In the following we shall use this test vector:
ch <- c('12:23:12','10:23:45','9:08:10')
1) To fix up the solution in the question we prepend a 0 and then replace any string of 3 digits with the last two:
num.substr <- function(...) as.numeric(substr(...))
time2dec <- function(time0) {
t0 <- sub("\\d(\\d\\d)", "\\1", paste0(0, time0))
num.substr(t0, 1, 2) + num.substr(t0, 4, 5) / 60 + num.substr(t0, 7, 8) / 3600
}
time2dec(ch)
## [1] 12.386667 10.395833 9.136111
2) Parsing the string is slightly easier with strapply
in the gsubfn package:
strapply(ch, "^(.?.):(..):(..)",
~ as.numeric(h) + as.numeric(m)/60 + as.numeric(s)/36000,
simplify = c)
## [1] 12.383667 10.384583 9.133611
3) We can reduce the string manipulation to just removing the colons and then convert the resulting character string to numeric so we can manipulate it numerically:
num <- as.numeric(gsub(":", "", ch))
num %/% 10000 + num %% 10000 %/% 100 / 60 + num %% 100 / 3600
## [1] 12.386667 10.395833 9.136111
4) The chron package has a "times"
class that internally represents times as fractions of a day. Converting that to hours gives an easy solution:
library(chron)
24 * as.numeric(times(ch))
## [1] 12.386667 10.395833 9.136111
ADDED Added more solutions.