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Curious
Posts:
1,982
Registered:
12/6/04


Re: time rounding issues
Posted:
Aug 20, 2014 4:51 PM


"Travis " <sinusoid2@hotmail.com> wrote in message <lt2u7a$npt$1@newscl01ah.mathworks.com>... > "someone" wrote in message <lt2q5u$cqd$1@newscl01ah.mathworks.com>... > > "Travis " <sinusoid2@hotmail.com> wrote in message <lt2n4e$3vv$1@newscl01ah.mathworks.com>... > > > I have a large amount of data that is timestamped by the instruments recording it. When processing the data I need a complete time series, but there are times when there are gaps in the record that I need to fill. I have converted the timestring to MATLAB time and created another full time variable using... > > > > > > time = t(1):1/48:t(end); %half hour data intervals > > > > > > When I use the intersect command though I do not get the desired result; not all the times are the same (even though when I put them back to datestrings they are identical). > > > > > > I set up a quick check... > > > > > > a(:,1) = 1:1/48:100; > > > b = 1; > > > for i = 2:4753; %length of a > > > b = [b;addtodate(b(i1),30,'minute')]; > > > end > > > > > > c = find(a ~= b); %3168 results. > > > > > > Is this a rounding issue with the 1/48? Is there a way to do this without using a forloop and the addtodate command (I have seen the code slow down a lot when working with larger datasets)? > > > > The MATLAB FAQ addresses this issue and provides a workaround: > > > > <<http://matlab.wikia.com/wiki/FAQ#Why_is_0.3__0.2__0.1_.28or_similar.29_not_equal_to_zero.3F>> > > That was a good read thank you. So using that new information is there a way to go about using the intersect command with two datasets that are not the same size? I have tried forcing rounding x = round(x*10000)/10000 and it seems to help in most cases, yet not in all.
See dpb's response above for a better workaround.
But to answer your question, you still have the same floating point problems (as you found out) using the round function (the way you are using it). This doesn't simply truncate the number to a certain decimal place as you might think it would. MATLAB still uses the full floatingpoint representation for the number. Look at the "rounded" numbers with the format long option and you will see it is not exactly what you would expect.



