# find the average of a vector with specific condition ?

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MUKESH KUMAR on 4 Sep 2018
Commented: MUKESH KUMAR on 4 Sep 2018
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 on 4 Sep 2018
Right, I figured that much :)
MUKESH KUMAR on 4 Sep 2018
So how can i take average of continuous values at starting point of each continuous group values and rest are zeros.
jonas on 4 Sep 2018
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.

jonas on 4 Sep 2018
Edited: jonas on 4 Sep 2018
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);

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Guillaume on 4 Sep 2018
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 on 4 Sep 2018
@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 on 4 Sep 2018