问题描述
我对参数方法在Julia中的工作方式感到困惑.我也希望参数化是在这里使用的正确词.我已经阅读了有关方法的文档,但仍不清楚为什么会发生以下错误.如果我定义如下的函数
I am confused about how parametric methods work in Julia. I also hope that parametric is the right word to use here. I've read the docs on methods but it is still not clear why the following error occurs. If I define a function as below
function Bazinga{T<:Real}(mat ::Union{Array{T,2},SparseMatrixCSC})
mat^4
end
并运行
Penny = sparse(randn(10,10))
Bazinga(Penny)
我知道
ERROR: MethodError: `Bazinga` has no method matching Bazinga(::SparseMatrixCSC{Float64,Int64})
Closest candidates are:
Bazinga{T<:Real}(::Union{Array{T<:Real,2},SparseMatrixCSC{Tv,Ti<:Integer}})
但是Int64<:Int
是正确的,那么这是什么问题呢?如果我将功能重新定义为
But Int64<:Int
is true so what is the problem here?If I redefine the function as
function Bazinga(mat ::Union{Array{Real,2},SparseMatrixCSC})
mat^4
end
没有错误产生并且该功能起作用.现在还有两种方法
no error is produced and the function works. Also there are now two methods of the function
Bazinga(mat::Union{Array{Real,2},SparseMatrixCSC{Tv,Ti<:Integer}}) at none:2
Bazinga{T<:Real}(mat::Union{Array{T<:Real,2},SparseMatrixCSC{Tv,Ti<:Integer}}) at none:2
这两个SparseMatrixCSC{Tv,Ti<:Integer}
都在其中的
,但仅在第一种情况下,它会导致错误.任何输入都很好!
in which both SparseMatrixCSC{Tv,Ti<:Integer}
is in, but only in the first case it causes an error.Any input is wellcome!
Edit1:为什么Array{Float64,1} <:Array{Real,1}
为false?Edit2:Edit1中的问题由 Julia中的数组中断字符串类型解决. .但是我不认为这可以解释该方法的结果.或者至少我看不到.
And why is Array{Float64,1} <:Array{Real,1}
false? The question in the Edit1 is solved by Arrays break string types in Julia . But I don't think it explains what happens with the method. Or at least I don't see it.
推荐答案
我不太清楚其背后的原因,但是对于您的用例,使用TypeVar
可能是一种解决方法:
I don't quite know the reason under the hood, but using a TypeVar
may be a workaround for your use case:
T = TypeVar(:T, Union{}, Real, false)
function Bazinga(mat::Union{Array{T,2},SparseMatrixCSC})
mat^4
end
julia> Bazinga(sparse(randn(2,2)))
2x2 sparse matrix with 4 Float64 entries:
[1, 1] = 0.840151
[2, 1] = -0.503551
[1, 2] = -0.437787
[2, 2] = 1.28652
julia> Bazinga(randn(2,2))
2x2 Array{Float64,2}:
0.203916 -0.261589
0.0107211 -0.0137373
更新:
经过一番调查,我发现解决此问题的规范"方法是定义typealias
:
typealias RealMatrix{T<:Real} Array{T, 2}
function Bazinga(mat::Union{RealMatrix, SparseMatrixCSC})
mat^4
end
我猜想原因是Julia无法在不知道所有参数类型到底是什么的情况下完成方法匹配任务.这是另一个示例:
I guess the reason is Julia can't complete method matching tasks without knowing what on earth all of the parametric types are. Here is another example:
Foo{R<:Real,C<:Complex}(x::Union{Array{R}, SparseMatrixCSC{C}}) = x
Foo(sparse([1+im 2; 3 4]))
和Foo(rand(2,2))
均不起作用.因此,编写参数联合方法的正确方法是通过typealias
约束那些不同的参数:
Neither Foo(sparse([1+im 2; 3 4]))
nor Foo(rand(2,2))
will work. So the correct way to write a parametric-union method is to constrain those different parameters via typealias
:
julia> typealias Bar{T<:Real} Array{T,2}
Array{T<:Real,2}
julia> typealias Baz{T<:Complex} SparseMatrixCSC{T}
SparseMatrixCSC{T<:Complex{T<:Real},Ti<:Integer}
julia> Foo(x::Union{Bar, Baz}) = x
Foo (generic function with 1 method)
julia> Foo(rand(2,2))
2x2 Array{Float64,2}:
0.000739447 0.713386
0.32024 0.705593
julia> Foo(sparse([1+im; 2im]))
2x1 sparse matrix with 2 Complex{Int64} entries:
[1, 1] = 1+1im
[2, 1] = 0+2im
如果联合的参数类型具有相同的类型参数T
,我们还可以如下使用typealias
:
If argument types of the union have the same type parameter T
, we can also use typealias
as follows:
julia> typealias MyUnion{T,N<:Integer} Union{Array{T,2}, SparseMatrixCSC{T,N}}
Union{Array{T,2},SparseMatrixCSC{T,N<:Integer}}
julia> Foo{T<:Real}(x::MyUnion{T}) = x
Foo (generic function with 1 method)
julia> Foo(rand(2,2))
2x2 Array{Float64,2}:
0.0172915 0.587518
0.234148 0.00543953
julia> Foo(rand(2))
ERROR: MethodError: `Foo` has no method matching Foo(::Array{Float64,1})
julia> Foo(sparse(rand(2)))
2x1 sparse matrix with 2 Float64 entries:
[1, 1] = 0.900029
[2, 1] = 0.0634815
julia> Foo(sparse(rand(2,2)))
2x2 sparse matrix with 4 Float64 entries:
[1, 1] = 0.592632
[2, 1] = 0.066563
[1, 2] = 0.805307
[2, 2] = 0.923422
julia> Foo(rand(2,2))
2x2 Array{Float64,2}:
0.167078 0.673194
0.681418 0.316017
更新2:
在Julia-v0.6 +中,无需定义类型别名:
UPDATE 2:
In Julia-v0.6+, defining the typealias is not necessary:
function Bazinga(mat::Union{Array{<:Real,2},SparseMatrixCSC})
mat^4
end
有关更多详细信息,请参考此答案.
Please refer to this answer for further details.
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