@distributed似乎可以正常工作，函数返回很奇怪 [英] @distributed seems to work, function return is wonky

问题描述

我正在学习如何在Julia中进行并行计算.我在3x嵌套的 for 循环的开始处使用 @sync @distributed 来并行化事物(请参见底部的代码).从行 println(errCmp [row，col])中，我可以看到数组 errCmp 的所有元素都已打印出来.例如

I'm just learning how to do parallel computing in Julia. I'm using @sync @distributed at the start of a 3x nested for loop to parallelize things (see code at bottom). From the line println(errCmp[row, col]) I can watch all the elements of the array errCmp be printed out. E.g.

From worker 3:    2.351134946074191e9
From worker 4:    2.3500830193505473e9
From worker 5:    2.3502416529551845e9
From worker 2:    2.3509105625656652e9
From worker 3:    2.3508352842971106e9
From worker 4:    2.3497049296121807e9
From worker 5:    2.35048428351797e9
From worker 2:    2.350742582031195e9
From worker 3:    2.350616273660934e9
From worker 4:    2.349709546599313e9

但是，当函数返回时， errCmp 是我在乞讨时预分配的零数组.

However, when the function returns, errCmp is the array of zeros I pre-allocate at the begging.

我想念一些收集所有东西的闭门词吗?

Am I missing some closing term to collect everything?

function optimizeDragCalc(df::DataFrame)
    paramGrid = [cd*AoM for cd = range(1e-3, stop = 0.01, length = 50), AoM = range(2e-4, stop = 0.0015, length = 50)]
    errCmp    = zeros(size(paramGrid))
    # totalSize = size(paramGrid, 1) * size(paramGrid, 2) * size(df.time, 1)
    @sync @distributed for row = 1:size(paramGrid, 1)
        for col = 1:size(paramGrid, 2)
            # Run the propagation here
            BC = 1/paramGrid[row, col]
            slns, _ = propWholeTraj(df, BC)
            for time = 1:size(df.time, 1)
                errDF = propError(slns[time], df, time)
                errCmp[row, col] += sum(errDF.totalErr)
            end # time
            # println("row: ", row, " of ",size(paramGrid, 1),"   col: ", col, " of ", size(paramGrid, 2))
            println(errCmp[row, col])
        end # col
    end # row
    # plot(heatmap(z = errCmp))
    return errCmp, paramGrid
end
errCmp, paramGrid = @time optimizeDragCalc(df)

推荐答案

您没有提供最低限度的工作示例，但我想这可能很难.这是我的MWE.让我们假设我们要使用 Distributed 来计算 Array 列的总和:

You did not provide a minimal working example but I guess it might be hard. So here is mine MWE. Let us assume that we want to use Distributed to calculate sums of Array's columns:

using Distributed
addprocs(2)
@everywhere using StatsBase
data = rand(1000,2000)
res = zeros(2000)
@sync @distributed for col = 1:size(data)[2]
    res[col] = StatsBase.mean(data[:,col])
    # does not work!
    # ... because data is created locally and never returned!
end

为了更正上面的代码，您需要提供一个聚合函数(我故意简化示例-可以进行进一步的优化).

In order to correct the above code you need to provide an aggregator function (I keep the example intentionally simplified - a further optimization is possible).

using Distributed
addprocs(2)
@everywhere using Distributed,StatsBase
data = rand(1000,2000)    
@everywhere function t2(d1,d2)
    append!(d1,d2)
    d1
end
res = @sync @distributed (t2) for col = 1:size(data)[2]
    [(myid(),col, StatsBase.mean(data[:,col]))]
end

现在让我们看一下输出.我们可以看到，有些值是在worker 2 上计算的，而其他值是在worker 3 上计算的:

Now let us see the output. We can see that some of the values have been calculated on worker 2 while others on worker 3:

julia> res
2000-element Array{Tuple{Int64,Int64,Float64},1}:
 (2, 1, 0.49703681326230276)
 (2, 2, 0.5035341367791002)
 (2, 3, 0.5050607022354537)
 ⋮
 (3, 1998, 0.4975699181976122)
 (3, 1999, 0.5009498778934444)
 (3, 2000, 0.499671315490524)

其他可能的改进/修改:

Further possible improvements/modifications:

• 使用 @spawnat 在远程进程(而不是主进程并发送它们)上生成值
• 使用 SharedArray -这可以在工作人员之间自动分配数据.根据我的经验，需要非常仔细的编程.
• 使用 ParallelDataTransfer.jl 在工作人员之间发送数据.非常易于使用，对于大量消息而言效率不高.
• 始终考虑使用Julia线程机制(在某些情况下，它使工作更轻松-再次取决于问题)

• use @spawnat to generate values at remote processes (instead of the master process and sending them)
• use SharedArray - this allows to automatically distribute data among workers. From my experience requires very careful programming.
• use ParallelDataTransfer.jl to send data among workers. Very easy to use, not efficient for huge number of messages.
• always consider Julia threading mechanism (in some scenarios it makes life easier - again depends on the problem)

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