I am trying to sample for two vectors 1000 times with replacement and calculate the ratio of means. Repeat this process 10,000 times.
I wrote a sample parallel code but it's taking much longer that using simple for loops on a single machine.
ratio_sim_par <- function(x1, x2, nrep = 1000) {
# Initiate cluster
cl <- makeCluster(detectCores() - 1) #Leave one core for other operations
clusterExport(cl, varlist=c("x1", "x2", "nrep"), envir=environment())
Tboot <- parLapply(cl, 1:nrep, function(x){
n1 <- length(x1)
n2 <- length(x2)
xx1 <- sample(x1, n1, replace = TRUE) # sample of size n1 with replacement from x1
xx2 <- sample(x2, n2, replace = TRUE) # sample of size n2 with replacement from x2
return(mean(xx1) / mean(xx2))
})
stopCluster(cl)
return(unlist(Tboot))
}
ratio_sim_par(x1, x2, 10000)
System time is unbearable. Can anyone help me understand the mistake I'm making? Thanks
Distributing tasks to different nodes takes a lot of computational overhead and can cancel out any gains you make from parallelizing your script. In your case, you're calling parLapply 10,000 times and probably spending more resources forking each task than actually doing the resampling. Try something like this with a non-parallel version of ratio_sim_par:
mclapply(1:10000, ratio_sim_par, x1, x2, nrep = 1000, mc.cores = n_cores)
mclapply will split the job into as many cores as you have available and fork it once. I'm using mclapply instead of parLapply because I'm used to it and doesn't require as much setup.
Related
I'm using the parallel package and mclapply() to parallel process simulations in R, using R Programming for Data Science (Chapter 22, Section 22.4.1) as a reference.
I'm setting the seed as instructed, however, when I change the number of cores used in the mclapply() function, I get different results even with the same seed.
A simple reprex:
# USING 2 CORES
library(parallel)
RNGkind("L'Ecuyer-CMRG")
set.seed(1)
x <- mclapply(1:100, function(i) {rnorm(1)}, mc.cores = 2)
y <- do.call(rbind, x)
z <- mean(y)
print(mean(z))
# returns 0.143
# USING 3 CORES
library(parallel)
RNGkind("L'Ecuyer-CMRG")
set.seed(1)
x <- mclapply(1:100, function(i) {rnorm(1)}, mc.cores = 3)
y <- do.call(rbind, x)
z <- mean(y)
print(mean(z))
# returns 0.035
How can I set the seed such that changing the number of cores used doesn't change the result? I feel like this should be a fairly simple thing to do - maintaining reproducibility irrespective of number of cores used.
Imagine I have two functions, one is a simple mean of sum of squares, and the other, a little more elaborated that computes a regression, that I want to apply to the lines of a "big" matrix or data frame.
In order to take advantage of multiple cores (on Windows) I tried the parallel package and got very different results for the two functions using the same sequence of commands.
For the apparently more complex function (regression) it appears that the time reduction is significant using more cores (Here I show a result from a PC with 3 cores and a PC with 12 cores, the behavior is similar with up to 11 cores, the time reduction decreases with more cores).
But for the "simple" function, mean of squares, the time of executions is very variable, almost erratic (also tested with up to 11 cores).
First, Is there a reason why this is happening? Second, I imagine there are other ways to do that task, can you suggest any?
Here is the code to generate the plots:
library(parallel)
nc=detectCores()-1 #number of cores
myFun =function(z) coef(lm(rep(1,length(z))~z)) #regression
myFun2 =function(z) sum(z^2)/length(z) # mean of squares
my.mat = matrix(rnorm(1000000,.01,0.4),ncol=100) #data
# using FUN = myFun
# Replicate 10 times
for(j in 1:10){
ncor=2:nc
timed=c()
for (i in seq_along(ncor)){
cl <- makeCluster(mc <- getOption("cl.cores", ncor[i]))
stime <- Sys.time()
res=parApply(cl = cl, X = my.mat, MARGIN = 1, FUN = myFun)
tm=Sys.time()-stime
timed[i]=tm
stopCluster(cl)
}
# no cores
stime <- Sys.time()
res=apply(my.mat, MARGIN = 1, FUN = myFun)
tm=Sys.time()-stime
(dr=data.frame(nc=c(1,ncor),ts=as.numeric(c(tm,timed))))
plot(dr,type="l",col=3,main=j)
#stopCluster(cl)
if (j==1)fres1=dr else fres1=merge(fres1,dr,by="nc")
}
plot(fres1[,1:2],type="l",col=2,ylim=range(fres1[,-1]))
for(i in 3:11)lines(fres1[,i],col=i+1)
# For the second plot use the same code but change FUN = myFun2
Firstly, I would like to say that I am new to this topic.
