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From: Yi Cao on 17 Jul 2010 12:52
One of many solutions is as follows:

[testNumZeros testIndexZeros] = max(testdecisionZZ(:,1,:)); % Find max col 1
a = diag(squeeze(testdecisionZZ(testIndexZeros,3,:)));
b = squeeze(testIndexZeros);

HTH
Yi

"Mike Bourassa" <bourassa-m(a)rmc.ca> wrote in message <i1slf3$5c1$1(a)fred.mathworks.com>...
> Below are first the code, then results and then the test data:
>
> [testNumZeros testIndexZeros] = max(testdecisionZZ(:,1,:)); % Find max col 1
>
> a = []; % Initialize
> b = [];
>
> for i = 1:size(testdecisionZZ,3) % Loop through each 'slice'
> a = [ a ; testdecisionZZ(testIndexZeros(:,:,i),3,i)];
> b = [ b ; testIndexZeros(i) ];
> end
>
> Should get:
>
> a =
>
> 1.3039
> 1.2937
> 1.3048
>
> b =
>
> 12
> 12
> 12
>
> Test data:
>
> testdecisionZZ
>
> testdecisionZZ(:,:,1) =
>
> 0 21.0000 0.7197
> 0 12.0000 -0.0178
> 0 10.0000 -0.3366
> 0 8.0000 -0.2939
> 0 7.0000 -0.2981
> 2.0000 17.0000 0.2763
> 0 3.0000 -0.5141
> 0 11.0000 -0.1186
> 0 2.0000 -0.5535
> 1.0000 19.0000 0.4276
> 1.0000 5.0000 -0.3463
> 21.0000 22.0000 1.3039
> 0 2.0000 -0.5169
> 0 20.0000 0.8102
> 0 5.0000 -0.5279
> 1.0000 18.0000 0.4675
> 0 14.0000 0.0179
> 1.0000 15.0000 0.2388
> 0 6.0000 -0.5043
> 0 7.0000 -0.4205
> 1.0000 16.0000 0.2636
> 0 13.0000 -0.0769
>
>
> testdecisionZZ(:,:,2) =
>
> 0 21.0000 0.7552
> 0 12.0000 -0.0284
> 0 10.0000 -0.3379
> 0 8.0000 -0.2994
> 0 7.0000 -0.2981
> 2.0000 17.0000 0.3471
> 0 4.0000 -0.5106
> 0 11.0000 -0.1280
> 0 1.0000 -0.5547
> 1.0000 19.0000 0.4958
> 1.0000 5.0000 -0.3533
> 21.0000 22.0000 1.2937
> 0 2.0000 -0.5208
> 0 20.0000 0.6416
> 0 5.0000 -0.5299
> 1.0000 18.0000 0.4976
> 0 14.0000 0.0096
> 1.0000 16.0000 0.2938
> 0 6.0000 -0.5063
> 0 7.0000 -0.4234
> 1.0000 15.0000 0.2438
> 0 13.0000 -0.0872
>
>
> testdecisionZZ(:,:,3) =
>
> 0 21.0000 0.7051
> 0 12.0000 -0.0153
> 0 10.0000 -0.3350
> 0 8.0000 -0.2921
> 0 7.0000 -0.2967
> 2.0000 17.0000 0.2716
> 0 3.0000 -0.5134
> 0 11.0000 -0.1162
> 0 2.0000 -0.5524
> 1.0000 19.0000 0.4232
> 1.0000 5.0000 -0.3445
> 21.0000 22.0000 1.3048
> 0 2.0000 -0.5154
> 0 20.0000 0.8086
> 0 5.0000 -0.5266
> 1.0000 18.0000 0.4669
> 0 14.0000 0.0204
> 1.0000 15.0000 0.2356
> 0 6.0000 -0.5030
> 0 7.0000 -0.4190
> 1.0000 16.0000 0.2676
> 0 13.0000 -0.0743
From: Jan Simon on 17 Jul 2010 13:00
Dear Mike,

> [testNumZeros testIndexZeros] = max(testdecisionZZ(:,1,:));
>
> a = []; % Initialize
> b = [];
> for i = 1:size(testdecisionZZ,3) % Loop through each 'slice'
> a = [ a ; testdecisionZZ(testIndexZeros(:,:,i),3,i)];
> b = [ b ; testIndexZeros(i) ];
> end

Ok. This shows a main problem of your code: Initializing a and b to the empty matrix and letting them grow in each step is horribly slow. It is recommended (again and again in this newsgroup), to pre-allocate the output array:
n = size(testdecisionZZ,3);
a = zeros(1, n);
for i = 1:n % Loop through each 'slice'
a(i) = testdecisionZZ(testIndexZeros(:,:,i),3,i);
end
And as far as I can see,
b = testIndexZeros;
can be determined outside the loop.

Anyhow, do you want to calculate this (I currently have no access to Matlab on my internet computer):
n = size(ZZ, 3);
[Num, Index] = max(ZZ(:,1,:))
a = ZZ(sub2ind(size(ZZ), Index(:), [3;3;3], transpose(1:n));

Kind regards, Jan
From: Mike Bourassa on 17 Jul 2010 13:52
Very many thanks to all who replied. I learned a lot! Cheers.
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