加快在R中龐大數據集上計算mann-kendall檢驗的並行過程

Question

假設每個月的時間步長都有一個龐大的氣候數據集，用於世界上的許多點。 然后將數據集定型為類型為data.frame的數據：

lon，lat，data_month_1_yr_1，...，data_month_12_yr_100

例：

set.seed(123)
data<- data.frame(cbind(runif(10000,-180,180), runif(10000,-90,90))
, replicate(1200, runif(10000,0,150)))

我想在每個空間點的每月時間序列上執行Mann-Kendall測試（使用trend::mk.test ），並在data.frame獲取主要統計信息。 為了加快這個非常漫長的過程，我並行化了我的代碼並編寫了如下內容：

coords<-data[,1:2] #get the coordinates out of the initial dataset
names(coords)<-c("lon","lat") 
data_t<- as.data.frame(t(data[,3:1202])) #each column is now the time series associated to a point
data_t$month<-rep(seq(1,12,1),100) # month index as last column of the data frame
# start the parallel processing

library(foreach)
library(doParallel)

cores=detectCores() #count cores
cl <- makeCluster(cores[1]-1) #take all the cores minus 1 not to overload the pc
registerDoParallel(cl)

mk_out<- foreach(m=1:12, .combine = rbind) %:%
         foreach (a =1:10000, .combine = rbind) %dopar% {

           data_m<-data_t[which(data_t$month==m),]
           library(trend) #need to load this all the times otherwise I get an error (don't know why)
           test<-mk.test(data_m[,a])
           mk_out_temp <- data.frame("lon"=coords[a,1],
                                     "lat"=coords[a,2],
                                     "p.value" = as.numeric(test$p.value),
                                     "z_stat" = as.numeric(test$statistic),
                                     "tau" = as.numeric(test$estimates[3]),
                                     "month"= as.numeric(m))
           mk_out_temp
}
stopCluster(cl)

head(mk_out)
         lon       lat    p.value     z_stat         tau month
1  -76.47209 -34.09350 0.57759040 -0.5569078 -0.03797980     1
2  103.78985 -31.58639 0.64436238  0.4616081  0.03151515     1
3  -32.76831  66.64575 0.11793238  1.5635113  0.10626263     1
4  137.88627 -30.83872 0.79096910  0.2650524  0.01818182     1
5  158.56822 -67.37378 0.09595919 -1.6647673 -0.11313131     1
6 -163.59966 -25.88014 0.82325630  0.2233588  0.01535354     1

這運行得很好，並且給我確切的信息：一個矩陣，報告每個坐標和月份組合的MK統計信息。 盡管該過程是並行的，但是計算仍需要花費大量時間。

有沒有辦法加快這個過程？ apply家庭有使用職能的空間嗎？

Answer 1

您注意到您已經解決了問題。 可通過以下步驟之一獲得：

1：使用.packages和.export將必要的對象復制到foreach循環中。 這樣可以確保在嘗試訪問同一內存時每個實例不會沖突。

2：利用諸如data.table的tidyverse之類的高性能庫來執行子集和計算。

后者稍微復雜一點，但極大地提高了我的微型筆記本電腦的性能。 （對整個數據集執行所有計算大約需要1.5分鍾。）

以下是我添加的代碼。 請注意，我用並行包中的單個parLapply函數替換了foreach。

set.seed(123)
data<- data.frame(cbind(runif(10000,-180,180), runif(10000,-90,90))
                  , replicate(1200, runif(10000,0,150)))

coords<-data[,1:2] #get the coordinates out of the initial dataset
names(coords)<-c("lon","lat") 
data_t<- as.data.frame(t(data[,3:1202])) #each column is now the time series associated to a point
data_t$month<-rep(seq(1,12,1),100) # month index as last column of the data frame
# start the parallel processing

library(data.table)
library(parallel)
library(trend)
setDT(data_t)
setDT(coords)
cores=detectCores() #count cores
cl <- makeCluster(cores[1]-1) #take all the cores minus 1 not to overload the pc

