Add size(::LinearIndices) method (#25541)

Fixes indexing LinearIndices with an array. This is useful in particular
to convert cartesian indices that find() now returns to linear indices.

NEWS.md

@@ -371,7 +371,7 @@ This section lists changes that do not have deprecation warnings.
  `AbstractDict`, `AbstractString`, `Tuple` and `NamedTuple` objects ([#24774]).
  In particular, this means that it returns `CartesianIndex` objects for matrices
  and higher-dimensional arrays instead of linear indices as was previously the case.
- Use `Int[LinearIndices(size(a))[i] for i in find(f, a)]` to compute linear indices.
+ Use `LinearIndices(a)[find(f, a)]` to compute linear indices.
 
  * `AbstractSet` objects are now considered equal by `==` and `isequal` if all of their
  elements are equal ([#25368]). This has required changing the hashing algorithm

base/multidimensional.jl

@@ -418,6 +418,7 @@ module IteratorsMD
  # AbstractArray implementation
  Base.IndexStyle(::Type{LinearIndices{N,R}}) where {N,R} = IndexCartesian()
  Base.axes(iter::LinearIndices{N,R}) where {N,R} = iter.indices
+ Base.size(iter::LinearIndices{N,R}) where {N,R} = length.(iter.indices)
  @inline function Base.getindex(iter::LinearIndices{N,R}, I::Vararg{Int, N}) where {N,R}
  dims = length.(iter.indices)
  #without the inbounds, this is slower than Base._sub2ind(iter.indices, I...)

test/abstractarray.jl

@@ -119,6 +119,8 @@ end
  @test LinearIndices()[1] == 1
  @test_throws BoundsError LinearIndices()[2]
  @test LinearIndices()[1,1] == 1
+ @test LinearIndices()[] == 1
+ @test size(LinearIndices()) == ()
  @test CartesianIndices()[1] == CartesianIndex()
  @test_throws BoundsError CartesianIndices()[2]
  end
@@ -129,6 +131,8 @@ end
  @test CartesianIndices((3,))[i] == CartesianIndex(i,)
  end
  @test LinearIndices((3,))[2,1] == 2
+ @test LinearIndices((3,))[[1]] == [1]
+ @test size(LinearIndices((3,))) == (3,)
  @test_throws BoundsError CartesianIndices((3,))[2,2]
  # ambiguity btw cartesian indexing and linear indexing in 1d when
  # indices may be nontraditional
@@ -140,22 +144,39 @@ end
  k = 0
  cartesian = CartesianIndices((4,3))
  linear = LinearIndices(cartesian)
+ @test size(cartesian) == size(linear) == (4, 3)
  for j = 1:3, i = 1:4
- @test linear[i,j] == (k+=1)
+ k += 1
+ @test linear[i,j] == linear[k] == k
  @test cartesian[k] == CartesianIndex(i,j)
  @test LinearIndices(0:3,3:5)[i-1,j+2] == k
  @test CartesianIndices(0:3,3:5)[k] == CartesianIndex(i-1,j+2)
  end
+ @test linear[linear] == linear
+ @test linear[vec(linear)] == vec(linear)
+ @test linear[cartesian] == linear
+ @test linear[vec(cartesian)] == vec(linear)
+ @test cartesian[linear] == cartesian
+ @test cartesian[vec(linear)] == vec(cartesian)
+ @test cartesian[cartesian] == cartesian
+ @test cartesian[vec(cartesian)] == vec(cartesian)
  end
 
  @testset "3-dimensional" begin
  l = 0
  for k = 1:2, j = 1:3, i = 1:4
- @test LinearIndices((4,3,2))[i,j,k] == (l+=1)
+ l += 1
+ @test LinearIndices((4,3,2))[i,j,k] == l
+ @test LinearIndices((4,3,2))[l] == l
+ @test CartesianIndices((4,3,2))[i,j,k] == CartesianIndex(i,j,k)
  @test CartesianIndices((4,3,2))[l] == CartesianIndex(i,j,k)
  @test LinearIndices(1:4,1:3,1:2)[i,j,k] == l
+ @test LinearIndices(1:4,1:3,1:2)[l] == l
+ @test CartesianIndices(1:4,1:3,1:2)[i,j,k] == CartesianIndex(i,j,k)
  @test CartesianIndices(1:4,1:3,1:2)[l] == CartesianIndex(i,j,k)
  @test LinearIndices(0:3,3:5,-101:-100)[i-1,j+2,k-102] == l
+ @test LinearIndices(0:3,3:5,-101:-100)[l] == l
+ @test CartesianIndices(0:3,3:5,-101:-100)[i,j,k] == CartesianIndex(i-1, j+2, k-102)
  @test CartesianIndices(0:3,3:5,-101:-100)[l] == CartesianIndex(i-1, j+2, k-102)
  end