how to select an interval of certain hours each day within a larger timetable using timerange
Show older comments
Hi, I have a timetable of minute by minute data running over multiple days. I wish to identify the times outside regular business hours (9 am to 5 pm) in order to (later) process that data differently from the rest. How do I do this using the timerange function? In other words, how do I select a timerange within each day, for every day, without selecting any full days? Thanks
Accepted Answer
The timerange function operates on timetable arrays, so to use it, the data must be in a timetable and the times must be duration arrays.
This uses table2timetable and duration array conversion —
DT = datetime('26-Mar-2022') + minutes(0:1440*2)'; % Two Days
Data = randn(numel(DT),3); % Create Other Values
T1 = [table(DT) array2table(Data)]
TT1 = table2timetable(T1);
TT1.DT = duration(hour(TT1.DT),minute(TT1.DT),second(TT1.DT))
BusinessHours = timerange(hours(9), hours(17), 'closed'); % 09:00 To 17:00
TT1_BusinessHours = TT1(BusinessHours,:)
OtherHours1 = timerange(hours(0),hours(9),'open'); % Midnight To 09:00
OtherHours2 = timerange(hours(17), hours(24), 'open'); % 17:00 To Midnight
TT1_OtherHours = [TT1(OtherHours1,:); TT1(OtherHours2,:)]
Using timerange is not as straightforward as I thought it would be, and I had to ‘fractionate’ it to get it to work.
% figure
% plot(TT1_BusinessHours.DT, TT1_BusinessHours.Data1)
% figure
% plot(TT1_OtherHours.DT, TT1_OtherHours.Data1)
.
6 Comments
Kevin Johnson
on 26 Mar 2022
That‘s correct.
It might be more straightforward to use the isbetween function. It creates a logical vector, and works with both datetime and duration arrays (according to the documentation) so using the conventions in my code, the two groups would be:
BusinessHours = isbetween(TT1,hours(9),hours(17),'open');
OtherHours = TT1(~BusinessHours,:);
DT = datetime('26-Mar-2022') + minutes(0:1440*2)'; % Two Days
Data = randn(numel(DT),3); % Create Other Values
T1 = [table(DT) array2table(Data)];
TT1 = table2timetable(T1)
TT1.DT = duration(hour(TT1.DT),minute(TT1.DT),second(TT1.DT))
BusinessHours = isbetween(TT1.DT,hours(9),hours(17),'open'); % 09:00 To 17:00
TT1_BusinessHours = TT1(BusinessHours,:)
TT1_OtherHours = TT1(~BusinessHours,:)
T1_BusinessHours = T1(BusinessHours,:)
T1_OtherHours = T1(~BusinessHours,:)
So using isbetween is likely more efficient (and much easier to code). Unfortunately, to have it work over multiple days, the time variable must be a duration. However, once it is defined, it can be used with the original table so that may be the best option. That required a bit of experimentation to fully appreciate. It would likely also work on the original timetable prior to converting the time variable to a duration array.
This may be the only way to make it work, so do the time segmentation on the duration array (that can probably be created specifically to use it with isbetween to create the logical vector), and do the rest of the calculations on the original timetable, after applying the isbetween logical vector to it.
.
Kevin Johnson
on 26 Mar 2022
Star Strider
on 26 Mar 2022
As always, my pleasure!
Kevin Johnson
on 26 Mar 2022
Star Strider
on 26 Mar 2022
Edited: Star Strider
on 26 Mar 2022
In MATLAB, indices must be integers greater than 0 or logical values.
I note that you are using R2019a and I (and the online Run feature here) are using R2022a. There very well could be version differences, however that line:
TT1.DT = duration(hour(TT1.DT),minute(TT1.DT),second(TT1.DT))
simply converts ‘TT1.DT’ into a duration array by extracting the hour, minute, and second from the datetime array to create it. Looking at the R2019a documentation, the duration and hour (and I assume the others) do not appear to have changed. I am at a loss to explain the inconsistency, or the error.
It runs correctly in my code, and since I can’t reproduce it, I can’t troubleshoot it.
Check to see that the arguments to the duration function are the simple integers it requires for each argument. Then, see if the duration call works with the same vectors. That’s the only approach I can think of to determine what the problem is.
EDIT — (26 Mar 2022 at 21:50)
The only other possibility that I can think of is that you have an ‘hour’, ‘minute’, or ‘second’ array that is overshadowing the MATLAB functions by those names.
Run:
which hour -all
and do the same for each of the others. If the first entry in the output of that call is:
hour is a variable.
or something similar, that is the problem. The solution is to re-name the variable to something that makes sense in the context of the code, and does not overshadow any MATLAB functions.
.
More Answers (1)
Do you accept not using function timerange as follows?
T = readtable('date.txt')
T.Var1(hour(T.Var1)<9 | hour(T.Var1)>=17 & ~(hour(T.Var1)>=17 & minute(T.Var1)==0 & second(T.Var1)==0))
1 Comment
Kevin Johnson
on 26 Mar 2022
Categories
Find more on Introduction to Installation and Licensing in Help Center and File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!
