syntax-error - 为什么我不能在命名元组中将名称的类型指定为 Julia 中的符号元组？

Question

有这个功能:

function Γ(state::Dict{Symbol, <:Tuple{Vararg{Symbol}}})::Vector{NamedTuple{<:Tuple{Vararg{Symbol}}, <:Tuple{Vararg{Symbol}}}}
    kwargs = Dict(kwargs)
    keys = [key for (key, value) ∈ state]
    instances_vec = [value for (key, value) ∈ state]
    data = Vector{Vector{Symbol}}()
    function recursive(data, current, depth = 1)
        if depth <= length(instances_vec)
            for instance ∈ instances_vec[depth]
                recursive(data, push!(copy(current), instance), depth + 1)
            end
        else
            push!(data, current)
        end
    end
    recursive(data, [])
    return [(;zip(keys,configuration)...) for configuration ∈ data]
end

当像这样运行时:

identifiers = Dict(:Id=>(:One, :Two, :Three, :Four, :Five), :Nationality=>(:Brit, :German), :Color=>(:Red, :Blue, :Green))

Γ(identifiers)

给出以下错误:

ERROR: MethodError: Cannot `convert` an object of type
NamedTuple{(:Id, :Nationality, :Color),Tuple{Symbol,Symbol,Symbol}} to an object of type
NamedTuple{#s589,#s588} where #s588<:Tuple{Vararg{Symbol,N} where N} where #s589<:Tuple{Vararg{Symbol,N} where N}

返回的类型为: Array{NamedTuple{(:Id, :Nationality, :Color),Tuple{Symbol,Symbol,Symbol}},1}和 (:Id, :Nationality, :Color) 的类型是 Tuple{Symbol,Symbol,Symbol}所以我不明白为什么在不需要转换时会出现转换错误，特别是因为:

typeof((:Id, :Nationality, :Color)) <: Tuple{Vararg{Symbol}} && Tuple{Symbol,Symbol,Symbol} <: Tuple{Vararg{Symbol}} = true

Answer 1

首先，转换来自这样一个事实，即您的函数声明
声明一个返回类型。如果您删除(或注释掉)此类型注释

function Γ(state::Dict{Symbol, <:Tuple{Vararg{Symbol}}}) #::Vector{NamedTuple{<:Tuple{Vararg{Symbol}}, <:Tuple{Vararg{Symbol}}}}
    kwargs = Dict() # <- there was a typo here in your minimal example
    keys = [key for (key, value) ∈ state]
    instances_vec = [value for (key, value) ∈ state]
    data = Vector{Vector{Symbol}}()
    function recursive(data, current, depth = 1)
        if depth <= length(instances_vec)
            for instance ∈ instances_vec[depth]
                recursive(data, push!(copy(current), instance), depth + 1)
            end
        else
            push!(data, current)
        end
    end
    recursive(data, [])
    return [(;zip(keys,configuration)...) for configuration ∈ data]
end

那么一切都很好:

julia> identifiers = Dict(:Id=>(:One, :Two, :Three, :Four, :Five), :Nationality=>(:Brit, :German), :Color=>(:Red, :Blue, :Green))
Dict{Symbol,Tuple{Symbol,Symbol,Vararg{Symbol,N} where N}} with 3 entries:
  :Id          => (:One, :Two, :Three, :Four, :Five)
  :Nationality => (:Brit, :German)
  :Color       => (:Red, :Blue, :Green)

julia> Γ(identifiers)
30-element Array{NamedTuple{(:Id, :Nationality, :Color),Tuple{Symbol,Symbol,Symbol}},1}:
 (Id = :One, Nationality = :Brit, Color = :Red)
 (Id = :One, Nationality = :Brit, Color = :Blue)
 (Id = :One, Nationality = :Brit, Color = :Green)
 [...]

请注意，在必要时键入函数/方法的参数被认为是一种很好的做法。然而，为返回值提供类型断言的代码在 Julia 中通常不那么惯用:编译器非常擅长确定返回类型是什么!

现在这个转换失败的原因是因为提供的元素类型
签名与返回的实际元素类型不一致
向量:

julia> elem = (Id = :One, Nationality = :Brit, Color = :Red)
(Id = :One, Nationality = :Brit, Color = :Red)

julia> typeof(elem)
NamedTuple{(:Id, :Nationality, :Color),Tuple{Symbol,Symbol,Symbol}}

julia> elem isa NamedTuple{<:Tuple{Vararg{Symbol}}, <:Tuple{Vararg{Symbol}}}
false

对于 NamedTuple{A,B}匹配 NamedTuple{<:Tuple{Vararg{Symbol}},
<:Tuple{Vararg{Symbol}}} ，以下两个条件都必须成立:

A <: Tuple{Vararg{Symbol}}

B <: Tuple{Vararg{Symbol}}

在这里，我们有:

A = (:Id, :Nationality, :Color)
B = Tuple{Symbol,Symbol,Symbol}

所以，就像你说的， B 的条件持有:

julia> B <: Tuple{Vararg{Symbol}}
true

但不是 A 的条件:

julia> A <: Tuple{Vararg{Symbol}}
ERROR: TypeError: in <:, expected Type, got Tuple{Symbol,Symbol,Symbol}
Stacktrace:
 [1] top-level scope at REPL[10]:1

       # A is a value of type Tuple{...}, not the type itself
julia> typeof(A) <: Tuple{Vararg{Symbol}}
true