datetime to samplenumber slow

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Lieke Numan
Lieke Numan 2019-5-23
评论： Lieke Numan 2019-5-23
I have a very long vector with datetime and corresponding heart rate value. Now I want to calculate median value for every 5 minutes (my sample frequency is 4 seconds). However, sometimes some samples are missing. To be able to calculate the median value of HR every 5 minutes, I need to know which samples these are. I did it like this:
t1=time(1)
t2=time(end)
time5min=t1:minutes(5:t2)
for j=[time5min]
[~,ind1]=min(abs(datenum(time)-datenum(j-1)));
[~,ind2]=min(abs(datenum(time)-datenum(j)));
HR_med(ind2)=nanmedian(HR(ind1:ind2));
end
This is, as my vector is very large, very time consuming. Is there a faster way to do this?

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回答（1 个）

Steven Lord
Steven Lord 2019-5-23
Do you want to compute the median for overlapping / sliding windows of 5 minutes, or do you want to compute it for each non-overlapping 5 minute window of your time period?
If you want overlapping bins, use movmedian specifying your datetime array as the SamplePoints and the K or NB/NF inputs as a duration of minutes(5).
If you want non-overlapping bins, take a look at the groupsummary function.
  1 个评论
Lieke Numan
Lieke Numan 2019-5-23
It is okay to have one median value for each 5 minutes, so no sliding window.
I have another function for which I need to use these timestamps. I have number of steps for each sample, and I want to calculate the number of steps per 5 minutes. So i take the difference in "steps(X)-steps(X-5minutes)".

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