如何在Julia中将Array {Array {Float64,1}},1}转换为Matrix？

Question

假设我输入了这种类型:

> [[0.8681299566762923,-0.3472589826095631], [3.2300860990307445,3.3731249077464946]]

如何将其转换为更令人愉快的类型,如Matrix(了解尺寸)？

Answer 1

你可以使用splatting(...)并hcat获得你想要的东西:

julia> a = Vector[[0.8681299566762923,-0.3472589826095631], [3.2300860990307445,3.3731249077464946]]
2-element Array{Array{T,1},1}:
 [0.8681299566762923,-0.3472589826095631]
 [3.2300860990307445,3.3731249077464946]

julia> hcat(a...)
2x2 Array{Float64,2}:
  0.86813   3.23009
 -0.347259  3.37312

或者,如果您希望将堆栈作为行而不是列,则可以执行以下操作:

julia> vcat(map(x->x', a)...)
2x2 Array{Float64,2}:
 0.86813  -0.347259
 3.23009   3.37312

我不建议Matrix逐行构建,因为这与Julia的列主阵列布局相冲突.对于较大的矩阵,实际上堆栈为列并转置输出更有效:

julia> a2 = Vector{Float64}[rand(10) for i=1:5000];

julia> stackrows1{T}(a::Vector{Vector{T}}) = vcat(map(transpose, a)...)::Matrix{T}
stackrows1 (generic function with 2 methods)

julia> stackrows2{T}(a::Vector{Vector{T}}) = hcat(a...)'::Matrix{T}
stackrows2 (generic function with 2 methods)

julia> stackrows1(a2) == stackrows2(a2)  # run once to compile and make sure functions do the same thing
true

julia> @time for i=1:100 stackrows1(a2); end
elapsed time: 0.142792896 seconds (149 MB allocated, 7.85% gc time in 7 pauses with 0 full sweep)

julia> @time for i=1:100 stackrows2(a2); end
elapsed time: 0.05213114 seconds (88 MB allocated, 12.60% gc time in 4 pauses with 0 full sweep)