select columns from one matrix according to values in a second matrix

Hi, I'm trying to teach myself matlab by converting some code I originally wrote in R. The idea is to select columns from one matrix using values stored in another matrix and average them together.

I think the code below should work but I'm not sure how to handle the NaN values. In R I could use a which() statement but I'm not sure how to address this in matlab.

Any advice (or more elegant solutions)?

Thanks tom

x=magic(4)

y=[1 NaN NaN;2 3 4; 5 NaN NaN] %artificial example but rows need to be different lengths

out=zeros(size(x,1),size(y,1))

for i=1:size(y,1)
    out(:,i)=mean(x(:,y(i,:)))

[Sean de Wolski]

Perhaps you could show the expected result for the above, since the code probably does not do what you want.

Answer 1

[Oleg Komarov]

That's the only way the averaging is consistent with the output matrix:

x = magic(5);

y = [1 NaN NaN
     2   3   4
     5 NaN NaN];

szY = size(y);
out = zeros(size(x,1),szY(1));

for i = 1:szY(1)
    idx = y(i,:);
    out(:,i) = mean(x(:,idx(~isnan(idx))),2);
end

EDIT

So, there's the accumarray alternative but also the arrayfun which is basically a hidden loop:

out2 = arrayfun(@(z) mean(x(:,y(z,~isnan(y(z,:)))),2), 1:size(y,1),'un',0);
[out2{:}]

[Tom Wright]

The use of the idx variable certainly make this more readable, I was wondering if there was a function similar to the apply() function in R that performs a function on all rows (or columns) of a matrix without using the for loop?

[Sean de Wolski]

Look at the dimensional argument to mean (the 2 at the end).

[Oleg Komarov]

@Tom: mean directly cannot be applied since you change the columns you select, one option is accumarray but you will need to build subs and vals which is not the most trivial thing here, considering that the loop would be IMHO the best solution in terms of performance (speed and memory) and readability.

Answer 2

[Sean de Wolski]

doc isnan

Might be of some use.

[Tom Wright]

y(i,(not(isnan(y(i,:)))))

may be of some use, it's a supremely ugly statement however.

[Sean de Wolski]

so

mean(x(:,y(i,~isnan(y(i,:))))

?

[Sean de Wolski]

We'll be able to see the operation more easily if you show us the expected result and logic.