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change the first digits when ends with 5959

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vicente Noguer
vicente Noguer 2021년 9월 20일
댓글: Stephen 2021년 9월 21일
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!!

채택된 답변

Stephen
Stephen 2021년 9월 21일
편집: Stephen 2021년 9월 21일
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"]
S = 9×1 string array
"000000" "000059" "000100" "000159" "005959" "000000" "005959" "000000" "005900"
D = duration(strcat(extractBefore(S,3),':',extractBetween(S,3,4),':',extractAfter(S,4)))
D = 9×1 duration array
00:00:00 00:00:59 00:01:00 00:01:59 00:59:59 00:00:00 00:59:59 00:00:00 00:59:00
D = D + hours(cumsum([0;diff(D)<0]))
D = 9×1 duration array
00:00:00 00:00:59 00:01:00 00:01:59 00:59:59 01:00:00 01:59:59 02:00:00 02:59:00
  댓글 수: 4
Stephen
Stephen 2021년 9월 21일
"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)))
D = 9×1 duration array
00:00:00 00:00:59 00:01:00 00:01:59 00:59:59 00:00:00 00:59:59 00:00:00 00:59:00
V = diff(D) % difference between adjacent durations.
V = 8×1 duration array
00:00:59 00:00:01 00:00:59 00:58:00 -00:59:59 00:59:59 -00:59:59 00:59:00
W = V<0 % locations where longer duration jumps back to shorter duration.
W = 8×1 logical array
0 0 0 0 1 0 1 0
X = [0;W] % pad with zero.
X = 9×1
0 0 0 0 0 1 0 1 0
Y = cumsum(X) % create numeric list of hourly jumps
Y = 9×1
0 0 0 0 0 1 1 2 2
Z = hours(Y) % convert to duration.
Z = 9×1 duration array
0 hr 0 hr 0 hr 0 hr 0 hr 1 hr 1 hr 2 hr 2 hr
D = D + Z % add to the original data
D = 9×1 duration array
00:00:00 00:00:59 00:01:00 00:01:59 00:59:59 01:00:00 01:59:59 02:00:00 02:59:00

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추가 답변(1개)

Steven Lord
Steven Lord 2021년 9월 20일
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)
ans = 1×5 duration array
00:00:00 00:00:15 00:00:30 00:00:45 00:01:00
d(end-4:end)
ans = 1×5 duration array
00:59:00 00:59:15 00:59:30 00:59:45 01:00:00
  댓글 수: 2
Stephen
Stephen 2021년 9월 21일
STR2DOUBLE is not really suitable for handling date/time values written as one string of digits.
You should be using duration objects to store time durations.

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