Working with a three dimensional matrix

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Pepa
Pepa el 24 de En. de 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

Respuesta aceptada

Oleg Komarov
Oleg Komarov el 25 de En. de 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 comentario
Pepa
Pepa el 25 de En. de 2011
In debug I found error. You have right. Matrix in kunn is same...

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Más respuestas (1)

Kenneth Eaton
Kenneth Eaton el 24 de En. de 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 comentarios
Oleg Komarov
Oleg Komarov el 25 de En. de 2011
isequal(kunn(:,:,1),kunn(:,:,2)) should be 0, can you confirm?
Pepa
Pepa el 25 de En. de 2011
No i can't. isequal(kunn(:,:,1),kunn(:,:,2)) result is logical 1. And this is my problem.

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