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

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Walter Roberson
Walter Roberson am 8 Feb. 2018

0 Stimmen

B = full( sparse(row, col, value) );
or
B = accumarray([row(:), col(:)], value);

2 Kommentare

GEORGIOS BEKAS
GEORGIOS BEKAS am 8 Feb. 2018
how about using arrayfun/bsxfun or something?
Walter Roberson
Walter Roberson am 8 Feb. 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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Weitere Antworten (1)

Birdman
Birdman am 8 Feb. 2018

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B(sub2ind([max(row) max(col)],row,col)) = value

8 Kommentare

GEORGIOS BEKAS
GEORGIOS BEKAS am 8 Feb. 2018
as usual; can you explain that?
Birdman
Birdman am 8 Feb. 2018
Bearbeitet: Birdman am 8 Feb. 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 am 8 Feb. 2018
It does not solve the problem. I want B(row(i), col(i)) = value, without using a for loop.
Birdman
Birdman am 8 Feb. 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 am 8 Feb. 2018
The reshape() is not needed.
Birdman
Birdman am 8 Feb. 2018
Doesn't my answer also work Walter?
GEORGIOS BEKAS
GEORGIOS BEKAS am 8 Feb. 2018
it did not :/

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