18

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 :

Mykola Zotko
  • 15,583
  • 3
  • 71
  • 73
Edouard
  • 183
  • 1
  • 1
  • 5

5 Answers5

15

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
miradulo
  • 28,857
  • 6
  • 80
  • 93
  • 1
    Thanks. Simple and clear. Would you know by any chance how to perform the same operation (groupby + check) but for all dates and not just the consecutive ones ? – Edouard Feb 25 '17 at 22:50
  • 1
    I'm not entirely sure what you mean... if the dates are sorted, what more would that accomplish? And I added an example of grouping by `id` for you. – miradulo Feb 25 '17 at 22:52
2

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)]
Adam Zeldin
  • 898
  • 4
  • 6
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
Mykola Zotko
  • 15,583
  • 3
  • 71
  • 73
1

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
farghal
  • 281
  • 1
  • 5
1

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
Guillaume Tréheux
  • 45
  • 6