This is not so hard. Yoy have to loop (lapply) through the list changing each member data.frame's columns of interest with another call to lapply.
First, some example dataset, since you have posted none.
set.seed(1234)
myfiles2 <- lapply(1:4, function(i){
data.frame(X = letters[1:5], Y = factor(sample(5)), Z = as.character(11:15))
})
str(myfiles2) # output omited
Now, the problem.
cols <- 2:3
myfiles3 <- lapply(myfiles2, function(DF){
DF[cols] <- lapply(DF[cols], as.numeric)
DF
})
To see that it worked any of the following two will do.
lapply(myfiles3, function(DF) sapply(DF, class))
str(myfiles3)
#List of 4
# $ :'data.frame': 5 obs. of 3 variables:
# ..$ X: Factor w/ 5 levels "a","b","c","d",..: 1 2 3 4 5
# ..$ Y: num [1:5] 1 3 2 4 5
# ..$ Z: num [1:5] 1 2 3 4 5
# $ :'data.frame': 5 obs. of 3 variables:
# ..$ X: Factor w/ 5 levels "a","b","c","d",..: 1 2 3 4 5
# ..$ Y: num [1:5] 4 1 5 2 3
# ..$ Z: num [1:5] 1 2 3 4 5
# $ :'data.frame': 5 obs. of 3 variables:
# ..$ X: Factor w/ 5 levels "a","b","c","d",..: 1 2 3 4 5
# ..$ Y: num [1:5] 4 3 1 2 5
# ..$ Z: num [1:5] 1 2 3 4 5
# $ :'data.frame': 5 obs. of 3 variables:
# ..$ X: Factor w/ 5 levels "a","b","c","d",..: 1 2 3 4 5
# ..$ Y: num [1:5] 5 2 1 3 4
# ..$ Z: num [1:5] 1 2 3 4 5