Convert column classes in data.table

For a single column:

dtnew <- dt[, Quarter:=as.character(Quarter)]
str(dtnew)

Classes ‘data.table’ and 'data.frame':  10 obs. of  3 variables:
 $ ID     : Factor w/ 2 levels "A","B": 1 1 1 1 1 2 2 2 2 2
 $ Quarter: chr  "1" "2" "3" "4" ...
 $ value  : num  -0.838 0.146 -1.059 -1.197 0.282 ...

---

Using lapply and as.character:

dtnew <- dt[, lapply(.SD, as.character), by=ID]
str(dtnew)

Classes ‘data.table’ and 'data.frame':  10 obs. of  3 variables:
 $ ID     : Factor w/ 2 levels "A","B": 1 1 1 1 1 2 2 2 2 2
 $ Quarter: chr  "1" "2" "3" "4" ...
 $ value  : chr  "1.487145280568" "-0.827845218358881" "0.028977182770002" "1.35392750102305" ...

Try this

DT <- data.table(X1 = c("a", "b"), X2 = c(1,2), X3 = c("hello", "you"))
changeCols <- colnames(DT)[which(as.vector(DT[,lapply(.SD, class)]) == "character")]

DT[,(changeCols):= lapply(.SD, as.factor), .SDcols = changeCols]

Raising Matt Dowle's comment to Geneorama's answer (https://stackoverflow.com/a/20808945/4241780) to make it more obvious (as encouraged), you can use for(...)set(...).

library(data.table)

DT = data.table(a = LETTERS[c(3L,1:3)], b = 4:7, c = letters[1:4])
DT1 <- copy(DT)
names_factors <- c("a", "c")

for(col in names_factors)
  set(DT, j = col, value = as.factor(DT[[col]]))

sapply(DT, class)
#>         a         b         c 
#>  "factor" "integer"  "factor"

^{Created on 2020-02-12 by the reprex package (v0.3.0)}

See another of Matt's comments at https://stackoverflow.com/a/33000778/4241780 for more info.

Edit.

As noted by Espen and in help(set), j may be "Column name(s) (character) or number(s) (integer) to be assigned value when column(s) already exist". So names_factors <- c(1L, 3L) will also work.