Question

I have a data.frame called mydata and a vector ids containing indices of the columns in the data.frame that I would like to convert to factors. Now the following code solves the problem

for(i in ids) mydata[, i]<-as.factor(mydata[, i])

Now I wanted to clean this code up by using apply instead of an explicit for-loop.

mydata[, ids]<-apply(mydata[, ids], 2, as.factor)

However, the last statement gives me a data.frame where the types are character instead of factors. I fail to see the distinction between these two lines of code. Why do they not produce the same result?

Kind regards,
Michael

Answer 1

The result of apply is a vector or array or list of values (see ?apply).

For your problem, you should use lapply instead:

data(iris)
iris[, 2:3] <- lapply(iris[, 2:3], as.factor)
str(iris)

'data.frame':   150 obs. of  5 variables:
 $ Sepal.Length: num  5.1 4.9 4.7 4.6 5 5.4 4.6 5 4.4 4.9 ...
 $ Sepal.Width : Factor w/ 23 levels "2","2.2","2.3",..: 15 10 12 11 16 19 14 14 9 11 ...
 $ Petal.Length: Factor w/ 43 levels "1","1.1","1.2",..: 5 5 4 6 5 8 5 6 5 6 ...
 $ Petal.Width : num  0.2 0.2 0.2 0.2 0.2 0.4 0.3 0.2 0.2 0.1 ...
 $ Species     : Factor w/ 3 levels "setosa","versicolor",..: 1 1 1 1 1 1 1 1 1 1 ...

---

Notice that this is one place where lapply will be much faster than a for loop. In general a loop and lapply will have similar performance, but the <-.data.frame operation is very slow. By using lapply one avoids the <- operation in each iteration, and replaces it with a single assign. This is much faster.