像expand.net这样的组合迭代器

是否有一种快速方法来迭代expand.grid或CJ(data.table)返回的组合.当有足够的组合时,这些太大而无法适应内存.在itertools2库(
Python的itertools的端口)中有iproduct,但它真的很慢(至少我正在使用它的方式 – 如下所示).还有其他选择吗?

下面是一个示例,其中的想法是将函数应用于来自两个data.frames(previous post)的每个行组合.

library(data.table)  # CJ
library(itertools2)  # iproduct iterator
library(doParallel)

## Dimensions of two data
dim1 <- 10
dim2 <- 100
df1 <- data.frame(a = 1:dim1, b = 1:dim1)
df2 <- data.frame(x= 1:dim2, y = 1:dim2, z = 1:dim2)

## function to apply to combinations
f <- function(...) sum(...)

## Too big to expand with bigger dimensions (ie, 1e6, 1e5) -> errors
## test <- expand.grid(seq.int(dim1), seq.int(dim2))
## test <- CJ(indx1 = seq.int(dim1), indx2 = seq.int(dim2))
## Error: cannot allocate vector of size 3.7 Gb

## Create an iterator over the cartesian product of the two dims
it <- iproduct(x=seq.int(dim1), y=seq.int(dim2))

## Setup the parallel backend
cl <- makeCluster(4)
registerDoParallel(cl)

## Run
res <- foreach(i=it, .combine=c, .packages=c("itertools2")) %dopar% {
  f(df1[i$x, ], df2[i$y, ])
}
stopCluster(cl)

## Expand.grid results (different ordering)
expgrid <- expand.grid(x=seq(dim1), y=seq(dim2))
test <- apply(expgrid, 1, function(i) f(df1[i[["x"]],], df2[i[["y"]],]))

all.equal(sort(test), sort(res))  # TRUE

最佳答案 如果你给每个工人一个数据框的一部分,让他们各自执行计算,然后合并结果,我认为你会得到更好的表现.这样可以提高计算效率并减少工作人员的内存使用量.

下面是一个使用itertools包中的isplitRow函数的示例:

library(doParallel)
library(itertools)
dim1 <- 10
dim2 <- 100
df1 <- data.frame(a = 1:dim1, b = 1:dim1)
df2 <- data.frame(x= 1:dim2, y = 1:dim2, z = 1:dim2)
f <- function(...) sum(...)

nw <- 4
cl <- makeCluster(nw)
registerDoParallel(cl)

res <- foreach(d2=isplitRows(df2, chunks=nw), .combine=c) %dopar% {
  expgrid <- expand.grid(x=seq(dim1), y=seq(nrow(d2)))
  apply(expgrid, 1, function(i) f(df1[i[["x"]],], d2[i[["y"]],]))
}

我拆分df2因为它有更多的行,但你可以选择其中之一.

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