Avoiding for loop with ismember

Hi, suppose N is a scalar, and c is a numerical vector with unique elements that are a subset of 1:N, i.e. its size is smaller or equal to N. For example, N=4, and c=[1 2 4]'; Then, is there a more elegant way to achieve this?

    cc = [];
for j = 1:N
    if ismember(j,c)
        cc      = [cc; N*(j-1)+c];
    end
end

Thanks!

[Stephen23]

What do you expect the output cc to be?

[Christian]

If N=4, and c=[1 2 4]'; then cc is going to be [1 2 4 5 6 8 13 14 16]'.

Answer 1

[Andrei Bobrov]

[Fixed]

N = 15;
c = randperm(N,4);

c1  = sort(c(:));
cc = reshape(bsxfun(@plus,N*(c1 - 1),c1')',[],1);

or

oc =  ones(numel(c1),1);
cc = N*(kron(c1,oc) - 1) + kron(oc,c1);

or

c = c(:);
nn = numel(c);
cc = N*(repelem(c,nn) - 1) + repmat(c,nn,1);

[Christian]

Thanks, but cc should be a column vector of size length(c)^2. You seem to be getting a matrix, and the elements are not right either. c is a column vector.

[Andrei Bobrov]

Vector?

>> N=4;
c=[1 2 4];
cc = [];
for j = 1:N
    if ismember(j,c)
        cc      = [cc; N*(j-1)+c];
    end
end
>> cc
cc =
     1     2     4
     5     6     8
    13    14    16
>>

[Christian]

c is a column vector, not a row vector. Then, cc will also be a column vector. And c is sorted to begin with.

[Andrei Bobrov]

I'm fixed my answer.

[Stephen23]

@Christian: With the data you have shown us cc will definitely be a matrix. If c is a column vector then you should specify this in your question.

[Christian]

I think that gives the right result. I wish it was more readable / elegant.