Linear Problem Matrix without For Loop
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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 个评论
Guillaume
2015-1-22
The number of columns is the max of K not the size of unique(K)
采纳的回答
Andreas
2015-1-22
4 个评论
Andreas
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.
Andreas
2015-1-22
更多回答（2 个）
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
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.
John D'Errico
2015-1-22
0 个投票
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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