list - l_ply : how to pass the list's name attribute into the function?

Question

假设我有一个这样的 R 列表:

> summary(data.list)
                                 Length Class      Mode
aug9104AP                        18     data.frame list
Aug17-10_acon_7pt_dil_series_01  18     data.frame list
Aug17-10_Picro_7pt_dil_series_01 18     data.frame list
Aug17-10_PTZ_7pt_dil_series_01   18     data.frame list
Aug17-10_Verat_7pt_dil_series_01 18     data.frame list

我想使用 l_ply 处理列表中的每个 data.frame ，但我还需要将名称(例如 aug9104AP)与 data.frame 一起传递到处理函数中。类似的东西:

l_ply(data.list,function(df,...) {

    cli.name<- arg_to_access_current_list_item_name

    #make plots with df, use cli.name in plot titles
    #save results in a file called cli.name

  }, arg_to_access_current_list_item_name
)

应该怎么做arg_to_access_current_list_item_name是？

Answer 1

如果您一一传递它们，您可以使用 deparse(substitute(arg)) ，例如:

test <- function(x){
       y <- deparse(substitute(x))
       print(y)
       print(x)
 }

 var <- c("one","two","three")
 test(var)
[1] "var"
[1] "one"   "two"   "three"

对于 l_ply，您必须求助于将属性添加到列表本身，例如:

for(i in 1:length(data.list)){
    attr(data.list[[i]],"name") <- names(data.list)[i]
}

然后你可以使用 attr :

cli <- attr(x,"name")

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