[R] A problem with string handling to make a time duration

[John Laing]

Regular expressions are the tool for this problem. This pattern
matches your input data:

t <- c("10H20M33S", "1H1M", "1M", "21M9S", "2H55S")
patt <- "^(([0-9]+)H)?(([0-9]+)M)?(([0-9]+)S)?$"
all(grepl(patt, t)) # TRUE

We can use the pattern to extract hour/minute/second components

hms <- lapply(c(h="\\2", m="\\4", s="\\6"), function(r) sub(patt, r, t))

And then just plug those components back into the desired format

formatted <- gsub(" ", "0", sprintf("%2s:%2s:%2s", hms$h, hms$m, hms$s))

In the last line we need the gsub because zero-padding with %02s seems
to be platform-dependent.

JL

On Mon, May 4, 2015 at 3:59 PM, gavinr <g.rudge at bham.ac.uk> wrote:
> I have a character string that represents a time duration. It has an hours
> minutes seconds structure(ish) but with letters denoting units (H,M or S) no
> leading zeros and no placeholder at all where one or other of the units are
> not required.
>
> It looks like this:
>
> t<-c("10H20M33S","1H1M","1M","21M9S","2H55S" ))
> df<-data.frame(t)
> df
>
> #ideally should look like:
> t2<-c("10:20:33","01:00:01","00:01:00","00:21:09","02:00:55")
> df2<-data.frame(t2)
> df2
>
> I need to get it into hours minutes and seconds either in time format or as
> a string with leading zeros and all three time units represented in each
> one, as in df2.  The data, part of a very large dataset, are for onward use
> and processing in a GIS application.  I’ve messed about with string handling
> statements in SQL to no avail, but wondered if R would be a better bet?
> I’ve had a look at some of the commands in stringr, but am unsure how to
> operationalise a solution using this package.  Any advice is welcome.
>
>
>
>
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