Efficient date range overlap calculation in python?

Question

I have two date ranges where each range is determined by a start and end date (obviously, datetime.date() instances). The two ranges can overlap or not. I need the number of

Answer 1

You can use the datetimerange package: https://pypi.org/project/DateTimeRange/

from datetimerange import DateTimeRange
time_range1 = DateTimeRange("2015-01-01T00:00:00+0900", "2015-01-04T00:20:00+0900") 
time_range2 = DateTimeRange("2015-01-01T00:00:10+0900", "2015-01-04T00:20:00+0900")
tem3 = time_range1.intersection(time_range2)
if tem3.NOT_A_TIME_STR == 'NaT':  # No overlap
    S_Time = 0
else: # Output the overlap seconds
    S_Time = tem3.timedelta.total_seconds()

"2015-01-01T00:00:00+0900" inside the DateTimeRange() can also be datetime format, like Timestamp('2017-08-30 20:36:25').

Answer 2

Ok my solution is a bit wonky because my df uses all series - but lets say you have the following columns, 2 of which are fixed which is your "Fiscal Year". PoP is "Period of performance" which is your variable data:

df['PoP_Start']
df['PoP_End']
df['FY19_Start'] = '10/1/2018'
df['FY19_End'] = '09/30/2019'

Assume all of the data is in datetime format ie -

df['FY19_Start'] = pd.to_datetime(df['FY19_Start'])
df['FY19_End'] = pd.to_datetime(df['FY19_End'])

Try the following equations to find the number of days overlap:

min1 = np.minimum(df['POP_End'], df['FY19_End'])
max2 = np.maximum(df['POP_Start'], df['FY19_Start'])

df['Overlap_2019'] = (min1 - max2) / np.timedelta64(1, 'D')
df['Overlap_2019'] = np.maximum(df['Overlap_2019']+1,0)

Answer 3

Function calls are more expensive than arithmetic operations.

The fastest way of doing this involves 2 subtractions and 1 min():

min(r1.end - r2.start, r2.end - r1.start).days + 1

compared with the next best which needs 1 subtraction, 1 min() and a max():

(min(r1.end, r2.end) - max(r1.start, r2.start)).days + 1

Of course with both expressions you still need to check for a positive overlap.

Answer 4

• Determine the latest of the two start dates and the earliest of the two end dates.
• Compute the timedelta by subtracting them.
• If the delta is positive, that is the number of days of overlap.

Here is an example calculation:

>>> from datetime import datetime
>>> from collections import namedtuple
>>> Range = namedtuple('Range', ['start', 'end'])

>>> r1 = Range(start=datetime(2012, 1, 15), end=datetime(2012, 5, 10))
>>> r2 = Range(start=datetime(2012, 3, 20), end=datetime(2012, 9, 15))
>>> latest_start = max(r1.start, r2.start)
>>> earliest_end = min(r1.end, r2.end)
>>> delta = (earliest_end - latest_start).days + 1
>>> overlap = max(0, delta)
>>> overlap
52

Answer 5

Pseudocode:

 1 + max( -1, min( a.dateEnd, b.dateEnd) - max( a.dateStart, b.dateStart) )

Answer 6

def get_overlap(r1,r2):
    latest_start=max(r1[0],r2[0])
    earliest_end=min(r1[1],r2[1])
    delta=(earliest_end-latest_start).days
    if delta>0:
        return delta+1
    else:
        return 0