4

I have a list of contracts with start and end dates.

How can I compute the number of overlapping contracts during the life span of a contracts?

df = pd.DataFrame({
    'contract': pd.Series(['A1', 'A2', 'A3', 'A4']),
    'start': pd.Series(['01/01/2015', '03/02/2015', '15/01/2015', '10/01/2015']),
    'end': pd.Series(['16/01/2015', '10/02/2015', '18/01/2015', '12/01/2015'])
})

which gives:

  contract         end       start
0       A1  16/01/2015  01/01/2015
1       A2  10/02/2015  03/02/2015
2       A3  18/01/2015  15/01/2015
3       A4  12/01/2015  10/01/2015

A1 overlaps with A3 and A4, therefore overlaps = 2. A2 overlaps with no contract, therefore overlaps = 0. A3 overlaps with A1, therefore overlaps = 1. A4 overlaps with A1, therefore overlaps = 1.

I could just compare each time span (start to end) but that is O(n**2) Any better idea?

I have the feeling an improvement could be gained by sorting and then looping through the sorted ranges

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1 Answers1

2

Here is a way to do it:

df = pd.DataFrame({
    'contract': pd.Series(['A1', 'A2', 'A3', 'A4']),
    'start': pd.Series(['01/01/2015', '03/02/2015', '15/01/2015', '10/01/2015']),
    'end': pd.Series(['16/01/2015', '10/02/2015', '18/01/2015', '12/01/2015'])
})
df['start'] = pd.to_datetime(df.start, dayfirst=True)
df['end'] = pd.to_datetime(df.end, dayfirst=True)

periods = df[['start', 'end']].apply(lambda x: (pd.date_range(x['start'], x['end']),), axis=1)
overlap = periods.apply(lambda col: periods.apply(lambda col_: col[0].isin(col_[0]).any()))
df['overlap_count'] = overlap[overlap].apply(lambda x: x.count() - 1, axis=1)
print df

Which yields:

  contract        end      start  overlap_count
0       A1 2015-01-16 2015-01-01              2
1       A2 2015-02-10 2015-02-03              0
2       A3 2015-01-18 2015-01-15              1
3       A4 2015-01-12 2015-01-10              1 

I have updated the code to output the count of overlaps and not the overlap in days.

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