not to use loops

x=[2,3,4,5;1,3,2,5;8,7,6,5];
y=[-1,2,3,6;8,6,7,5;10,11,12,18];
z=[1,2,3,4;5,18,7,8;12,16,35,-8];
[yuk,idx]=max(z) %as andrei says (http://www.mathworks.com/matlabcentral/answers/17409-without-using-loops)
here z is a function of x and y and the question is  to find which corresponds to the max z in each column in the matrices x and y

in short i need to find elements of matrices in each column using idx without using loops For example in the 1. colum max z is 12 and find x is 8 y is 10

[Daniel Shub]

Over the past 10 years or so the JIT accelerator has substantially sped up loops in MATLAB. While there are lots of cases where vectorized code is still faster, if you can solve your problems with a loop, there may be no reason to try and do it without a loop (i.e., the improvement in performance may not be worth it).

[osman]

but my advisor didnt accept this solution with 2 loops. our simulation take about 2 day and we need to find a way.

[Daniel Shub]

In some ways the loops are perfect. It means you can just borrow a bunch of computers one night and your simulation will be done. Way better than wasting your time trying to optimize code.

Answer 1

[Andrei Bobrov]

[zout,idx] = max(z);
[m n] = size(x) 
ind = sub2ind([m n],idx,1:n) 
xout = x(ind) 
yout = y(ind)

[osman]

it doesn't respond correct every time. false in id of maximum numbers.

maximum numbers id can be randomly disturbuted

[Matt Tearle]

I was about to say

[m,n] = size(z);

[~,idx] = max(z);

idx = idx + (0:(n-1))*m

x(idx)

y(idx)

but that's the same as Andrei's. What do you mean by "false in id of maximum numbers"? Compare zout and z(ind) -- they're the same. Try it with random x, y, z -- it's doing what you appeared to ask: find the maximum in each column of z, then return the x and y values from the corresponding locations.

[osman]

ok i misunderstood the output. it fits the problem. i gonna accept the answer