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need to remove some time or make time

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slowlearner on 22 May 2020
Commented: slowlearner on 23 May 2020
I have a datime matrice called tid with 8761 values such as tid = datetime(2015,1,1,00,00,00):hours(1):datetime(2016,1,1,00,00,00);
I'm trying to remove all the times exept the ones from 08-16 i wrote:
for o = 9:24:8761-23
worktime(o,:) = tid([o:o+8],:)
if o>32 %remove for whole year after test
break %this to
end%this aswell
now it alomst devides the timetable in the manner I needed however i was expecting a row and there are some other problems becouse i don't know how to remove the Not a Times (NaT) now (pic below). I'm sure there's a better way to do this than what i'm thinking with a loop but I also have a seperat matrice with values that needs to be seperated in the same manner later so I thought test with the time first. and i don't think i can use cells in the script i have becouse of the newbcoding on my part.
I hope things are clear on what i need to do..

  1 Comment

slowlearner on 22 May 2020
i think i could just remove them in the table menu however tidious that would be..

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Accepted Answer

Steven Lord
Steven Lord on 22 May 2020
Use logical indexing. The relational operators like <= and > work on datetime arrays.
tid = datetime(2015,1,1,00,00,00):hours(1):datetime(2016,1,1,00,00,00);
August16thMidnight = datetime(2015, 8, 16);
August17thMidnight = August16thMidnight + days(1);
onAugust16th = August16thMidnight <= tid & tid < August17thMidnight;


Show 2 older comments
slowlearner on 22 May 2020
Thank you! I didn't know this.
Peter Perkins
Peter Perkins on 22 May 2020
I would have thought an old-school guy like Steve would have suggested reshaping to 24x366 and deleting the first seven and last eight rows (you have to delete that last 2016 element first, though). But right, groupsummary etc. are worth looking at.
slowlearner on 23 May 2020
this is exactly what i did to get a quick fix but i'm gonna have to learn the grouping if i want to improve the code for general use purposes... but thank you

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