Matrix linear indexing: how to add more than once to the same index

Here is my question. Let's imagine we have this matrix:

A = [1 2 ; 3 4];

So if I use linear indexing to add 1 to, for instance, positions 1 and 3:

A([1 3]) = A([1 3]) + 1

A =

       2     3
       3     4

Now, let's imagine that, having the initial A matrix, I have an index vector in which some indices are repeated. For example, I need to add twice 1 to the first position, and once to the index 3:

A([1 3 1])=A([1 3 1])+1

A =

       2     3
       3     4

And surprisingly I get A(1,1)=2, instead of A(1,1)=3. How can I make it work?

The first idea was to put it inside a loop, but the matrix is normally big and this operation runs many times, and it becomes very slow. Any ideas?

Thank you very much in advance!

[per isakson]

"And surprisingly I get A(1,1)=2, instead of A(1,1)=3." &nbsp Matlab obviously doesn't work the way you think it should. I cannot find any support for this behavior in the documentation. Neither that Matlab should work this way, nor should not. In cases like this Matlab rules ;-)

>> B = 1
B =
     1
>> B([1,1,1]) = B([1,1,1])+1
B =
     2

[Guillaume]

The behaviour is not surprising at all and totally expected.

With

A = [10 20; 30 40];

A([1 3 1]) is the content of indices 1, 3 and 1, so: [10 30 10]. Therefore A([1 3 1]) + 1 is the vector [11 21 11].

You then ask matlab to assign that vector to A([1 3 1]), so A ends up being

[11 21; 30 40]

Answer 1

[Guillaume]

Another option, possibly faster than a loop or unique + histc:

A = [10 20;30 40];
idx = [1 3 1];

A(:) = A(:) + accumarray(idx(:), 1, [numel(A), 1]);

[dpb]

Yeah, but I thought it "cute"... :)

I always have a heckuva' time coming up w/ "the right stuff" with accumarray...

[Pablo García Auñón]

Thanks Guillaume! I speeded it up 40 times =D

Answer 2

[dpb]

>> ix=[1 3 1];
>> A(unique(ix))=A(unique(ix))+histc(ix,unique(ix))
A =
     3     3
     3     4
>>

Keep a temp for the unique index to eliminate the duplication; not sure if the JIT optimizer will find it or not if don't...

Answer 3

[Azzi Abdelmalek]

A = [1 2 ; 3 4]
idx=[1 3 1]
for k=1:numel(idx)
  A(idx(k))=A(idx(k))+1
end