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fix interval selection when keys are intervals #120

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fix interval selection when keys are intervals #120

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7 changes: 7 additions & 0 deletions src/selectors.jl
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Original file line number Diff line number Diff line change
Expand Up @@ -12,6 +12,13 @@ findindex(int::Interval, r::AbstractVector) =
findindex(int::Interval, r::AbstractRange{T}) where {T<:Union{Number,Char}} =
findall(in(int), r)

# find interval in a vector of intervals: same as generic findindex in lookup.jl
function findindex(int::Interval, r::AbstractVector{<:Interval})
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Hmm, I suppose this is a bit ambiguous as you could also want to find the first occurrence in which int overlaps or intersects with one of the intervals in r. I suppose that's what the functional interface is for though? @mcabbott do you have a preference?

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@aplavin aplavin Sep 1, 2022 •
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My motivation was that keeping invariants goes a long way. And "calling with a value of type eltype(axiskeys(A)) returns the array value at that specific axiskey value" is a nice invariant to follow whenever possible. Overlaps or whatever else can be done with separate selectors, potentially as simple as !isdisjoint(int)/⊆(int)/....

i = findfirst(isequal(int), r)
i === nothing && throw(ArgumentError("could not find key $(repr(a)) in vector $r"))
i
end

# Since that is now efficient for ranges, comparisons can go there:

findindex(eq::Base.Fix2{typeof(<=)}, r::AbstractRange{T}) where {T<:Union{Number,Char}} =
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4 changes: 4 additions & 0 deletions test/_basic.jl
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Expand Up @@ -77,6 +77,10 @@ end
@test V(Near(0.12)) == V(0.1) == V[2]
@test V(Interval(0.1, 0.3)) == V[2:4]

VI = KeyedArray([1, 2, 3], xs=[Interval(1, 2), Interval(2, 3), Interval(1, 3)])
@test VI(Interval(1, 3)) == VI[3]
@test VI([Interval(1, 2), Interval(1, 3)]) == VI[[1, 3]]

@test V(Index[1]) == V[1]
@test V(Index[2:3]) == V[2:3]
@test V(Index[end]) == V[end]
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