我想在一个命令中加载多个RData,正如Johua使用所解释的那样
> lapply(c(a_data, b_data, c_data, d_data), load, .GlobalEnv)
[[1]]
[1] "nRTC_Data"
[[2]]
[1] "RTA_Data"
[[3]]
[1] "RTC_Data"
[[4]]
[1] "RTA_Data"
> rm(a_data, b_data, c_data, d_data); ls()
[1] "nRTC_Data" "RTA_Data" "RTAC_data" "RTC_Data"
然而,由于我的RData很大,并且我发现lappy()
和多个load()
之间没有时间改进,所以我决定使用以下多核方法:
library(parallel)
mclapply(c(a_data, b_data, c_data, d_data),load,.GlobalEnv, mc.cores = parallel::detectCores())
虽然这显著改善了加载时间,但也返回了列表
[[1]]
[1] "nRTC_Data"
[[2]]
[1] "RTA_Data"
[[3]]
[1] "RTC_Data"
[[4]]
[1] "RTA_Data"
在我的工作区中,没有找到
> rm(a_data, b_data, c_data, d_data); ls()
character(0)
我也尝试用environment()
替换.GlobalEnv
,但仍然没有成功。
有人知道线索吗?
仅供参考,您可以尝试以下命令:
> a = "aa";save(a, file = "aa.RData")
> b = "bb";save(b, file = "bb.RData")
> c = "cc";save(c, file = "cc.RData")
> d = "dd";save(d, file = "dd.RData")
> # lapply approach
> rm(list = ls())
> a = "aa.RData"; b = "bb.RData"; c = "cc.RData"; d = "dd.RData"
> lapply(c(a, b, c, d), load, .GlobalEnv); rm(a, b, c, d)
> # mclapply approach
> rm(list = ls())
> a = "aa.RData"; b = "bb.RData"; c = "cc.RData"; d = "dd.RData"
> mclapply(c(a, b, c, d), load, .GlobalEnv, mc.cores = parallel::detectCores()); rm(a, b, c, d)
我认为这是因为当使用mclapply
时,底层的分叉会创建单独的进程。在下面的代码中,我使用mclapply
和myload
函数,该函数加载Rdata文件并返回加载的对象。与您的lapply
版本的不同之处在于,您在mclapply
返回的列表中有数据
myload <- function(x){
x <- load(x)
get(x)
}
a = "aa.RData"; b = "bb.RData"; c = "cc.RData"; d = "dd.RData"
res <- mclapply(c(a, b, c, d), myload, mc.cores = parallel::detectCores());