Linear Problem Matrix without For Loop

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!!

1 Commento

The number of columns is the max of K not the size of unique(K)

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 Risposta accettata

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!

4 Commenti

Guillaume
Guillaume il 22 Gen 2015
Modificato: Guillaume il 22 Gen 2015
That looks... overly complicated.
avoid using length, numel is better.
Your answer has a problem,there's a fifth column filled with zeros that is unecessary.Also,if there is a vector like this : K= [10 20 40 30 40 60], then there's another problem.! Thank you about numel!
Guillaume
Guillaume il 22 Gen 2015
Modificato: Guillaume il 22 Gen 2015
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.
amazing.!thank you very much!!

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Più risposte (2)

Guillaume
Guillaume il 22 Gen 2015
Modificato: Guillaume il 22 Gen 2015
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 Commento

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 il 22 Gen 2015
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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