这样的事情怎么样?它使用snowfall而不是foreach -library，但应给出相同的结果.

How about something like this? It uses snowfall instead of the foreach-library, but should give the same results.

library(snowfall)
library(ridge)

# for reproducability
set.seed(123)
num_of_cores <- parallel::detectCores()
mydata <- read.csv("http://www.ats.ucla.edu/stat/data/binary.csv")
data_per_core <- floor(nrow(mydata)/num_of_cores)

# we take random rows to each cluster, by sampleid
mydata$sampleid <- sample(1:num_of_cores, nrow(mydata), replace = T)

# create a small function that calculates the coefficients
regfun <- function(dat) {
  library(ridge) # this has to be in the function, otherwise snowfall doesnt know the logistic ridge function
  result <- logisticRidge(admit~ gre + gpa + rank, data = dat)
  coefs <- as.numeric(coefficients(result))
  return(coefs)
}

# prepare the data
datlist <- lapply(1:num_of_cores, function(i){
  dat <- mydata[mydata$sampleid == i, ]
})

# initiate the clusters
sfInit(parallel = T, cpus = num_of_cores)

# export the function and the data to the cluster
sfExport("regfun")

# calculate, (sfClusterApply is very similar to sapply)
res <- sfClusterApply(datlist, function(datlist.element) {
  regfun(dat = datlist.element)
})

#stop the cluster
sfStop()

# convert the list to a data.frame. data.table::rbindlist(list(res)) does the same job
res <- data.frame(t(matrix(unlist(res), ncol = num_of_cores)))
names(res) <- c("intercept", "gre", "gpa", "rank")
res
# res
# intercept          gre
# 1 -3.002592 1.558363e-03
# 2 -4.142939 1.060692e-03
# 3 -2.967130 2.315487e-03
# 4 -1.176943 4.786894e-05
# gpa         rank
# 1  0.7048146997 -0.382462408
# 2  0.9978841880 -0.314589628
# 3  0.6797382218 -0.464219036
# 4 -0.0004576679 -0.007618317