How do I assign a column of a matrix as another matrix?
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Hi, I want to assign columns from a matrix A(row,col), with an arbitrary number of rows and cols, as matrices themselves. I assign a cutoff value, x, with y > x = 0 and then I want to be discarding the zero elements such that the matrices shorten each time keeping only nonzero elements. It looks something like this:
[x x x x               [x [x [x
 x x x x                x  x  x
 x x x 0        --->    x  x  x] etc
 x x 0 0                x  x]   
 x 0 0 0                x]      
 0 0 0 0] etc
Is it possible to have the labeling scheme something like below where it will auto assign the string name, so that the matrices become A1, A2, A3 etc..., I'm not sure how to do this:
for k = 1:1:cols
    A'k'(:,k) = A(A>0)(:,k)
end
Thank you.
1 commentaire
Réponses (3)
  Guillaume
      
      
 le 28 Août 2015
        
      Modifié(e) : Guillaume
      
      
 le 28 Août 2015
  
      The proper way to do this is to store these new matrices into a cell array (certainly not to create numbered variable names using eval)
A simple one-liner does it:
z = [1 2 3; 4 5 6; 5 6 0; 7 0 0; 0 0 0]; %demo data
zcols = cellfun(@nonzeros, num2cell(z, 1), 'UniformOutput', false)
edit: typo pointed out by Star
1 commentaire
  Star Strider
      
      
 le 27 Août 2015
        
      Modifié(e) : Star Strider
      
      
 le 27 Août 2015
  
      There doesn’t seem to be an efficient, non-loop way to do this, so use a loop and a cell array to accommodate the different length vectors:
x = rand(5);
x = fliplr(triu(x));                % Create Matrix
for k1 = 1:size(x,2)
    xc{k1} = x(x(:,k1)>0,k1);       % ‘xc’ = Row Elements Of ‘x’>0
end
Here, ‘xc’ (‘x cell’) is the output.
2 commentaires
  Al Dente
      
 le 27 Août 2015
				you're looping over the rows and not the columns, if this was not a square matrix you would get an error if the number of rows is more than the number of columns
  Al Dente
      
 le 27 Août 2015
        if z was matrix that looks like this:
z = [1 2 3; 4 5 6; 5 6 0; 7 0 0; 0 0 0]
I could get all the columns (positive numbers only) in seperate arrays like this (can be better but if I'm in a hurry I wouldn't think too much about it):
% note that this is a terrible thing to do if you're creating a function -- I wouldn't use eval ever like this
for m = 1:size(z, 2)
    h = num2str(m); % m as a string
    eval(['col_' h '=z(z(:,' h ')>0, ' h ')'])
end
%this code will result in:
%col_1 =
%
%     1
%     4
%     5
%     7
%
%
%col_2 =
%
%     2
%     5
%     6
%
%
%col_3 =
%
%     3
%     6
if I want a better way to do it, I would go with a cell array of arrays:
for m = 1:size(z, 2)
    mycellarray{m} = z(z(:, m)>0, m);
end
2 commentaires
  Guillaume
      
      
 le 28 Août 2015
				This is not a good answer! This is actually poor practice, popping up new variables using eval, which is extremely discouraged. A simple search on this forum will give plenty of reasons why.
There are much cleaner options to achieve the same thing.
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