Remove extra string characters in decimal formatting

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how do I remove just the ".00" in a data frame column: so that 5.00% /GP per tonne becomes 5% / GP per tonne ? I am trying something like the below but it is saying invalid regexp.


sched$duty_rate <- c("5.00% /GP per tonne","10.00% /GP per tonne")

for(i in 1:nrow(sched)){
  sched$duty_rate[i] <- gsub("[.]?[0-9]*(?=%)","\\%",sched$duty_rate[i])
}
r regex
4 Answers

You can use

sched$duty_rate <- gsub("\\.0+(%)", "\\1", sched$duty_rate)

Details:

  • \. - a literal dot
  • 0+ - one or more 0 chars
  • (%) - Group 1: a % char (so that we do not remove zeros from other contexts, only between . and %)

See the R demo:

duty_rate <- c("5.00% /GP per tonne","10.00% /GP per tonne", "5.0% /GP per 2.000 tonne","10.0000% /GP per 20.000 tonne")
gsub("\\.0+(%)", "\\1", duty_rate)
## => [1] "5% /GP per tonne"         "10% /GP per tonne"       
##    [3] "5% /GP per 2.000 tonne"   "10% /GP per 20.000 tonne"

See the regex demo.

user_38329700

You can use sub, i.e.

sub('.00', '', sched$duty_rate, fixed = TRUE)
[1] "5% /GP per tonne"  "10% /GP per tonne"
user_56355800

Also, fairly simple and readable implementation can be derived with use of str_remove function offered via the stringr package.

x <- c("5.00% /GP per tonne", "10.00% /GP per tonne")
stringr::str_remove(string = x, pattern = ".00")
#> [1] "5% /GP per tonne"  "10% /GP per tonne"

Created on 2022-04-06 by the reprex package (v2.0.1)

user_16555670

Your pattern [.]?[0-9]*(?=%) matches an optional dot, optional digits and asserts % to the right which matches more than a mandatory dot and only zeroes.

You can make the positive lookahead work using perl=TRUE and match a dot and one or more zeroes using a more specific pattern [.]0+(?=%)

duty_rate <- c("5.00% /GP per tonne","10.00% /GP per tonne", "5.0% /GP per 2.000 tonne","10.0000% /GP per 20.000 tonne")
gsub("[.]0+(?=%)", "", duty_rate, perl=TRUE)

Output

[1] "5% /GP per tonne"         "10% /GP per tonne"       
[3] "5% /GP per 2.000 tonne"   "10% /GP per 20.000 tonne"

See an R demo.

user_54249880
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