Calculate the elapsed time in seconds from date with 'HH:MM:SS' format ( MATLAB)

Question

I'm trying to convert date in a string a vector of time, which as this format: 'HH:MM:SS', in a vector of time in seconds.It works well till a certain point and after it returns fake values.

Here is what I'm doing:

a={'596:30:18';'596:30:28';'596:30:38';'596:30:48';'596:30:58';'596:31:08';'596:31:18';'596:31:28';'596:31:38';'596:31:48';'596:31:58';'596:32:08';'596:32:18';'596:32:28';'596:32:38';'596:32:48';'596:32:58';'596:33:08'};

         formatIn = 'HH:MM:SS';
         DateVector = datevec(a(1:end,:),formatIn);
         time = etime(DateVector,DateVector(1,:));    

It returns:

DateVector =

2017    1   25  20  30  18
2017    1   25  20  30  28
2017    1   25  20  30  38
2017    1   25  20  30  48
2017    1   25  20  30  58
2017    1   25  20  31  8
2017    1   25  20  31  18
2016    12  7   3   28  40,7040000000000
2016    12  7   3   28  50,7040000000000
2016    12  7   3   29  0,704000000000000
2016    12  7   3   29  10,7040000000000
2016    12  7   3   29  20,7040000000000
2016    12  7   3   29  30,7040000000000
2016    12  7   3   29  40,7040000000000
2016    12  7   3   29  50,7040000000000
2016    12  7   3   30  0,704000000000000
2016    12  7   3   30  10,7040000000000
2016    12  7   3   30  20,7040000000000

and

time =

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-4294897,29600000
-4294887,29600000
-4294877,29600000
-4294867,29600000
-4294857,29600000
-4294847,29600000
-4294837,29600000
-4294827,29600000
-4294817,29600000
-4294807,29600000
-4294797,29600000

only the 7 firt value are correct. Anybody knows why is this happening?

Answer 1

here is another way without loops and way faster (in theory):

B = regexp(a, ':', 'split');
C = vertcat(B{:});
D = str2double(C);  %D = cellfun(@str2num, C); also works but slower

this translates your time stamps to Hours:Minutes:Seconds in 3 columns, you can then quite easily subtract and get the time:

E =  bsxfun(@minus,D,D(1,:));
F = E(:,1)*3600 + E(:,2)*60 + E(:,3)

>>F =

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btw, what a strange datetime format.

Answer 2

The only explanation I can think of is that the HH identifier is not intendet to be used with 3-digits hours. Actually the datevec documentation of datevec says

HH - Hour in two digits format

So maybe, '596:30:28' is not a valid date string and you need to convert it to a different format, eg 'dd:HH:MM:SS' .

This can be done like this (assuming that all hours are 3-digit numbers):

a={'596:30:18';'596:30:28';'596:30:38';'596:30:48';'596:30:58';'596:31:08';'596:31:18';'596:31:28';'596:31:38';'596:31:48';'596:31:58';'596:32:08';'596:32:18';'596:32:28';'596:32:38';'596:32:48';'596:32:58';'596:33:08'};

a_ddhhmmss = {};
for ii=1:length(a)
    datestring = a{ii};
    colon_indices = strfind(datestring, ':');  %check how many digits the hours have
    hours_digits = colon_indices(1)-1;
    hours = str2num(datestring(1:hours_digits));
    days = floor(hours/24) + 1; % start with day=1
    hours = rem(hours,24);
    a_ddhhmmss{end+1} = [num2str(days), ':', num2str(hours), datestring(4:end)];
end

formatIn = 'dd:HH:MM:SS';
DateVector = datevec(a_ddhhmmss(1:end,:),formatIn)
time = etime(DateVector,DateVector(1,:))

Of course, it still makes no sense that the behavior you observe just sets in between the dates '596:30:18' and '596:30:28' . Even if the datevec function is used out of scope, it should rather throw an error instead of returning meaningless output.

Answer 3

You can do this pretty easily by first converting a using datenum and datetime , subtracting the first time point, and converting to seconds :

D = datetime(datenum(a), 'ConvertFrom', 'datenum');  % A datetime array
secElapsed = seconds(D-D(1))

secElapsed =

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