{"id":2556,"date":"2010-05-14T10:50:08","date_gmt":"2010-05-14T10:50:08","guid":{"rendered":"https:\/\/blogs.mathworks.com\/pick\/2010\/05\/14\/progbar\/"},"modified":"2010-05-14T10:50:08","modified_gmt":"2010-05-14T10:50:08","slug":"progbar","status":"publish","type":"post","link":"https:\/\/blogs.mathworks.com\/pick\/2010\/05\/14\/progbar\/","title":{"rendered":"progbar"},"content":{"rendered":"
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Bob<\/a>'s pick this week is progbar<\/a> by Ben Mitch<\/a>.\r\n <\/p>\r\n <\/introduction>\r\n

Do your MATLAB programs take a long time to run? Maybe you have a large simulation that loops over many iterations. It's nice\r\n to monitor progress. You can do that with waitbar<\/a><\/tt>.\r\n <\/p>

h = waitbar(0,'Please wait...'<\/span>);\r\nfor<\/span> step = 1:1000\r\n  % computations take place here<\/span>\r\n  waitbar(step\/1000)\r\n  drawnow\r\nend<\/span>\r\nclose(h)<\/pre>
That works well for coarse loops with few iterations and considerable time spent during each. The instrumentation itself does\r\n      add some overhead, however. As the number of iterations increases, so does the cumulative time.\r\n   <\/p>
loopSizes = [1 10 100 1000 10e3];\r\nwaitbarTimes = zeros(size(loopSizes));\r\nfor<\/span> n=1:numel(loopSizes)\r\n  myBar = waitbar(0,num2str(loopSizes(n)));\r\n  myClock = tic;\r\n  for<\/span> k=1:loopSizes(n)\r\n    waitbar(k\/loopSizes(n),myBar)\r\n    drawnow\r\n  end<\/span> %loop simulation<\/span>\r\n  waitbarTimes(n) = toc(myClock);\r\n  close(myBar)\r\nend<\/span> %vary problem size<\/span>\r\nplot(loopSizes,waitbarTimes)\r\nxlabel 'Loop Size'<\/span>\r\nylabel 'Total Time (sec)'<\/span><\/pre> 
Since a loop of ten thousand iterations added that many seconds of overhead, how many minutes would a loop of one million\r\n      add?\r\n   <\/p>
100*waitbarTimes(end)\/60 % 100X bigger loop (sec->min)<\/span><\/pre>
ans =\r\n       41.332\r\n<\/pre>
Wow! If your whole simulation only takes a few minutes total to run then you probably don't want to bog it down that much\r\n      by over instrumenting it.\r\n   <\/p>\r\n   
You could add logic to your loop to only make updates every 10 iterations for example.<\/p>
myBar = waitbar(0,'Big loop with coarse updates'<\/span>);\r\nmyClock = tic;\r\nfor<\/span> step=1:1e4\r\n  % computations take place here<\/span>\r\n  if<\/span> mod(step,10)==0 %every 10-th iteration<\/span>\r\n    waitbar(step\/1e4,myBar)\r\n    drawnow\r\n  end<\/span>\r\nend<\/span>\r\ntoc(myClock)\r\nclose(myBar)<\/pre>
Elapsed time is 2.613291 seconds.\r\n<\/pre>
Of course now your scientific task involves some artistry to chose the right number of iterations to skip each time. However,\r\n      suppose the amount of time varies widely between each iteration. This approach falls apart in that case.\r\n   <\/p>\r\n   
What I like about progbar<\/tt> is that it features an update \"period\" setting you can control. The default is 0.1 seconds. So it will not update more than\r\n      10 times per second. Otherwise, it works (not exactly but) a lot like waitbar<\/tt>.\r\n   <\/p>
myBar = progbar;\r\nmyClock = tic;\r\nfor<\/span> step=1:1e4\r\n  % computations take place here<\/span>\r\n  progbar(myBar,100*step\/1e4)\r\n  drawnow\r\nend<\/span>\r\ntoc(myClock)\r\nclose(myBar)<\/pre>
Elapsed time is 3.375511 seconds.\r\n<\/pre>
If we vary the loop size again we can see where progbar<\/tt> begins to shine.\r\n   <\/p>
progbarTimes = zeros(size(loopSizes));\r\nfor<\/span> n=1:numel(loopSizes)\r\n  myBar = progbar;\r\n  myClock = tic;\r\n  for<\/span> k=1:loopSizes(n)\r\n    progbar(myBar,100*k\/loopSizes(n))\r\n    drawnow\r\n  end<\/span> %loop simulation<\/span>\r\n  progbarTimes(n) = toc(myClock);\r\n  close(myBar)\r\nend<\/span> %vary problem size<\/span>\r\nplot(loopSizes,[waitbarTimes' progbarTimes'])\r\nxlabel 'Loop Size'<\/span>\r\nylabel 'Total Time (sec)'<\/span>\r\nlegend waitbar<\/span> progbar<\/span> Location<\/span> Northwest<\/span><\/pre> 
Clearly progbar<\/tt> adds less overhead to our (hypothetical) simulation. Thanks, Ben!\r\n   <\/p>\r\n   
Comments?<\/a><\/p>