R character factor to numeric vector

I read in a csv file with "3:29" in one of the fields (without the quotation marks). This comes up as a factor. How can I convert this to a numeric vector e.g. c(3:29)? I tried as.vector() but this gives a string vector "3,4,5,6...29" (with the quotation marks, still character class).

EDIT Answer needs to be applicable to more general form, for exam开发者_如何学JAVAple, the column could contain 3:6,7,9:11, which needs to be converted to the equivalent c(3:6,7,9:11).

You can do:

> eval(parse(text='3:29'))
 [1]  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27
[26] 28 29

Split the string on : and convert to a vector of numerics and generate the call to seq() by hand:

> vars <- as.numeric(strsplit("3:29", ":")[[1]])
> seq(from = vars[1], to = vars[2], by = 1)
 [1]  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27
[26] 28 29

or slightly more elegantly by getting R to build the call to `:()` directly:

> do.call(`:`, as.list(as.numeric(strsplit("3:29", ":")[[1]])))
 [1]  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27
[26] 28 29

[Updated in light of Edit to original Q]

In the spirit of:

> require(fortunes)
> fortune(106)

If the answer is parse() you should usually rethink the question.
   -- Thomas Lumley
      R-help (February 2005)

this is as close as I can get without using parse():

unlist(lapply(strsplit(strsplit(txt, ",")[[1]], ":"),
       function(x) {
           x <- as.numeric(x)
           if(length(x) == 2) {
               seq(x[1], x[2], by = 1) ## `:`(x[1], x[2])
           } else { 
               x[1]
           }
       }))

yielding:

[1]  3  4  5  6  7  9 10 11

...but it makes me thing this might be one of those times when using parse() might make sense ;-)