sparse vector divide by float gives full vector

[augustinyiptong]

Hi

I have:

julia> s = sparsevec([1.0, 2.0, 3.0])
Sparse vector of length 3, with 3 Float64 nonzero entries:
  [1]  =  1.0
  [2]  =  2.0
  [3]  =  3.0

I do:

julia> s * 0.5
Sparse vector of length 3, with 3 Float64 nonzero entries:
  [1]  =  0.5
  [2]  =  1.0
  [3]  =  1.5

So far so good. If i do:

julia> s/2.0
 3-element Array{Float64,1}:
 0.5
 1.0
 1.5

Why does it get converted to a full vector?

I am using Julia Version 0.5.0-dev+1158

Thanks, Gus

[nalimilan]

Looks like an oversight, as sparsity is preserved for sparse matrices.

@augustinyiptong If you call @which s.*1.0 and @less s.*1.0, you'll see that multiplication is defined very simply by calling scale. This can be a good workaround at least, multiplying by 1/a.

A better fix would be to add a function like this:

import Base.SparseArrays.nonzeroinds
./(x::AbstractSparseVector, a::Number) = SparseVector(length(x), copy(nonzeroinds(x)), nonzeros(x)./a)

[tkelman]

be sure to check for a zero (edit: or NaN) denominator.

[ViralBShah]

We can do the obvious fix for now - but I think sparse solvers can generally do better with sparse RHS. I'm not sure what the state of the art is on this question today though.

[tkelman]

Unless your matrix is diagonal, solving a linear system of equations doesn't generally preserve sparsity in the right hand side. Simplex is the usual poster child algorithm where sparse vectors play a crucial role.

This issue is just about division by a scalar though, so "solvers" aren't so relevant here.

[tkelman]

closed by #15579