Question

我最近开始使用R中的parallel软件包，这对我来说是一个奇迹。不过，我遇到了一个找不到答案的问题。

我正在尝试重新格式化某些数据，为此，我在并行情况下使用sapply()或parSapply()。在正常情况下，我会去：

sapply(1:nrow(aux),function(x){
  r=which(M$Project==aux$Projecte[x] & M$Product==aux$Producte[x])
  c=which(names(M)==aux$Atribut[x])
  l=aux$meanss[x]
  M[r,c]<<-l
})

使用<<-为全球环境分配价值。对于并行情况，我去：

no_cores <- detectCores()-2
cl <- makeCluster(no_cores)
clusterExport(cl,c("aux","M"))
parSapply(cl,1:20,function(x){
  r=which(M$Project==aux$Projecte[x] & M$Product==aux$Producte[x])
  c=which(names(M)==aux$Atribut[x])
  l=aux$meanss[x]
  M[r,c]<<-l
})

我知道这些值正在计算（已打印），但是它们没有像使用M那样分配给sapply()数据帧。我环顾四周，但未找到任何有关此的信息。在并行应用函数内分配值时，是否应考虑任何特殊考虑？谢谢，请在下面找到可复制的示例。

M：

structure(list(Project = c("11I040119", "11I040119", "11I040119", 
"11I040119", "11I040119", "11I040119", "11I040119", "11I040119", 
"11I040119", "11I040119", "11I040119", "11I040119", "11I040119"
), Product = c("Brulerie St. Denis (BOLD)", "Ethical Beans (BOLD)", 
"Folgers (BOLD)", "Illy drip coffe (BOLD)", "Illy Espresso Coffee (BOLD)", 
"Just Us (BOLD)", "Lavazza caffè espresso (BOLD)", "Lavazza Crema e gusto (BOLD)", 
"Lavazza Tierra (BOLD)", "Medaglia d'Oro (BOLD)", "Seattle Best 4 (BOLD)", 
"Starbucks café Verona (BOLD)", "Tully's (BOLD)"), Thing1 = c(0, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0), Thing2 = c(0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0, 0)), row.names = c(NA, 13L), class = "data.frame")

辅助：

structure(list(Projecte = c("11I040119", "11I040119", "11I040119", 
"11I040119", "11I040119", "11I040119", "11I040119", "11I040119", 
"11I040119", "11I040119", "11I040119", "11I040119", "11I040119", 
"11I040119", "11I040119"), Producte = c("Brulerie St. Denis (BOLD)", 
"Ethical Beans (BOLD)", "Folgers (BOLD)", "Illy drip coffe (BOLD)", 
"Illy Espresso Coffee (BOLD)", "Just Us (BOLD)", "Lavazza caffè espresso (BOLD)", 
"Lavazza Crema e gusto (BOLD)", "Lavazza Tierra (BOLD)", "Medaglia d'Oro (BOLD)", 
"Seattle Best 4 (BOLD)", "Starbucks café Verona (BOLD)", "Tully's (BOLD)", 
"Brulerie St. Denis (BOLD)", "Ethical Beans (BOLD)"), Thing = c("Thing1", 
"Thing1", "Thing1", "Thing1", "Thing1", "Thing1", "Thing1", "Thing1", 
"Thing1", "Thing1", "Thing1", "Thing1", "Thing1", "Thing2", "Thing2"
), Value = c(0.142857142857143, 0.242857141154153, 0.614285715988704, 
0, 0, 0.0714285714285714, 1.01428570917674, 0, 0.971428564616612, 
0.5, 0.357142857142857, 0.642857142857143, 0.714285714285714, 
3, 5)), row.names = c(NA, 15L), class = "data.frame")

所需输出（M）：

     Project                       Product     Thing1 Thing2
1  11I040119     Brulerie St. Denis (BOLD) 0.14285714      3
2  11I040119          Ethical Beans (BOLD) 0.24285714      5
3  11I040119                Folgers (BOLD) 0.61428572      0
4  11I040119        Illy drip coffe (BOLD) 0.00000000      0
5  11I040119   Illy Espresso Coffee (BOLD) 0.00000000      0
6  11I040119                Just Us (BOLD) 0.07142857      0
7  11I040119 Lavazza caffè espresso (BOLD) 1.01428571      0
8  11I040119  Lavazza Crema e gusto (BOLD) 0.00000000      0
9  11I040119         Lavazza Tierra (BOLD) 0.97142856      0
10 11I040119         Medaglia d'Oro (BOLD) 0.50000000      0
11 11I040119         Seattle Best 4 (BOLD) 0.35714286      0
12 11I040119  Starbucks café Verona (BOLD) 0.64285714      0
13 11I040119                Tully's (BOLD) 0.71428571      0

Answer 1

这不是一个完整的解决方案，但是为了快速起见-并行化通过启动多个进程来工作（想象并排运行多个R会话）。这些进程中的每一个都有自己的全局环境.GlobalEnv，因此您的M[r,c] <<- l实际上是为每个进程分配其他位置。

一种可能的实现方式是，您以某种方式重写函数，例如，返回list(r, c, l)并使用parLapply，然后获得并行收集的索引和值的列表，并执行主要过程中的任务分配。

使用并行应用将值分配给全局环境变量

1 个答案: