Calculate time for count

i have a sequence of iteration i.e

seconds=5;

for i=0:seconds/100:seconds

% run task

end

how can i match 5 equal to 5 seconds? i do know 5 here is not exactly 5 seconds. is it depends on the processor speed? different computer, different time count?

so, next time if i want the calculation to be faster i just change 5 into 2 or 1.

thank you.

Answer 1

[Guillaume]

While theoretically, you could pace your loop to run at whichever speed you want by using tic, toc and pause, this wouldn't be very efficient and prone to drifting.

A better way of running a task at regular interval is to use a timer:

task = @(~,~) disp('event');
t = timer('Period', 5, 'ExecutionMode', 'fixedRate', 'TaskToExecute', 100, 'TimerFcn', task)
t.Run;

Your task must of course take less time to execute than the period of the timer. If not, you can set the 'BusyMode' property of the timer to decide what happens.

[m]

is there any easy way?

i've tried now

i.e

t1=now;
t2=now;
elapsed_days1 = t2 - t1;
elasped_second1 = elapsed_days1 *24*60*60;
% elapsed_seconds1

but, it only return total time the task is running not the time i control.

[EDITED, Jan, Code formatted]

[Guillaume]

It's only three lines of code to use a timer, you don't get any easier! timer is the easy way and more importantly is the reliable way. And it's designed for the purpose of running tasks at regular intervals.

As I said you can also regulate the pace of a loop with tic, toc, pause or some java thread sleep (see Jan's answer). I don't find that any easier, and I can guarantee you that after a while your period will have drifted slightly. It's also harder to monitor for task overrun.

Answer 2

[Jan]

Guillaume's suggestion is the most accurate and direct solution: Use a timer if you need a time event.

Another approach:

delay = 5;  % In seconds
for k = 1:100
   tic
   % You computations here
   disp(datestr(now, 0));

     pause(delay - toc);
  end

Note: This is more accurate than pause:

java.lang.Thread.sleep(delay * 1000);