change the first digits when ends with 5959
2 views (last 30 days)
Show older comments
Hey, I've got a doubt. I have a string of numbers that go from 0000 to 5959(First two are the minuts and the other two the seconds), when it reaches another time 0000 I wanted to add a unit that goes from 1 to 23(that indicates the hours). I have added this line of code:
T_Hour_1 = '00'+t1; % it gives us t1 with two 00 that indicates the hour
then I got this lines of code that changes the string of T_Hour_1 to dur.
DN1 = str2double(t1);
dur1 = minutes(floor(DN1/100)) + seconds(mod(DN1,100)); % separar en minutos y segundos
[~, M1, S1] = hms(dur1); % Dar tiempo en 2 variables, utilizar M1 para crear rangos
Before doing DN1 I would like to add the numbers from 0 to 23 in front of the others.
How could I do it?
Thank you for your help!!
0 Comments
Accepted Answer
Stephen23
on 21 Sep 2021
Edited: Stephen23
on 21 Sep 2021
Rather than messing around with strings you should probably just work with duration objects or numeric arrays.
S = ["000000";"000059";"000100";"000159";"005959";"000000";"005959";"000000";"005900"]
D = duration(strcat(extractBefore(S,3),':',extractBetween(S,3,4),':',extractAfter(S,4)))
D = D + hours(cumsum([0;diff(D)<0]))
4 Comments
Stephen23
on 21 Sep 2021
"could you explain to me the second line of code you gave me"
That line locates where the difference in duration is negative (i.e. the duration jumps from a larger value to a smaller value). It then uses CUMSUM on those jump points to create a vector of hourly increments.
If you do not understand how code works then take it apart and look at all of the intermediate results:
S = ["0000";"0059";"0100";"0159";"5959";"0000";"5959";"0000";"5900"];
D = duration(strcat('00:',extractBefore(S,3),':',extractAfter(S,2)))
V = diff(D) % difference between adjacent durations.
W = V<0 % locations where longer duration jumps back to shorter duration.
X = [0;W] % pad with zero.
Y = cumsum(X) % create numeric list of hourly jumps
Z = hours(Y) % convert to duration.
D = D + Z % add to the original data
More Answers (1)
Steven Lord
on 20 Sep 2021
Is there a reason you're starting with string data first rather than using a duration array from the start?
d = minutes(0:0.25:60);
d.Format = 'hh:mm:ss';
d(1:5)
d(end-4:end)
2 Comments
See Also
Categories
Find more on Dates and Time 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!