Question

我有以下Julia代码，我想将其并行化。

using DistributedArrays

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

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

Answer 1

您可以使用n维分布式数组理解：

首先，您需要添加一些本地或远程进程：

julia> addprocs(CPU_CORES - 1);

然后你必须在每个衍生进程中使用 DistributedArrays：

julia> @everywhere using DistributedArrays

最后，您可以使用@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 julian 最符合您的意图恕我直言的方法。

我们可以查看macroexpand输出，了解正在发生的事情：

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，

我是朱莉娅的新人，但面临同样的问题。尝试这种方法，看看它是否符合您的需求。

function f(x)
  return x^2;
end

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

println(y)

问候，RN

如何在朱莉娅写一个并行循环？

2 个答案: