使用lapply函数的并行版本保留列表名称

When I want to do some operation over the lists, and get the list with the same names as a result I use sapply function:

当我想对列表进行一些操作，并获得具有相同名称的列表时，我使用sapply函数：

newList <- sapply(oldList, someFunction, simplify = FALSE,USE.NAMES = TRUE)

Is it possible (and how) to do the same using parallel versions of function from apply family, eg mclapply from parallel package?

是否有可能（以及如何）使用apply系列的并行版本的函数来做同样的事情，例如来自并行包的mclapply？

2 个解决方案

#1

?mclapply help page says that this is possible (argument SIMPLIFY), although only for mcmapply. As you've already figured it out, (mc)mapply with only one object passed is a special case and is equivalent to (mc)lapply.

？mclapply帮助页面说这是可能的（参数SIMPLIFY），虽然仅适用于mcmapply。正如你已经想到的那样，（mc）mapply只传递一个对象是一个特例，相当于（mc）lapply。

#2

There is parSapply function in parallel package - https://stat.ethz.ch/R-manual/R-devel/library/parallel/html/clusterApply.html

并行包中有parSapply函数 - https://stat.ethz.ch/R-manual/R-devel/library/parallel/html/clusterApply.html

library("parallel")
cl <- makeCluster(getOption("cl.cores", 2))
newList <- parSapply(cl, oldList, someFunction, simplify = FALSE,USE.NAMES = TRUE)