Mean value of data with irregular intervals

Hello,

I am trying to obtain average y value for a data set (x,y). However, interval between data points is not regular and mean(y) is off from the expected value. For example,

x = [1 1.1 1.2 1.3 2 3 4 5.1 5.2 5.3 5.4]; y = [0 0.1 0.2 0.3 1 2 1 0.3 0.2 0.1 0];

Because x,y are more densely lying in x = [0 1] and [5 6] intervals, y values in those intervals are weighted more if I take mean(y). Do you know how to avoid this and obtain y value in uniform intervals?

Thanks, Bobo

Answer 1

[Iain]

Why not just take a weighted mean?

This method ignores y(1) - you could change it to ignore y(end), or add info and inclue it.

 totalweight = 0;
 weightedsum = 0;
 for i = 2:numel(x)
  weight = x(i)-x(i-1);
  weightedsum = weightedsum + weight*y(i);
  totalweight = totalweight + weight;
 end

 meanvalue = weightedsum / totalweight;

[dpb]

d=diff(xx);
m=sum(d.*yy(2:end))/sum(d);

which can also be written as

m=dot(diff(xx),yy(2:end))/sum(d);

Answer 2

[dpb]

One way to weight uniformly...

dx=min(diff(x));          % shortest difference in vector
xi=x(1):dx:x(end);        % fill in all values equally
yhat=interp1(x',y',x1');  % and get corresponding y's...
mean(yhat);               % obviously can combine steps ...

[bobo]

Thanks!!

[Andrew Sandeman]

In my opinion this is the correct answer, not the currently accepted answer.

• This method does not neglect y(1)
• Samples which are obtained more sparsely are weighted more accurately using this interpolation method