find the average of a vector with specific condition ?

I had A vector

   A=[1 0 0 2 3 2 3 5 0 0 0 0 1 2 1 0 0 0 0 3 4 0 0 0 0 0 0 1 2 3 2 0 0 0 0 2 0 0]

like this. Now I want a vector B having average of corresponding non-zero value, like

B(1)=average(A(1));
B(4)=average(A(4):A(8));
B(13)=average(A(13):A(15);
B(20)=average(A(20):A(21));
B(28)=average(A(28):A(31)) and so on .....rest values should be zeros.

hence

B=[1 0 0 3 0 0 0 0 0 0 0 0 1.33 0 0 0 0 0 0 3.5 0 0 0 0 0 0 0 2 0 0 0 0 0 0 0 2 0 0 ]

[jonas]

I've seen a number of your similar questions. Just out of curiosity, can I ask what the application is?

[MUKESH KUMAR]

I need this for some further more calculation in my problem

[jonas]

Right, I figured that much :)

[MUKESH KUMAR]

So how can i take average of continuous values at starting point of each continuous group values and rest are zeros.

[jonas]

I'll give it a try in an hour if no one gave you an answer. This one is not as straight-forward as your previous questions, at least to me.

Answer 1

[jonas]

Credits to Guillaume for cleaning up my original code.

A=[1 0 0 2 3 2 3 5 0 0 0 0 1 2 1 0 0 0 0 3 4 0 0 0 0 0 0 1 2 3 2 0 0 0 0 2 0 0];
B=zeros(size(A));
bounds = find(diff([0, A, 0] ~= 0));
starts = bounds(1:2:end);
ends = bounds(2:2:end) - 1;
B(starts) = arrayfun(@(s, e) mean(A(s:e)), starts, ends);

[MUKESH KUMAR]

error" Undefined function or variable 'id'." that should be P=id; its working and thanks for help and I also find an alternative of this

C=[1 0 0 5 0 0 0 0 0 0 0 0 3 0 0 0 0 0 0 2 0 0 0 0 0 0 0 4 0 0 0 0 0 0 0 1 0 0];

for i=1:38
  if C(i)==0
      AZ(i)=0;
  else
      BZ(i)=i+C(i)-1;
  B(i)= mean(A(1,i:BZ(i)));
  end
end

[Guillaume]

I would have

P = diff(A ~= 0)

so that it works with negative values as well. This is probably the most efficient solution. nanmean is not needed however. A slightly simpler version:

A = [1 0 0 2 3 2 3 5 0 0 0 0 1 2 1 0 0 0 0 3 4 0 0 0 0 0 0 1 2 3 2 0 0 0 0 2 0 0];
B = zeros(size(A));
bounds = find(diff([0, A, 0] ~= 0));
starts = bounds(1:2:end);
ends = bounds(2:2:end) - 1;
for i = 1:numel(starts)
  B(starts(i)) = mean(A(starts(i):ends(i)));
end

The loop can be replaced by the slightly slower but more compact:

B(starts) = arrayfun(@(s, e) mean(A(s:e)), starts, ends);

[jonas]

@Guillaume: Thank you sir, I'll update the code.

@Mukesh: The code should work now. There was a minor error in my original post. You were too quick to grab it :)

[MUKESH KUMAR]

thanks for your support :)