# How to speed up the computing time?

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Riccardo Tronconi on 19 Jul 2021
Commented: Steven Lord on 19 Jul 2021
Dear guys! My code is computationally to expensive and I'm looking for your help.
I have 3 input table with different legths in which each column represents:
1- position in X
2- position in Y
3- timestamp (in nanosec)
What I'm looking for is to create one table with the position in X and Y of the three tags with a common timestamp.
To do so I find the absolute maximum and minimum of the timestamps and then I create a coloumn vector as it follows A=min:seconds(5):max.
At this point, for each interval with the find function I check how many timestamp of the three tags belongs to it (es. find(A(i) > = tag3(:,3) & A(i+1)< tag3(.,3)). The output of the find thanks to the mean function allows me to calculate the mean position in X and Y inside this time interval.
I tested this procedure with a subset of my real data set and everything is as expected but unfortunately using my entire dataset It takes hours to be fully computed.
How could I speed up the entire running time?
Thanks to all of you in advance for your time ! :)
Julius Muschaweck on 19 Jul 2021
Did you check out the "Run and Time" feature, under "HOME" -> "CODE", to determine where your code really spends its time?

Steven Lord on 19 Jul 2021
Consider turning your table into a timetable and calling retime on the timetable.
Steven Lord on 19 Jul 2021
For question 1, it depends on which release you're using. I forget exactly when the 'regular' and 'TimeStep' input arguments were introduced but if your release has then you could use that syntax.
"TT2 = retime(TT1,'regular',method,'TimeStep',dt) calculates regularly spaced row times using the time step dt."
If not you could use the newTimes input argument.
For question 2, I'm not sure what specifically you mean.
rng default
initialTimes = minutes(randi(60, 7, 1));
rng(1234, 'twister')
x = randi([-10 10], 7, 1);
newtimes = minutes(0:10:60).';
myTimetable = timetable(initialTimes, x)
myTimetable = 7×1 timetable
initialTimes x ____________ __ 49 min -6 55 min 3 8 min -1 55 min 6 38 min 6 6 min -5 17 min -5
result = retime(myTimetable, newtimes, @mean)
result = 7×1 timetable
initialTimes x ____________ ___ 0 min -3 10 min -5 20 min NaN 30 min 6 40 min -6 50 min 4.5 60 min NaN
The first bin in result averages the values for x in rows 3 and 6 of myTimetable since both 6 min and 8 min are in the range [0 min, 10 min).
No row in myTimetable has an initialTimes value in the range [20 min, 30 min) and so the corresponding row of result has a NaN.

Walter Roberson on 19 Jul 2021
discretize() all of the A values at the same time, getting out group number G
Now you can
grpstats([X(:), Y(:)], G, @(x) mean(x,1))
and the result should be a max(G) by 2 array where first column is mean of X for the group and second column is mean of Y for the group.
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Walter Roberson on 19 Jul 2021
Assuming that A is a vector in monotonic increasing order:
G = discretize(tag3, A);
XYmean_per_group = grpstats([X(:), Y(:)], G, @(x) mean(x,1));
Xmean_per_group = XYmean(:,1);
Ymean_per_group = XYmean(:,2);