R：如何在foreach％dopar％中拆分数据帧

Question

This is a very simple example.

df = c("already ","miss you","haters","she's cool")
df = data.frame(df)

library(doParallel)
cl = makeCluster(4)
registerDoParallel(cl)    
foreach(i = df[1:4,1], .combine = rbind, .packages='tm')  %dopar% classification(i)
stopCluster(cl)

In real case I have dataframe with n=400000 rows. I don't know how to send nrow/ncluster data for each cluster in one step, i = ?

I tried with isplitRows from library(itertools) without success.

Answer 1

You should try to work with indices to create subsets of your data.

foreach(i = nrow(df), .combine = rbind, .packages='tm')  %dopar% {
  tmp <- df[i, ]
  classification(tmp)
}

This will take a new row of the data.frame each iteration.

Furthermore, you should notice that the result of a foreach loop will be written to a new variable. Thus, you should assign it like this:

res <- foreach(i = 1:10, .combine = c, ....) %dopar% {
  # things you want to do
  x <- someFancyFunction()

  # the last value will be returned and combined by the .combine function
  x 
}

Answer 2

Try using a combination of split and mclapply as proposed in Aproach 1 here: https://www.r-bloggers.com/trying-to-reduce-the-memory-overhead-when-using-mclapply/

split lets you split data into groups defined by a factor, or you can just use 1:nrow(df) if you want to do the operation on each row seperately.

Answer 3

My solution after your comments:

n = 8  #number of cluster
library(foreach)
library(doParallel)
cl = makeCluster(n)
registerDoParallel(cl)

z = nrow(df)
y = floor(z/n) 
x = nrow(df)%%n

ris = foreach(i = split(df[1:(z-x),],rep(1:n,each=y)), .combine = rbind, .packages='tm')  %dopar% someFancyFunction(i)

stopCluster(cl)

#sequential
if (x !=0 )
    ris = rbind(ris,someFancyFunction(df[(z-x+1):z,1]))

Note: I used the sequential esecution at the end, because if "x" is not zero, the function split put the rest of rows (z-(zx)) in the first cluster, and change the order of the result.