I would like to prune some unimportant rows and columns from a SparseArray to accelerate solving some eigenvalue problems.


sparse =
 KroneckerProduct[RandomReal[{-10, 10}, {30, 30}], 
   IdentityMatrix[30, SparseArray]] + 
  KroneckerProduct[IdentityMatrix[30, SparseArray], 
   RandomReal[{-10, 10}, {30, 30}]]

I would like to drop a selection of rows and the paired up columns, say 600 of the 900:

toDrop = RandomSample[Range@900, 600];

Currently I'm using Drop as suggested in the documentation. Unfortunately this will only drop a span of rows and columns at once, so first I have to build an accumulated set of spans, which I've been doing like:

dropSpans =
    {#[[1]] + #2[[2]] - #2[[1]] + 1, #2 - #[[1]]} &,
    {0, None},
    MinMax /@ Split[Sort@toDrop, (#2 - #) == 1 &]
    ][[2 ;;, 2]];

Then I have to Fold the Drop like:

Fold[Drop[#, #2, #2] &, sparse, dropSpans];

Now this is fine for the 30x30 case:

Fold[Drop[#, #2, #2] &, sparse, dropSpans]; // RepeatedTiming // First


But the since matrix size increases as $N^4$ for me here even by the time I'm feed in 120x120 base matrices it takes a full 120 seconds as I'm building a new SparseArray object for every step in Fold--and the number of steps also increases as $N^2$. If I could just drop all of the parts at once I assume this would be quite fast. I'm also hoping not to have to work with the internal SparseArray structure but that's fine if there's no other way.

  • 1
    $\begingroup$ I would use Part to extract the nondropped rows with the Complement of dropped indices as second argument. Delete might also work (with the full index set expanded; not with Spans). Quite likely, the first is what the latter does internally. So better try both. $\endgroup$ May 13, 2018 at 2:08
  • $\begingroup$ @HenrikSchumacher unfortunately Delete appears to convert the array to dense array. I had assumed Part would be less efficient since in some sense there is more work being done with that, but I suppose the construction overhead of this method will overwhelm that. $\endgroup$
    – b3m2a1
    May 13, 2018 at 2:12
  • $\begingroup$ Hm. I cannot confirm that, but anyways, A[[Complement[Range[Length[A]], idx]]] seems to be faster (A ist the matrix, idx the index set of vertices to delete). $\endgroup$ May 13, 2018 at 2:19
  • $\begingroup$ @HenrikSchumacher yeah using Part for both the rows and the columns simultaneously is actually quite fast. Like .01s to update a 14400x14400 matrix. $\endgroup$
    – b3m2a1
    May 13, 2018 at 2:20

1 Answer 1


My suggestion is

With[{toKeep = Complement[Range[Length[sparse]], toDrop]},
  sparse[[toKeep, toKeep]]
  • $\begingroup$ Works wonderfully. $\endgroup$
    – b3m2a1
    May 13, 2018 at 20:51

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