3
$\begingroup$

I would like to know if there is a way for increasing the speed of the following computation: I have a list, let's call it list, and another list of values, val. I wish to create a sparse matrix in the following way: The only nonzeros elements of the matrix are the positions of list (2 components) where the value is val[[i]]. I have generate this code: Mathematica

list = {{1, 3}, {1, 4}, {5, 5}, {5, 6}, {6, 2}, {6, 3}, {6, 4}, {6, 5}, {6, 6}, {7, 3}, {7, 4}, {7, 5}, {7, 6}, {8, 3}, {8, 5}, {8, 6}};

val = Table[i^2, {i, Length[list]}];

matrix = Table[0., {i, 13}, {j, 13}];(* 13 is just an example *)

For[i=1,i<= Length[list], i++, matrix[[list[[i]][[1]], list[[i]][[2]] ]] = val[[i]]];

Do you know a way for improving the speed? I ask this because if the length of list is long the loop requires much time to do.

$\endgroup$

2 Answers 2

5
$\begingroup$
sa = SparseArray[list -> val]

enter image description here

TeXForm @ MatrixForm @ sa

$\left( \begin{array}{cccccc} 0 & 0 & 1 & 4 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 9 & 16 \\ 0 & 25 & 36 & 49 & 64 & 81 \\ 0 & 0 & 100 & 121 & 144 & 169 \\ 0 & 0 & 196 & 0 & 225 & 256 \\ \end{array} \right)$

Use the second argument to specify the desired dimensions: For example, to get a sparse matrix with 10 rows and 9 columns, use

sa2 = SparseArray[list -> val, {10, 9}];

TeXForm @ MatrixForm @ sa2

$\left( \begin{array}{ccccccccc} 0 & 0 & 1 & 4 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 9 & 16 & 0 & 0 & 0 \\ 0 & 25 & 36 & 49 & 64 & 81 & 0 & 0 & 0 \\ 0 & 0 & 100 & 121 & 144 & 169 & 0 & 0 & 0 \\ 0 & 0 & 196 & 0 & 225 & 256 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ \end{array} \right)$

$\endgroup$
3
  • $\begingroup$ Wow! Many thanks, kglr. One more question: What about if val is a matrix instead a list, call it M, and I want to assign matrix[[list[[i]][[1]], list[[i]][[2]] ]] = M[[list[[i]][[1]], list[[i]][[2]] ]] ? $\endgroup$
    – Benigno
    Commented Apr 24, 2023 at 14:59
  • $\begingroup$ @Benigno, use SparseArray[list -> Extract[M, list]]? $\endgroup$
    – kglr
    Commented Apr 24, 2023 at 15:06
  • $\begingroup$ I think we can close this question with the two great answers of kglr. $\endgroup$
    – Benigno
    Commented Apr 24, 2023 at 15:16
2
$\begingroup$
m = SparseArray[MapIndexed[# -> #2[[1]]^2 &, list], {13, 13}];

m // MatrixForm

$$ \left( \begin{array}{ccccccccccccc} 0 & 0 & 1 & 4 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 9 & 16 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 25 & 36 & 49 & 64 & 81 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 100 & 121 & 144 & 169 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 196 & 0 & 225 & 256 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ \end{array} \right) $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.