How can I take this
d <- "3:10"
and magically turn it into this
[1] 3 4 5 6 7 8 9 10
I've seen this done before, and searched SO extensively, but was unable to recall anything.
Here's what I've tried.
> eval(d)
# [1] "3:10"
> eval(noquote(d))
# [1] 3:10
> noquote(eval(d))
# [1] 3:10
> evalq(d)
# [1] "3:10"
> substitute(d)
# d
I think you are looking for the dreaded eval(parse(...)) construct.
Try:
d <- "3:10"
eval(parse(text = d))
# [1] 3 4 5 6 7 8 9 10
An interesting approach to this could be:
Reduce(":", as.numeric(strsplit(d, ":", TRUE)[[1]]))
# [1] 3 4 5 6 7 8 9 10
And, whaddyaknow! I wouldnathunkit, but it's faster than eval(parse(...)) (but your approach in the comment is faster yet).
fun1 <- function() Reduce(":", as.numeric(strsplit(d, ":", TRUE)[[1]]))
fun2 <- function() eval(parse(text = d))
fun3 <- function() {
s <- as.numeric(strsplit(d, ":", fixed = TRUE)[[1]])
s[1]:s[2]
}
library(microbenchmark)
microbenchmark(fun1(), fun2(), fun3())
# Unit: microseconds
# expr min lq median uq max neval
# fun1() 24.375 26.0865 32.2865 55.8070 113.751 100
# fun2() 108.192 112.4680 121.4490 204.8375 453.720 100
# fun3() 8.553 10.6920 12.8300 20.9550 40.198 100
eval especially. Is it better to go with s <- as.numeric(strsplit(d, ":")[[1]]); s[1]:s[2] ?fortune(106)? I'll dig up a couple of links. I've never had any problems with it myself though.eval(parse(...)) construct. Regarding your subquestion, how did the data get in this form?d <- "12:18" with your approach.