Secondly, although I read a lot about Parallel processing in R, I'm still not confident about it.
I just invented simulation in R. So can someone help me with this invented code to understand Parallel processing? (I can see how it works)
My code as follows (Large Random numbers)
SimulateFn<-function(B,n){
M1=list()
for (i in 1:B){
M1[i]=(n^2)}
return(M1)}
SimulateFn(100000000,300000)
Could you please help me?
First of all, parallelization is the procedure of dividing a task into sub tasks, which are simultaneously processed by multiple processors or cores and can be independent or share some dependency between them - the latter case needs more planning and attention.
This procedure has some overhead to shedule subtasks - like copying data to each processor. That said, parallelization is worthless for fast computations. In your example, the threee main procedures are indexing ([), assignment (<-), and a (fast) math operation (^). The overhead for paralellization may be greater than the time to execute the subtask, so in that case parallelization can result in poorer performance!
Despite that, simple parallelization in R is fairly easy. An approach to parallelize your task is provided below, using the doParallel package. Other approachs include using packages as parallel.
library(doParallel)
## choose number of processors/cores
cl <- makeCluster(2)
registerDoParallel(cl)
## register elapsed time to evaluate code snippet
## %dopar% execute code in parallale
B <- 100000; n <- 300000
ptime <- system.time({
M1=list()
foreach(i=1:B) %dopar% {
M1[i]=(n^2)
}
})
## %do% execute sequentially
stime <- system.time({
M1=list()
foreach(i=1:B) %do% {
M1[i]=(n^2)
}
})
The elapsed times on my computer (2 core) were 59.472 and 44.932, respectively. Clearly, there were no improvement by parallelization: indeed, performance was worse!
A better example is shown below, where the main task is much more expensive in terms of computation need:
x <- iris[which(iris[,5] != "setosa"), c(1,5)]
trials <- 10000
ptime <- system.time({
r <- foreach(icount(trials), .combine=cbind) %dopar% {
ind <- sample(100, 100, replace=TRUE)
result1 <- glm(x[ind,2]~x[ind,1], family=binomial(logit))
coefficients(result1)
}
})
stime <- system.time({
r <- foreach(icount(trials), .combine=cbind) %do% {
ind <- sample(100, 100, replace=TRUE)
result1 <- glm(x[ind,2]~x[ind,1], family=binomial(logit))
coefficients(result1)
}
})
And elapsed times were 24.709 and 34.502: a gain of 28%.
I'm attempting to "map" a function onto an array. However when trying both simple and complex functions, the parallel version is always slower than the serial version. How can I improve the performance of a parallel computation in R?
Simple parallel example:
library(parallel)
# Number of elements
arrayLength = 100
# Create data
input = 1:arrayLength
# A simple computation
foo = function(x, y) x^y - x^(y-1)
# Add complexity
iterations = 5 * 1000 * 1000
# Perform complex computation on each element
compute = function (x) {
y = x
for (i in 1:iterations) {
x = foo(x, y)
}
return(x)
}
# Parallelized compute
computeParallel = function(x) {
# Create a cluster with 1 fewer cores than are available.
cl <- makeCluster(detectCores() - 1) # 8-1 cores
# Send static vars & funcs to all cores
clusterExport(cl, c('foo', 'iterations'))
# Map
out = parSapply(cl, x, compute)
# Clean up
stopCluster(cl)
return(out)
}
system.time(out <- compute(input)) # 12 seconds using 25% of cpu
system.time(out <- computeParallel(input)) # 160 seconds using 100% of cpu
The problem is that you traded off all of the vectorization for parallelization, and that's a bad trade. You need to keep as much vectorization as possible to have any hope of getting an improvement with parallelization for this kind of problem.
