From timespan (for example “15 min” or “2 sec”) to

Question

I am searching all over help for R function that would convert timespan, for example "15 min" or "1 hour" or "6 sec" or "1 day" into datetime object like "00:15:00" or "01:00:00" or "00:00:06" or "1960-01-02 00:00:00" (not sure for this one). I am sure a function like this exists or there is a neat way to avoid programming it...

To be more specific I would like to do something like this (using made up function name transform.span.to.time):

library(chron)

times(transform.span.to.time("15 min"))

which should yield the same result as

times("00:15:00")

Does a function like transform.span.to.time("15 min") which returns something like "00:15:00" exists or does there exists a trick how to do that?

Answer 1

We will assume a single space separating the numbers and units, and also no trailing space after "secs" unit. This will handle mixed units:

test <- "0 hours 15 min 0 secs"

 transform.span <- function(test){ 
     testh <-         if(!grepl( " hour | hours ", "0 hours 15 min 0 secs")){ 
             # First consequent if no hours
                                      sub("^", "0:", test)} else { 
                                      sub(" hour | hours ", ":", test)}
     testm <- if(!grepl( " min | minutes ", testh)) {    
             #  first consequent if no minutes
                                      sub(" min | minutes ", "0:", testh)} else{
                                      sub(" min | minutes ", ":", testh)  }

     test.s <- if(!grepl( " sec| secs| seconds", testm)) { 
              # first consequent if no seconds
                                       sub(" sec| secs| seconds", "0", testm)} else{ 
                                       sub(" sec| secs| seconds", "", testm)}

     return(times(test.s)) }
###  Use
> transform.span(test)
[1] 00:15:00
> test2 <- "21 hours 15 min 38 secs"
> transform.span(test2)
[1] 21:15:38

Answer 2

@DWin: thank you.

Based on DWin example I rearranged a bit and here is the result:

transform.span<-function(timeSpan) { 
    timeSpanH <- if(!grepl(" hour | hours | hour| hours|hour |hours |hour|hours", timeSpan)) { 
                # First consequent if no hours
                sub("^", "00:", timeSpan)
            } else { 
                sub(" hour | hours | hour| hours|hour |hours |hour|hours", ":", timeSpan)
            }
    timeSpanM <- if(!grepl( " min | minutes | min| minutes|min |minutes |min|minutes", timeSpanH)) {    
                #  first consequent if no minutes
                paste("00:", timeSpanH, sep="")
            } else{
                sub(" min | minutes | min| minutes|min |minutes |min|minutes", ":", timeSpanH)  
            }
    timeSpanS <- if(!grepl( " sec| secs| seconds|sec|secs|seconds", timeSpanM)) { 
                # first consequent if no seconds
                paste(timeSpanM, "00", sep="")
            } else{ 
                sub(" sec| secs| seconds|sec|secs|seconds", "", timeSpanM)
            }

    return(timeSpanS) 
}
###  Use
test <- "1 hour 2 min 1 sec"
times(transform.span(test))
test1hour <- "1 hour"
times(transform.span(test1hour))
test15min <- "15 min"
times(transform.span(test15min))
test4sec <- "4 sec"
times(transform.span(test4sec))

Answer 3

You can define the time span with difftime:

span2time <- function(span,  units = c('mins', 'secs', 'hours')) {
  span.dt <- as.difftime(span,  units = match.arg(units))
  format(as.POSIXct("1970-01-01") + span.dt, "%H:%M:%S")
}

For example:

> span2time(15)
[1] "00:15:00"

EDIT: modified to produce character string acceptable to chron's times.

Answer 4

The first solution uses strapply in the gsubfn package and transforms to days, e.g. 1 hour is 1/24th of a day. The second solution transforms to an R expression which calculates the number of days and then evaluates it.

library(gsubfn)
library(chron)

unit2days <- function(d, u) 
     as.numeric(d) * switch(tolower(u), s = 1, m = 60, h = 3600)/(24 * 3600)
transform.span.to.time <- function(x) 
     sapply(strapply(x, "(\\d+) *(\\w)", unit2days), sum)

Here is a second solution:

library(chron)

transform.span.to.time2 <- function(x) {
    x <- paste(x, 0)
    x <- sub("h\\w*", "*3600+", x, ignore.case = TRUE)
    x <- sub("m\\w*", "*60+", x, ignore.case = TRUE)
    x <- sub("s\\w*", "+", x, ignore.case = TRUE)
    unname(sapply(x, function(x) eval(parse(text = x)))/(24*3600))
}

Tests:

> x <- c("12 hours 3 min 1 sec", "22h", "18 MINUTES 23 SECONDS")
>
> times(transform.span.to.time(x))
[1] 12:03:01 22:00:00 00:18:23
>
> times(transform.span.to.time2(x))
[1] 12:03:01 22:00:00 00:18:23

Answer 5

The base function ?cut.POSIXt does this work for a specified set of values for breaks:

breaks: a vector of cut points _or_ number giving the number of
      intervals which ‘x’ is to be cut into *_or_ an interval
      specification, one of ‘"sec"’, ‘"min"’, ‘"hour"’, ‘"day"’,
      ‘"DSTday"’, ‘"week"’, ‘"month"’, ‘"quarter"’ or ‘"year"’,
      optionally preceded by an integer and a space, or followed by
      ‘"s"’.  For ‘"Date"’ objects only ‘"day"’, ‘"week"’,
      ‘"month"’, ‘"quarter"’ and ‘"year"’ are allowed.*

See the source code by typing in cut.POSIXt, the relevant section starts with this:

else if (is.character(breaks) && length(breaks) == 1L) {

You could adopt the code in this section to work for your needs.