Question

我在R中运行以下脚本如果我使用％do％而不是％dopar％，脚本运行正常。但是，如果在外部循环中我使用％dopar％，则循环将永远运行而不会抛出任何错误（内存使用量会不断增加，直到内存不足为止）。我正在使用16个核心。

library(parallel)
library(foreach)
library(doSNOW)
library(dplyr)


NumberOfCluster <- 16 
cl <- makeCluster(NumberOfCluster) 
registerDoSNOW(cl) 


foreach(i = UNSPSC_list, .packages = c('data.table', 'dplyr'), .verbose = TRUE) %dopar% 
    { 
      terms <- as.data.table(unique(gsub(" ", "", unlist(terms_list_by_UNSPSC$Terms[which(substr(terms_list_by_UNSPSC$UNSPSC,1,6) == i)])))) 
      temp <- inner_join(N_of_UNSPSCs_by_Term, terms, on = 'V1') 
      temp$V2 <- 1/as.numeric(temp$V2)
      temp <- temp[order(temp$V2, decreasing = TRUE),]
      names(temp) <- c('Term','Imp')
      ABNs <- unique(UNSPSCs_per_ABN[which(substr(UNSPSCs_per_ABN$UNSPSC,1,4) == substr(i,1,4)), 1])

      predictions <- as.numeric(vector()) 
      predictions <- foreach (j = seq(1 : nrow(train)), .combine = 'c', .packages = 'dplyr')  %do% 
      { 
        descr <- names(which(!is.na(train[j,]) == TRUE)) 
        if(unlist(predict_all[j,1]) %in% unlist(ABNs) || !unlist(predict_all[j,1]) %in% unlist(suppliers)) {union_all(predictions, sum(temp$Imp[which(temp$Term %in% descr)]))} else {union_all(predictions, 0)}    

      } 
    save(predictions, file = paste("Predictions", i,".rda", sep = "_")) 
    }

Answer 1

嵌套foreach循环的正确方法是使用%:%运算符。查看示例。我在Windows上测试过它。

library(foreach)
library(doSNOW)

NumberOfCluster <- 4
cl <- makeCluster(NumberOfCluster) 
registerDoSNOW(cl) 

N <- 1e6

system.time(foreach(i = 1:10, .combine = rbind) %:%
              foreach(j = 1:10, .combine = c) %do% mean(rnorm(N, i, j)))

system.time(foreach(i = 1:10, .combine = rbind) %:%
              foreach(j = 1:10, .combine = c) %dopar% mean(rnorm(N, i, j)))

输出：

> system.time(foreach(i = 1:10, .combine = rbind) %:%
+               foreach(j = 1:10, .combine = c) %do% mean(rnorm(N, i, j)))
   user  system elapsed 
   7.38    0.23    7.64 
> system.time(foreach(i = 1:10, .combine = rbind) %:%
+               foreach(j = 1:10, .combine = c) %dopar% mean(rnorm(N, i, j)))
   user  system elapsed 
   0.09    0.00    2.14

使用嵌套循环的方案如下：

foreach(i) %:% foreach(j) {foo(i, j)}

运算符%:%用于嵌套多个foreach循环。你无法在嵌套之间进行计算。在您的情况下，您必须执行两个循环，例如：

# Loop over i
x <- foreach(i = 1:10, .combine = c) %dopar% 2 ^ i

# Nested loop over i and j
foreach(i = 1:10, .combine = rbind) %:% foreach(j = 1:10, .combine = c) %dopar% {x[i] + j}

未经测试的代码：

library(data.table)
library(foreach)
library(doSNOW)

NumberOfCluster <- 2
cl <- makeCluster(NumberOfCluster)
registerDoSNOW(cl)

# Create ABNs as list
ABNs <- foreach(i = UNSPSC_list, .packages = c('data.table', 'dplyr'), .verbose = TRUE) %dopar% {
  terms <- as.data.table(unique(gsub(" ", "", unlist(terms_list_by_UNSPSC$Terms[which(substr(terms_list_by_UNSPSC$UNSPSC, 1, 6) == i)]))))
  temp <- inner_join(N_of_UNSPSCs_by_Term, terms, on = 'V1')
  temp$V2 <- 1 / as.numeric(temp$V2)
  temp <- temp[order(temp$V2, decreasing = TRUE), ]
  names(temp) <- c('Term', 'Imp')
  unique(UNSPSCs_per_ABN[which(substr(UNSPSCs_per_ABN$UNSPSC,1,4) == substr(i,1,4)), 1])
}

# Nested loop
predictions <- foreach(i = UNSPSC_list, .packages = c('data.table', 'dplyr'), .verbose = TRUE) %:%
  foreach(j = seq(1:nrow(train)), .combine = 'c', .packages = 'dplyr') %dopar% {
    descr <- names(which(!is.na(train[j, ]) == TRUE))
    if (unlist(predict_all[j, 1]) %in% unlist(ABNs[[i]]) || !unlist(predict_all[j, 1]) %in% unlist(suppliers)) {
      sum(temp$Imp[which(temp$Term %in% descr)])
    } else 0
  }

for (i in seq_along(predictions)) save(predictions[[i]], file = paste("Predictions", i, ".rda", sep = "_"))

R在外环中嵌套foreach％dopar％，在内环中嵌套％do％

1 个答案: