如何在朱莉娅写一个并行循环？

Question

我有以下Julia代码,我想并行化它.

using DistributedArrays

function f(x)
    return x^2;
end
y = DArray[]
@parallel for i in 1:100
    y[i] = f(i)
end
println(y)

输出是DistributedArrays.DArray[].我希望y的值如下:y=[1,4,9,16,...,10000]

Answer 1

您可以使用n维分布式数组解析:

首先,您需要添加更多进程,本地或远程:

julia> addprocs(CPU_CORES - 1);

然后,你必须使用 DistributedArrays在衍生过程的每一个:

julia> @everywhere using DistributedArrays

最后你可以使用@DArray宏,如下所示:

julia> x = @DArray [@show x^2 for x = 1:10];
        From worker 2:  x ^ 2 = 1
        From worker 2:  x ^ 2 = 4
        From worker 4:  x ^ 2 = 64
        From worker 2:  x ^ 2 = 9
        From worker 4:  x ^ 2 = 81
        From worker 4:  x ^ 2 = 100
        From worker 3:  x ^ 2 = 16
        From worker 3:  x ^ 2 = 25
        From worker 3:  x ^ 2 = 36
        From worker 3:  x ^ 2 = 49

你可以看到它符合你的期望:

julia> x
10-element DistributedArrays.DArray{Int64,1,Array{Int64,1}}:
   1
   4
   9
  16
  25
  36
  49
  64
  81
 100

请记住,它适用于任意数量的维度:

julia> y = @DArray [@show i + j for i = 1:3, j = 4:6];
        From worker 4:  i + j = 7
        From worker 4:  i + j = 8
        From worker 4:  i + j = 9
        From worker 2:  i + j = 5
        From worker 2:  i + j = 6
        From worker 2:  i + j = 7
        From worker 3:  i + j = 6
        From worker 3:  i + j = 7
        From worker 3:  i + j = 8

julia> y
3x3 DistributedArrays.DArray{Int64,2,Array{Int64,2}}:
 5  6  7
 6  7  8
 7  8  9

julia>

这是最愚蠢的方式做你想要的恕我直言.

我们可以查看macroexpand输出以查看发生了什么:

注意:为了便于阅读,此输出已经过轻微编辑,T代表:

DistributedArrays.Tuple{DistributedArrays.Vararg{DistributedArrays.UnitRange{DistributedArrays.Int}}}

julia> macroexpand(:(@DArray [i^2 for i = 1:10]))
  :(
    DistributedArrays.DArray(
      (
        #231#I::T -> begin
          [i ^ 2 for i = (1:10)[#231#I[1]]]
        end
      ),
      DistributedArrays.tuple(DistributedArrays.length(1:10))
    )
  )

这与手动输入基本相同:

julia> n = 10; dims = (n,);

julia> DArray(x -> [i^2 for i = (1:n)[x[1]]], dims)
10-element DistributedArrays.DArray{Any,1,Array{Any,1}}:
   1
   4
   9
  16
  25
  36
  49
  64
  81
 100

julia>

Answer 2

嗨基拉，

我是 Julia 的新手，但面临着同样的问题。尝试这种方法，看看它是否适合您的需求。

function f(x)
  return x^2;
end

y=@parallel vcat for i= 1:100
 f(i);
end;

println(y)

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