You are analyzing a train timetable (you have some time to kill since you just missed your train!) What is the average daily trip duration?
Example:
DepTime = datetime(2016,11,1,[6 8 9],[22 51 05],0)';
ArrTime = datetime(2016,11,1,[9 10 12],[17 32 09],0)';
Destination = categorical({'BOS';'NYC';'BOS'});
tt = timetable(DepTime,Destination,ArrTime)
tt =
DepTime Destination ArrTime
____________________ ___________ ____________________
01-Nov-2016 06:22:00 BOS 01-Nov-2016 09:17:00
01-Nov-2016 08:51:00 NYC 01-Nov-2016 10:32:00
01-Nov-2016 09:05:00 BOS 01-Nov-2016 12:09:00
The result should be a duration variable:
Dur = [duration(3,9,05);duration(2,55,15);duration(3,0,24)]
Dur =
3×1 duration array
03:09:50
02:55:15
03:00:24
Feature Tip: R2016b introduces timetables with related functions which may be helpful. To learn more see MATLAB Timetables.
Related Problems:
Use a timetable to analyze a train timetable (Part 4)
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Could you be more specific on the underlying relationship between the input (timetable) and output (mean duration)? That is, among which group are you calculating the mean duration? From your example, there are only 2 groups of destinations, i.e., BOS and NYC, but why there are 3 values in the output? What's the mapping behind? Also, your problems including Part 1, 4, 5 are all based on single test, which allows hard-coded solution to lead.
Peng do share the grouping method cause it eludes me :)
There seems to be an error with the test suite: the first mean time should be 03:09:50 and not 03:09:05. And the test code should probably be abs(ttDur - Dur) <= duration([0 0 1]) (not 0.01).
The test suite has been updated to use 03:09:50. Also, the check for each of the three average durations is made against 0.0001 now, instead of 0.01, to ensure 03:09:05 will not pass (as that was incorrect). Finally, the test suite now accommodates a row vector (in addition to the expected column vector) as the function output.