Convert integer to logical array in MATLAB

Question

I want to convert an integer i to a logical vector with an i-th non-zero element. That can de done with 1:10 == 2, which returns

0     1     0     0     0     0     0     0     0     0

Now, I want to vectorize this process for each row. Writing repmat(1:10, 2, 1) == [2 5]' I expect to get

0     1     0     0     0     0     0     0     0     0
0     0     0     0     1     0     0     0     0     0

But instead, this error occurs:

Error using ==
Matrix dimensions must agree.

Can I vectorize this process, or is a for loop the only option?

@johnny5 It's not, but it's very ill-formed and unclear. The answer by beaker below clarifies what OP wants, and it's even on-topic. — Andras Deak -- Слава Україні
– Andras Deak -- Слава Україні, Commented Jan 31, 2016 at 16:00
In general, be careful about expecting 1s and 0s to be treated as logical. Consider: >> foo = 1:10; >> foo([1,0,1,0,1,0,1,0,1,0]) Subscript indices must either be real positive integers or logicals. >> foo(logical([1,0,1,0,1,0,1,0,1,0])) ans = 1 3 5 7 9 — Carl Witthoft
– Carl Witthoft, Commented Feb 1, 2016 at 14:17

beaker · Accepted Answer · 2016-01-31 15:58:28Z

9

You can use bsxfun:

>> bsxfun(@eq, 1:10, [2 5].')
ans =

   0   1   0   0   0   0   0   0   0   0
   0   0   0   0   1   0   0   0   0   0

Note the transpose .' on the second vector; it's important.

answered Jan 31, 2016 at 15:58

beaker

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4 Comments

Andras Deak -- Слава Україні Over a year ago

Mentalist badge, +1 ;)

beaker Over a year ago

@AndrasDeak I've been working out with my Jedi Force Trainer.

ederag Over a year ago

For people using octave, broadcasting allows to write simply (1:10) == [2 5].'

beaker Over a year ago

@ederag Very true, and I actually generated the sample above on Octave. But I didn't want to confuse things as the question is only tagged MATLAB and not Octave.

rayryeng · Accepted Answer · 2016-01-31 16:32:28Z

4

Another way is to use eye and create a logical matrix that is n x n long, then use the indices to index into the rows of this matrix:

n = 10;
ind = [2 5];

E = eye(n,n) == 1;
out = E(ind, :);

We get:

>> out

out =

     0     1     0     0     0     0     0     0     0     0
     0     0     0     0     1     0     0     0     0     0

answered Jan 31, 2016 at 16:32

rayryeng

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Comments

Daniel · Accepted Answer · 2016-01-31 16:48:26Z

2

Just another possibility using indexing:

n = 10;
ind = [2 5];
x=zeros(numel(ind),n);
x(sub2ind([numel(ind),n],1:numel(ind),ind))=1;

answered Jan 31, 2016 at 16:48

Daniel

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