formulating elapsed time in hours for input data...

I have some code that I need to change from a string into a vector of numbers representing elapsed hours. My input data looks like:

    date_str =     '2015-06-05 14:17:36'
      '2015-06-05 14:17:36'
      '2015-06-05 14:17:36'
      '2015-06-05 14:17:36'
      '2015-06-05 14:17:36'
      '2015-06-05 14:17:39'
      '2015-06-05 14:17:39'
      '2015-06-05 14:17:39'
      '2015-06-05 14:17:39'
      '2015-06-05 14:17:39'
      '2015-06-05 14:17:39'
      '2015-06-05 14:17:40'
      '2015-06-05 14:20:09'
      '2015-06-05 14:20:09'
      '2015-06-05 14:20:09'
      '2015-06-05 14:20:09'
      '2015-06-05 14:23:00'
      '2015-06-05 14:23:00'
      '2015-06-05 14:23:34'
      '2015-06-05 14:23:34'
      '2015-06-05 14:23:35'
      '2015-06-05 14:23:35'
      '2015-06-05 14:23:35'
      '2015-06-05 14:23:35'
      '2015-06-05 14:23:35'
      '2015-06-05 14:26:47';

This is a rather short list of the data, it is actually made up of over 1000000 date strings, the function I use now is:

function [e_time] = date2time(date_str)
date_format = 'mm/dd/yyyy HH:MM:SS';

t_datevec = zeros(size(date_str,1),6);
e_time = zeros(1,length(date_str));

for i = 1:length(date_str)
    t_str = datestr(date_str(i),date_format);
    t_datevec(i,:) = datevec(t_str);
    e_time(i) = etime(t_datevec(i,:),t_datevec(1,:))/3600;
end
end

So it takes a long time to actually use this function, I'm sure that is due to the for loop, any suggestions or ideas to speed this up? Thank You!

Answer 1

[Peter Perkins]

In MATLAB R2014b or later, use datetime. With your date_str variable converted to a cell array of strings:

>> d = datetime(cellstr(date_str))
d = 
   05-Jun-2015 14:17:36
   05-Jun-2015 14:17:36
   05-Jun-2015 14:17:36
   05-Jun-2015 14:17:36
   05-Jun-2015 14:17:36
   05-Jun-2015 14:17:39
   05-Jun-2015 14:17:39
   ...
   05-Jun-2015 14:26:47
>> t1 = d - d(1); t1.Format = 'h' % elapsed since first
t1 = 
            0 hrs
            0 hrs
            0 hrs
            0 hrs
            0 hrs
   0.00083333 hrs
   0.00083333 hrs
   ...
      0.15306 hrs
>> t2 = diff(d); t2.Format = 'h' % successive diffs
t2 = 
            0 hrs
            0 hrs
            0 hrs
            0 hrs
   0.00083333 hrs
            0 hrs
            0 hrs
   ...
     0.053333 hrs

Those create durations. If you need double values, wrap the hours function around them. Hope this helps.