Calculation of conditional expectations

Suppose I have a vector A and I would like to calculate the sum of the elements in A divided by the number of positive elements of A. Please advise.

[Greg]

Sounds like the mean of A after you set all negative values of A to 0. What have you tried, and where are you stuck?

[Walter Roberson]

No, if you set the negative values to 0 then the count of those 0 get included in the mean. Furthermore, the negative values are to be included in the sum even if the count is only to be for the positive values.

[Walter Roberson]

http://www.mathworks.com/matlabcentral/answers/6200-tutorial-how-to-ask-a-question-on-answers-and-get-a-fast-answer

[alpedhuez]

sum(A(:,1))./sum(A(:,1)~=0);

[Walter Roberson]

That code assumes you have a row vector.

That code counts the number of non-zeros in the vector, instead of counting the number of positive values.

[Greg]

Touche, I wasn't thinking very hard. Was trying to spark some thinking in the poster, rather than hand out an answer.

And unless I'm really out of it, A(:,1) is a column vector.

Answer 1

[Stephen23]

sum(A)./nnz(A>=0)

[Greg]

Is nnz more efficient than sum? Since A >= 0 returns logicals, sum should be same as nnz, correct?

[Walter Roberson]

Is nnz more efficient than sum?

In my timing tests, Yes.

[Greg]

Fascinating. I'll have to start implementing where I can. Normally, I just skip those ones since the difference in typical cases is going to be negligible.