How can create a function with floating median?

Hello, I have a data with 24 hour continuous RR interval (distance between consecutive heart beats)data and i am trying to create a function that would obtain floating averages for different time durations. Ideally i would like to have a floating median rather than average since the data is non normally distributed. I am having hard time for some reason to create median float. I was wondering if some one would be able to help me out in changing the average float function to median float function. Thanks Tamil

function [avge,stdeviation,pnts,timesampl]=avgfloat(data,timestep,tstart)
% timestep is a vector containing the series of time intervals of interest
% in hr (converted to seconds)
timestep=timestep*3600;
% data is the data
% figure out how many timesteps were entered
l=length(timestep);
% calculate total time elapsed (s) during test (ignoring first data point)
time=zeros(length(data),1);
for i=2:length(data)
    time(i)=time(i-1)+data(i);
end
% calcuate number of given time intervals in test
pntstop=floor(time(length(time))./timestep);
for i=1:l
    % initialize start variable
    strt=1;
    for j=1:pntstop(i)
        % find index of end of next timestep
        n=find(time>=timestep(i)*j,1);
        % find number of points that includes
        pnts{i}(j)=n-strt+1;
        % find average over that time interval
        avge{i}(j)=sum(data(strt:n))/length(data(strt:n));
        % find standard deviation over that time interval
        stdeviation{i}(j)=std(data(strt:n));
        % define what time range the sample represents
        timesampl{i}(j)=time(floor((n+strt)/2))+tstart*3600;
        % update start variable
        strt=n+1;
        % shift data to align with hours of day
        if timesampl{i}(j)>=24*3600
            timesampl{i}(j)=timesampl{i}(j)-24*3600;
        elseif timesampl{i}(j)>=48*3600
            timesampl{i}(j)=timesampl{i}(j)-48*3600;
        end
    end
end

[Wolfgang]

can you give a more detailed explanation and a short example of your input values

[Guillaume]

Tamilselvam's comment moved from Answer to here:

The input is a single column variable of RR intervals starting at different times of the day. about 180,000 or so data points. I have attached an example data for one subject.

[Tamilselvam Gunasekaran]

Basically it is holter recording HR information. Starts at some part of the days. My goal was to align the data to the time of the day and from the 24 hour RR interval i am trying to pull consecutive medians for different durations for example (5min, 30min, 1hour and so on).

[Guillaume]

Bear in mind that most of us don't work in the same field as you do, so expressions like 'holter recording HR information' may not mean anything to us.

As input to your function you appear to have two vectors, data that I assume represents some signal, and timestep that obviously represents some time. Yet, early on, you have

time(i) = time(i-1) + data(i);

Is that correct? The data also represents some time? What is its relationship with timestep?

Note that the above could be achieved more simply without a loop as:

time = cumsum([0 data(2:end)]);

It also begs the question, why is data(1) discarded?

[Tamilselvam Gunasekaran]

Sorry about the explanation. I will try to do better in explaining next time time step is calcultated from the data itself. For example if the time is interval between Heart rate is below for 9 consecutive heart beats.

0.337

0.352

0.307

0.302

0.318

0.365

0.405

0.425

And if the starting time is 14:00.000 time will be calculated as

14:00.000+.33700

00:14:00.337

00:14:00.689

00:14:00.996

00:14:01.298

00:14:01.616

00:14:01.981

00:14:02.386

00:14:02.806

Answer 1

[Jos (10584)]

% find MEDIAN over that time interval
MyMedian{i}(j)=median(data(strt:n));

[Tamilselvam Gunasekaran]

IT worked for me. It was trying just Median instead of MyMedian. Thank you very much