Linear Problem Matrix without For Loop

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Andreas
Andreas 2015-1-22
评论： Andreas 2015-1-22
Hello all, I have a problem. Suppose I have a matrix like this:
K= [1 2 4 3 4 6 ]
And I want to end with a matrix like this
L=[1 0 0 0 0 ; 0 1 0 0 0 ; 0 0 0 1 0 ; 0 0 1 0 0 ; 0 0 0 1 0 ; 0 0 0 0 1]
As you can see the number of columns is the size of unique(K) and row is the length(K) How can I create a matrix like L WITHOUT using for.? I tried ind2sub but I failed.* * * * Maybe a solution could be to have a matrix like this
U= [1 2 4 3 4 5 ; 1 2 4 3 4 6]
where the first row is the index. Thank you all!!

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Andreas
Andreas 2015-1-22
Hello all.! thank you for your answers!! I think I found it,with your help.
K = [1 2 4 3 4 6 ];
[~,n]=ismember((K),unique(K));
L=zeros(length(K),length(unique(K)))
L(sub2ind(size(L), 1:length(K(1,1:end)),n))= 1
if there's something better I'll be glad to see it.! Thank you all!
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Guillaume
Guillaume 2015-1-22
编辑：Guillaume 2015-1-22
Well, I'd assume 6 means 6th column, 10 mean 10th column, etc.
If the numbers don't mean anything other than specifying an ordering, then do:
[~, ~, K] = unique(K);
before any of my answers.
K = 'ACPIPZ'; %you can even use letters since the numbers have no meaning.
%challenge: find an English word that gives the same result.
[~, ~, K] = unique(K);
L = full(sparse(1:numel(K), K, 1))
is still a lot simpler.

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Guillaume
Guillaume 2015-1-22
编辑：Guillaume 2015-1-22
K = [1 2 4 3 4 6 ];
L = zeros(numel(K), max(K));
L(sub2ind(size(L), 1:numel(K), K)) = 1
or using sparse matrices:
L = full(sparse(1:numel(K), K, 1))
or using accumarray:
L = accumarray([1:numel(K); K]', ones(numel(K), 1))
  1 个评论
John D'Errico
John D'Errico 2015-1-22
Those would have been my solutions, preferring either sparse or accumarray, depending on whether the result is desired to be full or sparse in the end.

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John D'Errico
John D'Errico 2015-1-22
Time to learn how to use sparse.
Or, if you prefer to work with full matrices, learn to use accumarray.
Or, if those options are not to your liking, learn to use use subsindx. That will take slightly more effort though.

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