#user  system elapsed 
#17.80   35.12   98.72
system.time({
  test <- data_t[,parLapply(cl, 
                            .SD, function(x){
                              (
                                unlist(
                                  trend::mk.test(x)[c("p.value","statistic","estimates")]
                                )
                               )
                              }
                            ), by = month] #Perform the calculations across each month
  #create a column that indicates what each row is measuring
  rows <- rep(c("p.value","statistic.z","estimates.S","estimates.var","estimates.tau"),12)

  final_tests <- dcast( #Cast the melted structure to a nice form
                      melt(cbind(test,rowname = rows), #Melt the data for a better structure
                        id.vars = c("rowname","month"), #Grouping variables
                        measure.vars = paste0("V",seq.int(1,10000))), #variable names
                      month + variable ~ rowname, #LHS groups the data along rows, RHS decides the value columns
                      value.var = "value", #Which column contain values? 
                      drop = TRUE) #should we drop unused columns? (doesnt matter here)
  #rename the columns as desired
  names(final_tests) <- c("month","variable","S","tau","var","p.value","z_stat")
  #finally add the coordinates
  final_tests <- cbind(final_form,coords) 
})

Answer 2

最后，可以通過用lapply函數替換第二個循環來輕松解決此問題 （受此答案啟發）。 現在，執行時間僅為幾秒鍾。 向量化仍然是解決R中執行時間的最佳解決方案（請參閱此職位和this ）

我在下面共享最終代碼以供參考：

set.seed(123)
data<- data.frame(cbind(runif(10000,-180,180), runif(10000,-90,90)), replicate(1200, runif(10000,0,150)))
coords<-data[,1:2]
names(coords)<-c("lon","lat")
data_t<- as.data.frame(t(data[,3:1202]))
data_t$month<-rep(seq(1,12,1),100)


library(foreach)
library(doParallel)

cores=detectCores()
cl <- makeCluster(cores[1]-1) #take all the cores minus 1
registerDoParallel(cl)

mk_out<- foreach(m=1:12, .combine = rbind) %dopar% {
    data_m<-data_t[which(data_t$month==m),]
    library(trend)
    mk_out_temp <- do.call(rbind,lapply(data_m[1:100],function(x)unlist(mk.test(x))))
    mk_out_temp <-cbind(coords,mk_out_temp,rep(m,dim(coords)[1]))
    mk_out_temp
  }
stopCluster(cl)


head(mk_out)

head(mk_out)
         lon       lat data.name            p.value        statistic.z null.value.S parameter.n estimates.S estimates.varS
1  -76.47209 -34.09350         x  0.577590398263635 -0.556907839290681            0         100        -188         112750
2  103.78985 -31.58639         x  0.644362383361713  0.461608102085858            0         100         156         112750
3  -32.76831  66.64575         x  0.117932376736468   1.56351131351662            0         100         526         112750
4  137.88627 -30.83872         x   0.79096910003836  0.265052394100912            0         100          90         112750
5  158.56822 -67.37378         x 0.0959591933285242  -1.66476728429674            0         100        -560         112750
6 -163.59966 -25.88014         x  0.823256299016955  0.223358759073802            0         100          76         112750
       estimates.tau alternative                  method              pvalg rep(m, dim(coords)[1])
1 -0.037979797979798   two.sided Mann-Kendall trend test  0.577590398263635                      1
2 0.0315151515151515   two.sided Mann-Kendall trend test  0.644362383361713                      1
3  0.106262626262626   two.sided Mann-Kendall trend test  0.117932376736468                      1
4 0.0181818181818182   two.sided Mann-Kendall trend test   0.79096910003836                      1
5 -0.113131313131313   two.sided Mann-Kendall trend test 0.0959591933285242                      1
6 0.0153535353535354   two.sided Mann-Kendall trend test  0.823256299016955                      1