并行化 foreach 循环 R 中种子设置的问题

Question

我有一些要并行运行的测试。当我使用 foreach() 执行此操作时，我得到了 20 个测试迭代对的预期输出：

## Without seed
require(data.table)
require(foreach)
require(iterators)
require(doParallel)
require(doRNG)

numCores = 2
registerDoParallel(numCores)

iterations = 5
num_tests = 2:5

foreach( i = 1:iterations, .combine = 'rbind', .multicombine = TRUE, .inorder = FALSE ) %:%
  foreach( n = num_tests, .combine = 'rbind', .multicombine = TRUE, .inorder = FALSE ) %dopar% {
    
    ## Print iteration
    print(paste('Tests =',n,'Iteration =',i))
    
  }

输出：

result.1 "Tests = 2 Iteration = 1"
result.2 "Tests = 3 Iteration = 1"
result.3 "Tests = 4 Iteration = 1"
result.4 "Tests = 5 Iteration = 1"
result.1 "Tests = 2 Iteration = 2"
result.2 "Tests = 3 Iteration = 2"
result.3 "Tests = 4 Iteration = 2"
result.4 "Tests = 5 Iteration = 2"
result.1 "Tests = 2 Iteration = 3"
result.2 "Tests = 3 Iteration = 3"
result.3 "Tests = 4 Iteration = 3"
result.4 "Tests = 5 Iteration = 3"
result.1 "Tests = 2 Iteration = 4"
result.2 "Tests = 3 Iteration = 4"
result.3 "Tests = 4 Iteration = 4"
result.4 "Tests = 5 Iteration = 4"
result.1 "Tests = 2 Iteration = 5"
result.2 "Tests = 3 Iteration = 5"
result.3 "Tests = 4 Iteration = 5"
result.4 "Tests = 5 Iteration = 5"

但是，当我尝试向此循环添加一个步骤以设置随机种子时，如 doRNG 小插图中所述，我在每次迭代中得到不同数量的测试（14 个测试迭代对）：

## With seed
numCores = 2
registerDoParallel(numCores)

iterations = 5
num_tests = 2:5
rng <- RNGseq( iterations * (iterations+1) / 2, 1234)

foreach( i = 1:iterations, .combine = 'rbind', .multicombine = TRUE, .inorder = FALSE ) %:%
  foreach( n = num_tests, r = rng[(i-1)*i/2 + 1:i], .combine = 'rbind', .multicombine = TRUE, .inorder = FALSE ) %dopar% {
    
    ##Set seed
    rngtools::setRNG(r)
    
    ## Print iteration
    print(paste('Tests =',n,'Iteration =',i))
    
  }

输出：

result.1 "Tests = 2 Iteration = 1"
result.1 "Tests = 2 Iteration = 2"
result.2 "Tests = 3 Iteration = 2"
result.1 "Tests = 2 Iteration = 3"
result.2 "Tests = 3 Iteration = 3"
result.3 "Tests = 4 Iteration = 3"
result.1 "Tests = 2 Iteration = 4"
result.2 "Tests = 3 Iteration = 4"
result.3 "Tests = 4 Iteration = 4"
result.4 "Tests = 5 Iteration = 4"
result.1 "Tests = 2 Iteration = 5"
result.2 "Tests = 3 Iteration = 5"
result.3 "Tests = 4 Iteration = 5"
result.4 "Tests = 5 Iteration = 5"

我正在寻找一种方法来并行设置种子，嵌套的 foreach() 循环具有正确的迭代次数。有任何想法吗？

Answer 1

在我看来，小插图中的这个例子过于聪明。它在 foreach 循环中建立 r 作为 rng 的子列表作为终止条件。您使用了一些错误的索引/列表边界，这是导致您混淆输出的原因。

每次迭代都会为r 提供rng 子列表的一个元素。它摆脱了奇怪的子列表数学，但它令人难以置信的混乱，并且该行将在代码审查中获得“WTF”10 次中的 11 次。

至于最接近小插图示例的立即修复：

## With seed
numCores = 2
registerDoParallel(numCores)

iterations = 5
num_tests = 2:5
rng <- RNGseq( iterations * length(num_tests), 1234)

foreach( i = 1:iterations, .combine = 'rbind', .multicombine = TRUE, .inorder = FALSE ) %:%
  foreach( n = num_tests, r = rng[(i-1)*length(num_tests) + 1:length(num_tests)], .combine = 'rbind', .multicombine = TRUE, .inorder = FALSE ) %dopar% {
    
    ##Set seed
    rngtools::setRNG(r)
    
    ## Print iteration
    print(paste('Tests =',n,'Iteration =',i))
    
  }

这改变了

rng <- RNGseq( iterations * (iterations+1) / 2, 1234)

至

rng <- RNGseq( iterations * length(num_tests), 1234)

和变化

r = rng[(i-1)*i/2 + 1:i]

至

r = rng[(i-1)*length(num_tests) + 1:length(num_tests)]

可能更具可读性/可理解性的替代方案

rng 是一个RNGSeq 的列表（它们只是自己列出），我们要做的就是从该列表中为每个并行操作选择一个新的单个RNGSeq。我们需要为每个独立的并行迭代创建一个元素。

# one RNGseq for each inner+outer loop combination
iterations = 5
num_tests = 2:5
rng <- RNGseq(iterations * length(num_tests), 1234)

foreach( i = 1:iterations, .combine = 'rbind', .multicombine = TRUE, .inorder = FALSE ) %:%
    foreach( n = num_tests, .combine = 'rbind', .multicombine = TRUE, .inorder = FALSE ) %dopar% {
        
        # this is moved out of the loop termination conditions
        # and gets a new list element for each iteration
        # note the [[ ]]s to get a single element instead of a sublist!
        #         [-----------------------] conditions that change based on outer loop
        #                                     [---]  conditions that change based on inner loop
        r = rng[[ ((i-1)*length(num_tests)) + (n-1) ]]

        ##Set seed
        rngtools::setRNG(r)
        
        ## Print iteration
        print(paste('Tests =',n,'Iteration =',i))
        
    }

r的赋值需要(n-1)，因为num_tests从2开始。使用当前n在num_tests中的索引会更加灵活：r = rng[ ((i-1)*i) + (which(num_tests == n)) ]