What wrong with the indexing code?

If I want to fill in a matrix, of which I have its indices without a for loop, how could I do it? Suppose that the (row(i),col(i)) elements to be filled are given.
value = 189;
row = [ 1 3 6 7 8 ];
col = [ 1 1 4 4 5];
B(row,col) = value

 Respuesta aceptada

Walter Roberson
Walter Roberson el 8 de Feb. de 2018
B = full( sparse(row, col, value) );
or
B = accumarray([row(:), col(:)], value);

2 comentarios

GEORGIOS BEKAS
GEORGIOS BEKAS el 8 de Feb. de 2018
how about using arrayfun/bsxfun or something?
Walter Roberson
Walter Roberson el 8 de Feb. de 2018
Not at all easily with arrayfun or bsxfun as assignment through function calls is difficult and never less than ugly.

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Más respuestas (1)

Birdman
Birdman el 8 de Feb. de 2018
B(sub2ind([max(row) max(col)],row,col)) = value

8 comentarios

GEORGIOS BEKAS
GEORGIOS BEKAS el 8 de Feb. de 2018
as usual; can you explain that?
Birdman
Birdman el 8 de Feb. de 2018
Editada: Birdman el 8 de Feb. de 2018
sub2ind function converts the equivalent indexes of given rows and columns and it should also know the size of your matrix. The inputs are accordingly given. For instance, the linear index representation for a matrix having 2 rows and 5 columns, the index of the element at
row=2;
col=3;
is
6
GEORGIOS BEKAS
GEORGIOS BEKAS el 8 de Feb. de 2018
It does not solve the problem. I want B(row(i), col(i)) = value, without using a for loop.
Birdman
Birdman el 8 de Feb. de 2018
I did not use for loop here,as you can see.
sz = [max(row), max(col)]
B(sub2ind(sz, row, col)) = value
B = reshape(B, sz)
Walter Roberson
Walter Roberson el 8 de Feb. de 2018
The reshape() is not needed.
Birdman
Birdman el 8 de Feb. de 2018
Doesn't my answer also work Walter?
GEORGIOS BEKAS
GEORGIOS BEKAS el 8 de Feb. de 2018
it did not :/

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