matrix indexing and creating a new matrix with new size

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Dear experts, I hope somebody can help me. I have a n*4 matrix. In a loop I want to index some rows (the range is not fixed) from column 1 and put them in a new matrix with separate columns. I guess since the size of rows are not the same, Matlab gives me the mismatch dimension error for new matrix! Any solution? Thanks in advance Sobhan

Risposta accettata

Wayne King
Wayne King il 6 Giu 2012
How about
clear output
For f = 1:nfiles
IndexStart= RHs(RHs<peak);
IndexEnd= RHs(RHs>peak);
output{f} = Mydata (Index1:Index2,1);
end
If you know how big output is going to be in advance you can preallocate the cell array.
  1 Commento
Sobhan
Sobhan il 6 Giu 2012
thanks Wayne,
your solution only gives me a cell array with the size of output and not a matrix.
Cheers

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Più risposte (3)

Wayne King
Wayne King il 6 Giu 2012
How about using a cell array? Beyond that I think you will need to give us a little MATLAB example to illustrate your issue.
  1 Commento
Sobhan
Sobhan il 6 Giu 2012
Hi Wayne,
This is a simplified version of my loop.
For f = 1:nfiles
IndexStart= RHs(RHs<peak);
IndexEnd= RHs(RHs>peak);
output (:, f) = Mydata (Index1:Index2,1);
End
only the first file can be saved in output but it gives error for the second file because it has different rows. Is it clear now? Sorry I am really new to matlab.
Cheers
Sobahn

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Christoph
Christoph il 6 Giu 2012
Hi Sobhan,
as Wayne said, an example might be helpful. Nevertheless you might use:
X_new(:,end+1) = [X_old(:,i); zeros(size(X_new, 1) - size(X_old, 1),1)];
in your loop as long as the length of the old matrix is shorter or equal as the length of the new one. But waynes answer is the "politcally correct" one.
  2 Commenti
Sobhan
Sobhan il 6 Giu 2012
Hi,
Thanks for the answer. I gave an example above.
The length of the new matrix is sometimes bigger sometimes smaller. I think I need something to add a new column with the size of new file to previously created matrix. I am trying to figure it out.
Cheers
Christoph
Christoph il 6 Giu 2012
ok, try to implement the next few nasty codelines:
if length(X_new) < length(X_old)
X_new(end+1,:) = zeros(length(X_old)-length(X_new), size(X_new, 2))
end

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Sobhan
Sobhan il 8 Giu 2012
Cell array was the solution: output{f} = Mydata (Index1:Index2,1)

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