how to convert hours to 'yyyy-MM-dd hh:mm:ss' format?

I have time in hours from 0-72 hours (i.e. 0, 1, 2, 3,..., 72). i.e. a = 0:72;
I would like to convert this hours into datestr with 'yyyy-MM-dd hh:mm:ss' format.
I know initial yyyy-MM-dd i.e. 2012-04-24.
So how can i set my output in 'yyyy-MM-dd hh:mm:ss' format?
I want o/p to be like;
2012-04-24 00:00:00, 2012-04-24 01:00:00, ...., 2012-04-25 00:00:00, 2012-04-25 01:00:00, ...., 2012-04-26 00:00:00, 2012-04-26 01:00:00
date should change with each 24 hours.
Any help will be greatly appriciated.

 Réponse acceptée

dtm = datetime(2012,4,24,'Format','yyyy-MM-dd HH:mm:ss') + hours(0:72).'
dtm = 73×1 datetime array
2012-04-24 00:00:00 2012-04-24 01:00:00 2012-04-24 02:00:00 2012-04-24 03:00:00 2012-04-24 04:00:00 2012-04-24 05:00:00 2012-04-24 06:00:00 2012-04-24 07:00:00 2012-04-24 08:00:00 2012-04-24 09:00:00 2012-04-24 10:00:00 2012-04-24 11:00:00 2012-04-24 12:00:00 2012-04-24 13:00:00 2012-04-24 14:00:00 2012-04-24 15:00:00 2012-04-24 16:00:00 2012-04-24 17:00:00 2012-04-24 18:00:00 2012-04-24 19:00:00 2012-04-24 20:00:00 2012-04-24 21:00:00 2012-04-24 22:00:00 2012-04-24 23:00:00 2012-04-25 00:00:00 2012-04-25 01:00:00 2012-04-25 02:00:00 2012-04-25 03:00:00 2012-04-25 04:00:00 2012-04-25 05:00:00

4 commentaires

MP
MP le 15 Juil 2022
Modifié(e) : MP le 15 Juil 2022
Thank you so much for your efforts.
This is datetime array. Could you please tell me how to get a string. I want to put xticklabels.
"something like datestr"
MP
MP le 15 Juil 2022
Ok I got it
b = char(dtm); will get me desired output.
Thanks a lot.
Stephen23
Stephen23 le 15 Juil 2022
Modifié(e) : Stephen23 le 15 Juil 2022
str = string(dtm) % string array
txt = cellstr(dtm) % cell array of character vectors
I want to put xticklabels.
If you're plotting using this datetime data as one of the inputs in your call to plot, the tick labels will automatically be generated from the plotted datetime array. You don't need to convert to a string array in this case.
v = 0:10;
x = datetime('today') + days(v);
y = v.^2;
plot(x, y)

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Plus de réponses (1)

MP
MP le 15 Juil 2022

0 votes

Yes, that did a great help.... :)
Thank you so much @Stephen23

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