Names of variables inside the 'for loop' [duplicate] - r

Question

This question already has answers here:
Dynamically select data frame columns using $ and a character value
(10 answers)
Closed 6 years ago.
I am trying to create a function that allows the conversion of selected columns of a data frame to categorical data type (factor) before running a regression analysis.
Question is how do I slice a particular column from a data frame using a string (character).
Example:
strColumnNames <- "Admit,Rank"
strDelimiter <- ","
strSplittedColumnNames <- strsplit(strColumnNames, strDelimiter)
for( strColName in strSplittedColumnNames[[1]] ){
dfData$as.name(strColName) <- factor(dfData$get(strColName))
}
Tried:
dfData$as.name()
dfData$get(as.name())
dfData$get()
Error Msg:
Error: attempt to apply non-function
Any help would be greatly appreciated! Thank you!!!

Answer 1

You need to change
dfData$as.name(strColName) <- factor(dfData$get(strColName))
to
dfData[[strColName]] <- factor(dfData[[strColName]])
You may read ?"[[" for more.
In your case, column names are generated programmingly, [[]] is the only way to go. Maybe this example will be clear enough to illustrate the problem of $:
dat <- data.frame(x = 1:5, y = 2:6)
z <- "x"
dat$z
# [1] NULL
dat[[z]]
# [1] 1 2 3 4 5
Regarding the other answer
apply definitely does not work, because the function you apply is as.factor or factor. apply always works on a matrix (if you feed it a data frame, it will convert it into a matrix first) and returns a matrix, while you can't have factor data class in matrix. Consider this example:
x <- data.frame(x1 = letters[1:4], x2 = LETTERS[1:4], x3 = 1:4, stringsAsFactors = FALSE)
x[, 1:2] <- apply(x[, 1:2], 2, as.factor)
str(x)
#'data.frame': 4 obs. of 3 variables:
# $ x1: chr "a" "b" "c" "d"
# $ x2: chr "A" "B" "C" "D"
# $ x3: int 1 2 3 4
Note, you still have character variable rather than factor. As I said, we have to use lapply:
x[1:2] <- lapply(x[1:2], as.factor)
str(x)
#'data.frame': 4 obs. of 3 variables:
# $ x1: Factor w/ 4 levels "a","b","c","d": 1 2 3 4
# $ x2: Factor w/ 4 levels "A","B","C","D": 1 2 3 4
# $ x3: int 1 2 3 4
Now we see the factor class in x1 and x2.
Using apply for a data frame is never a good idea. If you read the source code of apply:
dl <- length(dim(X))
if (is.object(X))
X <- if (dl == 2L)
as.matrix(X)
else as.array(X)
You see that a data frame (which has 2 dimension) will be coerced to matrix first. This is very slow. If your data frame columns have multiple different class, the resulting matrix will have only 1 class. Who knows what the result of such coercion would be.
Yet apply is written in R not C, with an ordinary for loop:
for (i in 1L:d2) {
tmp <- forceAndCall(1, FUN, newX[, i], ...)
if (!is.null(tmp))
ans[[i]] <- tmp
so it is no better than an explicit for loop you write yourself.

Answer 2

I would use a different method.
Create a vector of column names you want to change to factors:
factorCols <- c("Admit", "Rank")
Then extract these columns by index:
myCols <- which(names(dfData) %in% factorCols)
Finally, use apply to change these columns to factors:
dfData[,myCols] <- lapply(dfData[,myCols],as.factor)