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Working with a three dimensional matrix

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Pepa
Pepa on 24 Jan 2011
I have a matrix with three dimensions, and I need to take two dimensions through the use of cycle FOR.
For example:
Name variable Value
k <39x17x60>
And I need this:
a = 60;
for i = 1:60;
l = k...; %size of l <39x17>
xlswrite...;
end

Accepted Answer

Oleg Komarov
Oleg Komarov on 25 Jan 2011
The problem is how you construct kunn, in general there are so many nested loops that my suspect is you end up overwriting the data, go in debug mode and see what happens each time you increase the frist loop and so on:
function[F0mean,F0max,F0min,F0max_min,F0std,F0median,relF0VR,relF0SD,...
Emean,Emax,Emin,Emax_min,Estd,Emedian,TEO,Jmean,...
Jmax,Jmin,Jmax_min,Jstd,Jmedian,Smean,Smax,Smin,...
Smax_min,Sstd,Smedian,F1mean,F1max,F1min,F1max_min,F1std,...
F1median,F2mean,F2max,F2min,F2max_min,F2std,F2median]...
= HRA()
[wav,lab]=databaze;
wav = wav';
lab = lab';
L = length(wav);
for j = 2:L+1
brd = process_lab_file(lab{j-1});
label = brd(:,1);
K = length(label);
for i = 1:K
[F0mean,F0max,F0min,F0max_min,F0std,F0median,relF0VR,relF0SD,...
Emean,Emax, Emin,Emax_min, Estd, Emedian, TEO,...
Jmean,Jmax,Jmin,Jmax_min,Jstd,Jmedian,...
Smean,Smax,Smin,Smax_min,Sstd,Smedian,...
F1mean,F1max,F1min,F1max_min,F1std,F1median,...
F2mean,F2max,F2min,F2max_min,F2std,F2median]=extr_part_of_sig(wav{j-1},label(i),lab{j-1});
priznaky =...
[F0mean,F0max,F0min,F0max_min,F0std,F0median,relF0VR,relF0SD,...
Emean,Emax, Emin,Emax_min, Estd, Emedian, TEO,...
Jmean,Jmax,Jmin,Jmax_min,Jstd,Jmedian,...
Smean,Smax,Smin,Smax_min,Sstd,Smedian,...
F1mean,F1max,F1min,F1max_min,F1std,F1median,...
F2mean,F2max,F2min,F2max_min,F2std,F2median];
O = length(priznaky);
for o = 1:O
zap(o,i) = priznaky(o);
end
end
F = length(priznaky);
for i = 1:K
for a = 1:F
kunn(a,j-1,i) = zap(a);
end
end
end
S = length(label);
jaj = zeros(39,17);
for f = 1:S
jaj = kunn(:,:,2);
koj = jaj';
%xlswrite('ZK.xls', koj, label{i}, 'B2');
end
  1 Comment
Pepa
Pepa on 25 Jan 2011
In debug I found error. You have right. Matrix in kunn is same...

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More Answers (1)

Kenneth Eaton
Kenneth Eaton on 24 Jan 2011
If you're asking how you would index k in your loop to get a 2-D submatrix, you would do the following:
l = k(:,:,i);
  14 Comments
Pepa
Pepa on 25 Jan 2011
No i can't. isequal(kunn(:,:,1),kunn(:,:,2)) result is logical 1. And this is my problem.

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