Replace values in sparse matrix

Question

I am interested in updating the values of the sparse matrix. I have a list of matrix element which I am interested in updating and a list of new values. Currently, I do it in the following manner:

s = SparseArray[{{i_, i_} -> i}, {100, 100}];
newP = RandomInteger[{1, 100}, {5, 2}];
newV = RandomReal[1, {5}];
MapThread[(s[[#1[[1]], #1[[2]]]] = #2) &, {newP, newV}]

Any suggestions on how to do it more efficiently?

In you actual use case: Are the new positions among the nonzero positions of the old array? — Henrik Schumacher, Aug 01 '18 at 09:15
Module[{nzp = Join[newP, s["NonzeroPositions"]], nzv = Join[newV, s["NonzeroValues"]]}, SparseArray[nzp->nzv, Dimensions[s]]] seems to be faster. — kglr, Aug 01 '18 at 09:23
Related, but not duplicate: (777) – that question deals with element-by-element updates, whereas this question permits bulk replacement. — Mr.Wizard, Aug 01 '18 at 10:27

kglr · Accepted Answer · 2018-08-01T12:09:52.257

7

s = SparseArray[{{i_, i_} -> i}, {10000, 10000}];

newP = RandomInteger[{1, 10000}, {4000, 2}];
newV = RandomReal[1, {4000}];

s1 = s;
MapThread[(s1[[#1[[1]], #1[[2]]]] = #2) &, {newP, newV}] // 
  RepeatedTiming // First

1.93

(s2 = With[{nzp = Join[newP, s["NonzeroPositions"]],
      nzv = Join[newV, s["NonzeroValues"]]}, 
     SparseArray[nzp -> nzv, Dimensions[s]]]) // 
  RepeatedTiming  // First

0.0019

(s4 = With[{mask = SparseArray[newP -> 1, Dimensions[s]], 
      ns = SparseArray[newP -> newV, Dimensions[s]]}, (1 - mask) s + 
      mask ns]) // RepeatedTiming // First

0.0012

s1 == s2 == s4

True

Versus Mr.Wizard's method:

RepeatedTiming[
    s3 = s +  SparseArray[newP -> newV - Extract[s, newP], Dimensions[s]];
  ] // First

0.0016

edited Aug 01 '18 at 12:09

answered Aug 01 '18 at 09:34

kglr

394,356
18
477
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Interesting; s4 is significantly slower than the others on my system, timing 0.00270. A variation of that was the first thing I tried. – Mr.Wizard Aug 01 '18 at 11:17
@Mr.Wizard, please note that i had changed 5000 to 4000 to avoid memory limits of free cloud plan. The timings i posted are obtained in Wolfram Cloud (v11.3). – kglr Aug 01 '18 at 11:35
@Mr.Wizard, in version 9 s4 is slower than both s2 and s3. – kglr Aug 01 '18 at 11:42

score 2 · Answer 2 · answered Aug 01 '18 at 10:52

Compared to kglr's answer this is marginally faster on my system (version 10.1), and a little simpler. It may be acceptable in many cases, however it will not work if you are trying to update Real to Integer values for example, because by way of numeric operations the Integers will be cast to Real.

s + SparseArray[newP -> newV - Extract[s, newP], Dimensions[s]]

Timings:

(* example code from kglr's answer *)

s = SparseArray[{{i_, i_} -> i}, {10000, 10000}];
newP = RandomInteger[{1, 10000}, {5000, 2}];
newV = RandomReal[1, {5000}];

(* his method again for comparative timing *)

(s2 = 
    Module[{nzp = Join[newP, s["NonzeroPositions"]], 
      nzv = Join[newV, s["NonzeroValues"]]}, 
     SparseArray[nzp -> nzv, Dimensions[s]]]) // RepeatedTiming // First

(* my method *)

RepeatedTiming[
  s3 = s + SparseArray[newP -> newV - Extract[s, newP], Dimensions[s]];
] // First

(* confirm equivalence *)

s2 == s3

0.00162
0.00145
True

Replace values in sparse matrix

2 Answers2