从for (i in range)传递mclapply()参数

Question

我正在尝试这样做：

nmf.sub <- function(n){
sub.data.matrix <- data.matrix[, (index[n, ])] ## the index is a permutation of the original matrix at a 0.8 resampling proportion (doesn't really matter)
temp.result <- nmf(sub.data.matrix, rank = 2, seed = 12345) ## want to change 2 to i
return(temp.result)
}

class.list <- list()
for (i in nmf.rank){ ## nmf.rank is 2:4
results.list <- mclapply(mc.cores = 16, 1:resamp.iterations, function(n) nmf.sub(n)) ## resamp.iterations is 10, nmf.sub is defined above
}

但是，在temp.result的nmf中，我不希望rank =2，而是希望rank =i

知道我怎么才能把这个参数传给它吗？仅仅将其作为函数(n，i)传递给mclapply是行不通的。

Answer 1

您似乎有两个循环:一个用于nmf.rank中的i，另一个用于1:resamp.iterations中的n。因此，您需要将i和n都传递给nmf.sub，例如：

nmf.sub <- function(n, i){
    ## the index is a permutation of the original matrix at a 0.8
    ## resampling proportion (doesn't really matter)
    sub.data.matrix <- data.matrix[, (index[n, ])] 
    ## want to change 2 to i
    temp.result <- nmf(sub.data.matrix, rank = i, seed = 12345)
    return(temp.result)
}


resamp.iterations <- 10
nmf.rank <- 2:4

res <- lapply(nmf.rank, function(i){
    results.list <- mclapply(mc.cores = 16, 1:resamp.iterations,
                             function(n) nmf.sub(n,i))
})
## then you can flatten/reshape res

关于你关于效率的评论(下面)：大部分数值计算是在nmf()函数中执行的，因此循环是正确设置的，从这个意义上讲，每个进程/核心都有一个数值密集型作业。但是，为了加快计算速度，您可以考虑使用以前计算的结果，而不是种子12345 (除非由于与您的问题相关的某种原因，必须使用后一个种子)。在下面的示例中，我的执行时间减少了30-40%：

library(NMF)
RNGkind("L'Ecuyer-CMRG") ## always use this when using mclapply()
nr <- 19
nc <- 2e2
set.seed(123)
data.matrix <- matrix(rexp(nc*nr),nr,nc)

resamp.iterations <- 10
nmf.rank <- 2:4

index <- t(sapply(1:resamp.iterations, function(n) sample.int(nc,nc*0.8)))


nmf.sub <- function(n, i){
    sub.data.matrix <- data.matrix[ ,index[n, ]] 
    temp.result <- nmf(sub.data.matrix, rank = i, seed = 12345)
    return(temp.result)
}

## version 1
system.time({
    res <- lapply(nmf.rank, function(i){
        results.list <- mclapply(mc.cores = 16, 1:resamp.iterations,
                                 function(n) nmf.sub(n,i))
    })
})

## version 2: swap internal and external loops
system.time({
    res <- 
        mclapply(mc.cores=16, 1:resamp.iterations, function(n){
            res2 <- nmf(data.matrix[ ,index[n, ]], rank=2, seed = 12345)
            res3 <- nmf(data.matrix[ ,index[n, ]], rank=3, seed = 12345)
            res4 <- nmf(data.matrix[ ,index[n, ]], rank=4, seed = 12345)
            list(res2,res3,res4)
        })
})

## version 3: use previous calculation as starting point
##   ==> 30-40% reduction in computing time
system.time({
    res <- 
        mclapply(mc.cores=16, 1:resamp.iterations, function(n){
            res2 <- nmf(data.matrix[ ,index[n, ]], rank=2, seed = 12345)
            res3 <- nmf(data.matrix[ ,index[n, ]], rank=3, seed = res2)
            res4 <- nmf(data.matrix[ ,index[n, ]], rank=4, seed = res3)
            list(res2,res3,res4)
        })
})