Effective way to apply a data list function

Question

I have a list of data frames in R. What I need to do is apply a function to each data file, in this case delete special characters and return a list of data.

Using lapplyand as.data.frame, the following works perfectly and provides exactly what I need:

my_df =data.frame(names = seq(1,10), chars = c("abcabc!!", "abcabc234234!!"))
my_list = list(my_df, my_df, my_df)

#str(my_list)
List of 3
 $ :'data.frame':   10 obs. of  2 variables: ...

new_list <- lapply(my_list, function(y) as.data.frame(lapply(y, function(x) gsub("[^[:alnum:][:space:]']", "", x))))

# str(new_list)
List of 3
 $ :'data.frame':   10 obs. of  2 variables:
  ..$ names: Factor w/ 10 levels "1","10","2","3",..: 1 3 4 5 6 7 8 9 10 2
  ..$ chars: Factor w/ 2 levels "abcabc","abcabc234234": 1 2 1 2 1 2 1 2 1 2
 $ :'data.frame':   10 obs. of  2 variables:
  ..$ names: Factor w/ 10 levels "1","10","2","3",..: 1 3 4 5 6 7 8 9 10 2
  ..$ chars: Factor w/ 2 levels "abcabc","abcabc234234": 1 2 1 2 1 2 1 2 1 2
 $ :'data.frame':   10 obs. of  2 variables:
  ..$ names: Factor w/ 10 levels "1","10","2","3",..: 1 3 4 5 6 7 8 9 10 2
  ..$ chars: Factor w/ 2 levels "abcabc","abcabc234234": 1 2 1 2 1 2 1 2 1 2

But I am wondering if there is a more efficient way that does not require nested lapply. Perhaps another apply-family function that returns elements as a data frame?

Answer 1

We do not need nested lapply, only one lapplywith transformdoes it

lapply(my_list, transform, chars = gsub("[^[:alnum:][:space:]']", "", chars))

"[^[[:alnum:] ']"

Answer 2

@akrun , lapply , , , , , .

, as.data.frame, lapply. lapply , .

, data.table. , .

library(data.table)

f <- function(x) gsub("[^[:alnum:][:space:]']", "", x)

my_df <- as.data.frame(matrix(paste0(sample(c(letters,'!'), size=1000000, replace=T),
                                 sample(c(letters,'!'), size=1000000, replace=T)), 
                                 ncol=250), stringsAsFactors = FALSE)
my_list = list(my_df, my_df, my_df)

system.time(lapply(my_list, function(y) as.data.frame(lapply(y, f))))
# 2.256 seconds

my_dt <- as.data.table(my_df)
my_list2 = list(my_dt, my_dt, my_dt)

system.time(lapply(my_list2, function(y) y[,lapply(.SD,f)]))
# 1.180 seconds