change the first digits when ends with 5959
2 次查看(过去 30 天)
显示 更早的评论
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 个评论
采纳的回答
Stephen23
2021-9-21
编辑:Stephen23
2021-9-21
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 个评论
Stephen23
2021-9-21
"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
更多回答(1 个)
Steven Lord
2021-9-20
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 个评论
另请参阅
类别
在 Help Center 和 File Exchange 中查找有关 Dates and Time 的更多信息
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!