively. I ran into this solution
this solution.
It uses the function overlaps from pandas, which is documented in detail here:
here.
def function(df):
timeintervals = pd.IntervalIndex.from_arrays(df.from,df.to,closed='both')
index = np.arange(timeintervals.size)
index_to_keep=[]
for intervals in timeintervals:
index_to_keep.append([0])
control = timeintervals[index].overlaps(timeintervals[index[0]])
if control.any():
index = index[~control]
else:
break
if index.size==0:
break
temp = df.index[index_to_keep]
output = df.loc[temp]
return output
Sources
This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.
Source: Stack Overflow
| Solution | Source |
|---|---|
| Solution 1 | Adam Zeldin |
| Solution 2 | Mykola Zotko |
| Solution 3 | |
| Solution 4 |