I am stuck with something simple: I have vector of minutes and seconds like, e.g., myvec=c("10:56","2:32","0:00","0:00","62:59","60:40","46:23"). Here the number before the : notes the minutes and after the colon we have the seconds.
I tried
difftime(myvec,format='%M:%S')
Error in difftime(myvec, format = "%M:%S") :
unused argument (format = "%M:%S")
however my problem seems to be that some minutes values are bigger than 60. I am a bit startled that I can't think of any straightforward solution. Maybe I just misunderstood the difftime documentation. Sorry if this creates a duplicate.
EDIT:
Since some people asked:
as.difftime(myvec,format='%M:%S',units='mins')
10.933333 2.533333 0.000000 0.000000 NA NA 46.383333
Instead of the two NAs I want the command to return something like 62.9999 60.6666. At this point I am just curious whether there is any solution without using regex.
Here are some alternatives. Except for (2) they use no packages.
1) sub Extract the numbers, convert to numeric and compute the seconds.
60 * as.numeric(sub(":.*", "", myvec)) + as.numeric(sub(".*:", "", myvec))
## [1] 656 152 0 0 3779 3640 2783
2) strapply We could also use strapply in gsubfn:
library(gsubfn)
strapply(myvec, "(.+):(.+)", ~ 60 * as.numeric(min) + as.numeric(sec), simplify = TRUE)
## [1] 656 152 0 0 3779 3640 2783
3) scan Another solution can be based on scan and matrix multiplication:
c(c(60, 1) %*% matrix(scan(text = myvec, sep = ":", quiet = TRUE), 2))
## [1] 656 152 0 0 3779 3640 2783
4) eval eval has a bad reputation but if you don't mind using it then it is the shortest of the solutions here.
sapply(parse(text = paste0("60*", sub(":", "+", myvec))), eval)
## [1] 656 152 0 0 3779 3640 2783