Element of a list use as a variable in R?-CodePudding

I have a list of elements:

E.list <- list("ELP","ARC","DEG","UEG","BAC")

And I want to use them as a variable as well as a pattern in grepl such as:

ELP <- df[, grepl("ELP", names(df))]
ELP_META <- dplyr::filter(Meta, grepl("ELP", CASE_Name))
ARC <- df[, grepl("ARC", names(df))]
ARC_META <- dplyr::filter(Meta, grepl("ARC", CASE_Name))
.
.
.

How I can do it using for loop or foreach?

CodePudding user response：

We may extract the element of the list with [[

df[, grepl(E.list[[1]], names(df))]

If we need to loop

out <- lapply(E.list, function(nm) {
       tmp <- df[, grepl(nm, names(df))]
       tmp_META <- dplyr::filter(Meta, grepl(nm, CASE_Name))
        return(setNames(list(tmp, tmp_META), c(nm, paste0(nm, "_META"))))

       })

In a for loop we can do

out <- vector('list', length(E.list))
out_META <- vector('list', length(E.list))
names(out) <- E.list
names(out_META) <- paste0(E.list, "_META")
for(nm in E.list) {
     out[[nm]] <- df[, grepl(nm, names(df))]
     out_META[[paste0(nm, "_META")]] <- dplyr::filter(Meta, grepl(nm, CASE_Name))
}

If the intention is to create multiple objects in the global env, it is not recommended, instead store it in a list as in the above solutions. But, if we use assign it would create those objects

for(nm in E.list) {
    assign(nm, df[, grepl(nm, names(df))]
   assign(paste0(nm, "_META"), dplyr::filter(Meta, grepl(nm, CASE_Name)))
}