The pvec function in the parallel package can be a good solution to this kind of problem, but it isn't supported in parallel on Windows. A more general solution which works on Windows is to use foreach with the itertools package which contains functions which are useful for iterating over various objects. Here's an example that uses the "isplitVector" function to create one subvector for each worker:
library(doParallel)
library(itertools)
cl <- makeCluster(detectCores() - 1)
registerDoParallel(cl)
computeChunk <- function(x) {
foreach(xc=isplitVector(x, chunks=getDoParWorkers()),
.export=c('foo', 'iterations', 'compute'),
.combine='c') %dopar% {
compute(xc)
}
}
This still may not compare very well to the pure vector version, but it should get better as the value of "iterations" increases. It may actually help to decrease the number of workers unless the value of "iterations" is very large.
parSapply will run the function on each element of input separately, which means you are giving up the speed you gained from writing foo and compute in a vectorized fashion.
pvec will run a vectorized function on multiple cores by chunks. Try this:
system.time(out <- pvec(input, compute, mc.cores=4))
I try to compare up to thousands of estimated beta distributions. Each beta distribution is characterized by the two shape parameters alpha & beta.
I now draw 100,000 samples of every distribution. As a final result I want to get an order of the distributions with the highest Probability in every sample draw.
My first approach was to use lapply for generating a matrix of N * NDRAWS numeric values which was consuming too much memory as N gets beyond 10,000. (10,000 * 100,000 * 8 Bytes)
So I decided to use a sequential approach of ordering every single draw, then cumsum the order of all draws and get the final order as shown in the example below:
set.seed(12345)
N=100
NDRAWS=100000
df <- data.frame(alpha=sample(1:20, N, replace=T), beta=sample(1:200, N, replace=T))
vec <- vector(mode = "integer", length = N )
for(i in 1:NDRAWS){
# order probabilities after a single draw for every theta
pos <- order(rbeta(N, shape1=df$alpha, shape2=df$beta) )
# sum up winning positions for every theta
vec[pos] <- vec[pos] + 1:N
}
# order thetas
ord <- order(-vec)
df[ord,]
This is only consuming N * 4 Bytes of memory, as there is no giant matrix but a single vector of length N. My Question now is, how to speed up this operation using snowfall (or any other multicore package) by taking advantage of my 4 CPU Cores, instead of using just one core???
# parallelize using snowfall pckg
library(snowfall)
sfInit( parallel=TRUE, cpus=4, type="SOCK")
sfLapply( 1:NDRAWS, function(x) ?????? )
sfStop()
Any help is appreciated!
This can be parallelized in the same way that one would parallelize random forest or bootstrapping. You just perform the sequential code on each of the workers but with each using a smaller number of iterations. That is much more efficient than splitting each iteration of the for loop into a separate parallel task.
Here's your complete example converted to use the foreach package with the doParallel backend:
set.seed(12345)
N=100
NDRAWS=100000
df <- data.frame(alpha=sample(1:20, N, replace=T),
beta=sample(1:200, N, replace=T))
library(doParallel)
nworkers <- detectCores()
cl <- makePSOCKcluster(nworkers)
clusterSetRNGStream(cl, c(1,2,3,4,5,6,7))
registerDoParallel(cl)
vec <- foreach(ndraws=rep(ceiling(NDRAWS/nworkers), nworkers),
.combine='+') %dopar% {
v <- integer(N)
for(i in 1:ndraws) {
pos <- order(rbeta(N, shape1=df$alpha, shape2=df$beta) )
v[pos] <- v[pos] + 1:N
}
v
}
ord <- order(-vec)
df[ord,]
Note that this gives different results than the sequential version because different random numbers are generated by the workers. I used the parallel random number support provided by the parallel package since that is good practice.
Well, the functionality is there. I'm not sure though what you'd be returning with each iteration.
Perhaps try this?
myFunc <- function(xx, N) {
pos <- order(rbeta(N, shape1=df$alpha, shape2=df$beta) )
vec[pos] + 1:N
}
Using doParallel will allow you to add results:
require(doParallel)
registerDoParallel(cores=4)
foreach(i=1:NDRAWS, .combine='+') %dopar% myFunc(i, N)