加速计算 R 中庞大数据集上的 mann-kendall 测试的并行过程

Question

让我们假设在世界上的大量点的每月时间步长上有一个大型气候数据集。然后数据集被塑造成data.frame的类型：

经度、纬度、data_month_1_yr_1、...、data_month_12_yr_100

例子：

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

我想对每个空间点的每月时间序列执行 Mann-Kendall 检验（使用trend::mk.test），并在data.frame 中获取主要统计数据。为了加快这个漫长的过程，我并行化了我的代码并编写了如下内容：

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

这运行得很好，并给出了我所追求的：一个矩阵报告每个坐标和月份组合的 M-K 统计数据。尽管过程是并行的，但是计算仍然需要相当长的时间。

有没有办法加快这个过程？有使用applyfamily 功能的空间吗？

Answer 1

您注意到您已经解决了您的问题。可通过以下步骤之一获得：

1：使用.packages 和.export 将必要的对象复制到foreach 循环中。这样可以确保每个实例在尝试访问同一内存时不会发生冲突。

2：利用data.table的tidyverse等高性能库进行子集和计算。

后者稍微复杂一些，但极大地提升了我的微型笔记本电脑的性能。 （对整个数据集执行所有计算大约需要 1.5 分钟。）

下面是我添加的代码。请注意，我用并行包中的一个 parLapply 函数替换了 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，通过将第二个循环替换为 lapply 函数（受此 answer 启发），问题很容易解决。执行时间现在被控制在几秒钟内。向量化仍然是 R 中执行时间的最佳解决方案（请参阅 this post 和 this）

我在下面分享最终代码以供参考：

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