问题
Suppose I have list of time interval like
a = [datetime.time(0,0),datetime.time(8,0)]
Now I Have lacs of intervals in list like given below.
b = [[datetime.time(0,0),datetime.time(8,0)], [datetime.time(0,0),datetime.time(10,0)], [datetime.time(0,0),datetime.time(23,59,59)], [datetime.time(15,0),datetime.time(9,0)], [datetime.time(9,0),datetime.time(15,0)]]
We have to filter list b with intervals containing interval a. like in example result will be.
b = [[datetime.time(0,0),datetime.time(8,0)], [datetime.time(0,0),datetime.time(10,0)], [datetime.time(0,0),datetime.time(23,59,59)], [datetime.time(15,0),datetime.time(9,0)]]
Note: I have changed end time from 00 to 23:59:59 but the case remain persistent as we need to understand that daily interval of 00:00 to 08:00 is contained in interval of 15:00 to 09:00
Hint: I have divided 15:00 to 09:00 into two intervals: 00:00-09:00 and 15:00-23:59:59
回答1:
Your code has errors (usage of datetime/datetime.time).
This code will filter out from b everything, that does not overlap with a:
b = [x for x in b if a[0] < x[1] and x[0] < a[1]]
来源：https://stackoverflow.com/questions/12390698/time-interval-overlap-in-python