not to use loops

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osman
osman 2011-10-4
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
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osman
osman 2011-10-4
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
Daniel Shub 2011-10-4
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.

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Andrei Bobrov
Andrei Bobrov 2011-10-4
[zout,idx] = max(z);
[m n] = size(x)
ind = sub2ind([m n],idx,1:n)
xout = x(ind)
yout = y(ind)
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Matt Tearle
Matt Tearle 2011-10-4
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
osman 2011-10-5
ok i misunderstood the output. it fits the problem. i gonna accept the answer

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