Fix view of VectorOfArray with heterogeneous array sizes

Fixes #453

The issue was that `view(f, :, i)` for a VectorOfArray with
heterogeneous inner array sizes returned incorrect results. The
problem was that `to_indices` uses `axes`, which is based on the
first element's size, causing views of columns with different sizes
to have incorrect dimensions.

Solution: Added special handling for `view(A, :, i)` where `i` is
an Int, to use the actual size of the specific column being viewed
rather than relying on `to_indices`.

Changes:
- Modified Base.view for AbstractVectorOfArray to handle heterogeneous
  arrays by computing the correct indices based on the actual column size
- Added comprehensive tests for views of heterogeneous arrays including:
  * Testing correct lengths of views
  * Testing that views match regular indexing
  * Testing that views can be modified

All existing tests pass.

🤖 Generated with [Claude Code](https://claude.com/claude-code)

Co-Authored-By: Claude <[email protected]>

Author: ChrisRackauckas

src/vector_of_array.jl

@@ -675,6 +675,15 @@ function Base.view(A::AbstractVectorOfArray{T, N, <:AbstractVector{T}},
 end
 function Base.view(A::AbstractVectorOfArray, I::Vararg{Any, M}) where {M}
     @inline
+    # Special handling for heterogeneous arrays when viewing a single column
+    # The issue is that to_indices uses axes, which is based on the first element's size
+    # For heterogeneous arrays, we need to use the actual size of the specific column
+    if length(I) == 2 && I[1] == Colon() && I[2] isa Int
+        @boundscheck checkbounds(A.u, I[2])
+        # Use the actual size of the specific column instead of relying on axes/to_indices
+        J = (Base.OneTo(length(A.u[I[2]])), I[2])
+        return SubArray(A, J)
+    end
     J = map(i -> Base.unalias(A, i), to_indices(A, I))
     @boundscheck checkbounds(A, J...)
     SubArray(A, J)

test/basic_indexing.jl

@@ -145,6 +145,31 @@ diffeq = DiffEqArray(recs, t)
 @test diffeq[:, 1] == recs[1]
 @test diffeq[1:2, 1:2] == [1 3; 2 5]
 
+# Test views of heterogeneous arrays (issue #453)
+f = VectorOfArray([[1.0], [2.0, 3.0]])
+@test length(view(f, :, 1)) == 1
+@test length(view(f, :, 2)) == 2
+@test view(f, :, 1) == [1.0]
+@test view(f, :, 2) == [2.0, 3.0]
+@test collect(view(f, :, 1)) == f[:, 1]
+@test collect(view(f, :, 2)) == f[:, 2]
+
+f2 = VectorOfArray([[1.0, 2.0], [3.0]])
+@test length(view(f2, :, 1)) == 2
+@test length(view(f2, :, 2)) == 1
+@test view(f2, :, 1) == [1.0, 2.0]
+@test view(f2, :, 2) == [3.0]
+@test collect(view(f2, :, 1)) == f2[:, 1]
+@test collect(view(f2, :, 2)) == f2[:, 2]
+
+# Test that views can be modified
+f3 = VectorOfArray([[1.0, 2.0], [3.0, 4.0, 5.0]])
+v = view(f3, :, 2)
+@test length(v) == 3
+v[1] = 10.0
+@test f3[1, 2] == 10.0
+@test f3.u[2][1] == 10.0
+
 t = 1:5
 recs = [rand(2, 2) for i in 1:5]
 testva = VectorOfArray(recs)