Reverting to Factor Codes R - r

Question

Let's say I have a data.frame that looks like this:
df.test <- data.frame(1:26, 1:26)
colnames(df.test) <- c("a","b")
and I apply a factor:
df.test$a <- factor(df.test$a, levels=c(1:26), labels=letters)
Now, how I would like to convert it back the integer codes:
as.numeric(df.test[1])## replies with an error code.
But this works:
as.numeric(df.test$a)
Why is that?

Answer 1

Actually Joshua's link are not applicable here because the task is not coverting from a factor with levels that have numeric interpretation. Your original effort that produced an error was almost correct. It was missing only a comma before the 1:
df.test <- data.frame(1:26, 1:26)
colnames(df.test) <- c("a","b")
df.test$a <- factor(df.test$a, levels=c(1:26), labels=letters)
as.numeric(df.test[,1])
# [1] 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18
# [19] 19 20 21 22 23 24 25 26
Or you could have used "[["
> as.numeric(df.test[[1]])
[1] 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18
[19] 19 20 21 22 23 24 25 26

Answer 2

as.numeric will convert a factor to numeric:
as.numeric(df.test$a)

Answer 3

Accessing a column by name gives you a factor vector, which can be converted to numeric.
However, a data frame is a list (of columns), and when you use the single bracket operator and a single number on a list, you get a list of length one. The same applies for data frames, so df.test[1] gets you column one as a new data frame, which cannot be coerced by as.numeric(). I did not know this!
> str(df.test$a)
Factor w/ 26 levels "a","b","c","d",..: 1 2 3 4 5 6 7 8 9 10 ...
> str(df.test[1])
'data.frame': 26 obs. of 1 variable:
$ a: Factor w/ 26 levels "a","b","c","d",..: 1 2 3 4 5 6 7 8 9 10 ...

Answer 4

To respond to your edit: Keep in mind that a factor has two parts: 1) the labels, and 2) the underlying integer codes. The two answers I linked to in my comment were to convert the labels to numeric. If you just want to get the internal codes, use as.integer(df.test$a) as demonstrated in the examples section of ?factor. aL3xa answered your question about why as.numeric(df.test[1]) throws an error.