How to format a number as percentage in R?

Question

One of the things that used to perplex me as a newby to R was how to format a number as a percentage for printing.

For example, display 0.12345 as 12.345%. I have a number of workarounds for this, but none of these seem to be "newby friendly". For example:

set.seed(1)
m <- runif(5)

paste(round(100*m, 2), "%", sep="")
[1] "26.55%" "37.21%" "57.29%" "90.82%" "20.17%"

sprintf("%1.2f%%", 100*m)
[1] "26.55%" "37.21%" "57.29%" "90.82%" "20.17%"

Question: Is there a base R function to do this? Alternatively, is there a widely used package that provides a convenient wrapper?

---

Despite searching for something like this in ?format, ?formatC and ?prettyNum, I have yet to find a suitably convenient wrapper in base R. ??"percent" didn't yield anything useful. library(sos); findFn("format percent") returns 1250 hits - so again not useful. ggplot2 has a function percent but this gives no control over rounding accuracy.

Answer 1

You can use the scales package just for this operation (without loading it with require or library)

scales::percent(m)

Answer 2

try this~

data_format <- function(data,digit=2,type='%'){
if(type=='d') {
    type = 'f';
    digit = 0;
}
switch(type,
    '%' = {format <- paste("%.", digit, "f%", type, sep='');num <- 100},
    'f' = {format <- paste("%.", digit, type, sep='');num <- 1},
    cat(type, "is not a recognized type\n")
)
sprintf(format, num * data)
}

Answer 3

Seeing how scalable::percent had already been shown to be slowest and Liliana Pacheco offering up another solution, I went ahead and tried to benchmark it against some of the other options based on the example Michael set:

library(microbenchmark)
library(scales)
library(formattable)

x<-runif(1e5)

lilip <- function() formattable::percent(x,2)
krlmlr <- function() scales::percent(x)
andrie1 <- function() paste0(round(x,4) * 100, '%')

microbenchmark(times=100L,lilip(), krlmlr(), andrie1())

These are the results I got:

Unit: microseconds
      expr        min          lq        mean      median          uq        max neval
   lilip()    194.562    373.7335    772.5663    889.7045    950.4035   1611.537   100
  krlmlr() 226270.845 237985.6560 260194.9269 251581.0235 280704.2320 373022.180   100
 andrie1()  87916.021  90437.4820  92791.8923  92636.8420  94448.7040 102543.252   100

I have no idea, though, why my krlmlr() and andrie1() performed so much worse than in MichaelChirico's example. Any clues?