how to convert numeric column to factor in R

Question

I'm trying to use the softImpute command (from the softImpute package) for filling in missing values, and I'm trying to turn categorical variables in a large data frame into factor type before using the softImpute.

I've used as.factor command and factor command but they all yield the following

train[a]=factor(train[a])

Error in sort.list(y) : 'x' must be atomic for 'sort.list'
Have you called 'sort' on a list?

a here is a vector like: c(1:92)

I tried as.character too but the softImpute command would not recognize the variables as character and would treat them as numeric, resulting in decimal values for categorical/indicator variables.

Are you looking for like this: x <- factor(x), where x <- c(1:92). Output format will help us more. — vamosrafa
– vamosrafa, Commented Dec 2, 2013 at 6:07
No, I'm trying to turn 1st to 92st column in the data frame "train" into factors, because they are categorical variables. — Asteroid098
– Asteroid098, Commented Dec 2, 2013 at 6:09

IRTFM · Accepted Answer · 2013-12-02 06:16:05Z

3

Try:

train[[a]]=factor(train[[a]])

This does assume, of course that ,a is an object with either a numerical value in the range 1:length(train) or is one of the values in the names(train) vector. If you reference a dataframe using "[" you get a list with one element which happens to be the vector you were hoping to "factorize" but it isn't really a vector but is rather a one element list. The "[[" function instead gives you just the vector.

answered Dec 2, 2013 at 6:16

IRTFM

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4 Comments

Asteroid098 Over a year ago

I get the following error:Error in .subset2(x, i, exact = exact) : recursive indexing failed at level 2.

Asteroid098 Over a year ago

a is a numerical value in the range. The structure of a is integer and a is a vector of certain numbers. I'm trying to select the ith, jth, and etc. columns in the train data frame and turn them into factors.

vamosrafa Over a year ago

So you want individual elements to be listed as factors? Is my understanding correct.

Asteroid098 Over a year ago

Yes. I searched the site and used train[,a] <- lapply(train[,a] , factor) and this worked! Thank you for your refs.

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