Computing percentile rank in R How can I add new variable into data frame which will be percentile rank of one of the variables? I can do this in Excel easily, but I really want to do that in R.
Thanks
 A: If your original data.frame is called dfr and the variable of interest is called myvar, you can use dfr$myrank<-rank(dfr$myvar) for normal ranks, or dfr$myrank<-rank(dfr$myvar)/length(myvar) for percentile ranks.
Oh well. If you really want it the Excel way (may not be the simplest solution, but I had some fun using new (to me) functions and avoiding loops):
percentilerank<-function(x){
  rx<-rle(sort(x))
  smaller<-cumsum(c(0, rx$lengths))[seq(length(rx$lengths))]
  larger<-rev(cumsum(c(0, rev(rx$lengths))))[-1]
  rxpr<-smaller/(smaller+larger)
  rxpr[match(x, rx$values)]
}

so now you can use dfr$myrank<-percentilerank(dfr$myvar)
HTH.
A: Given a vector of raw data values, a simple function might look like
perc.rank <- function(x, xo)  length(x[x <= xo])/length(x)*100

where x0 is the value for which we want the percentile rank, given the vector x, as suggested on R-bloggers. 
However, it might easily be vectorized as
perc.rank <- function(x) trunc(rank(x))/length(x)

which has the advantage of not having to pass each value. So, here is an example of use:
my.df <- data.frame(x=rnorm(200))
my.df <- within(my.df, xr <- perc.rank(x))

A: A problem with the presented answer is that it will not work properly, when you have NAs.    
In this case, another possibility (inspired by the function from chl♦) is:
perc.rank <- function(x) trunc(rank(x,na.last = NA))/sum(!is.na(x))
quant <- function (x, p.ile) {
      x[which.min(x = abs(perc.rank(x-(p.ile/100))))]
}

Here, x is the vector of values, and p.ile is the percentile by rank. 2.5 percentile by rank of (arbitrary) coef.mat may be calculated by:  
quant(coef.mat[,3], 2.5)  
[1] 0.00025  

or as a single function:
quant <- function (x, p.ile) {
   perc.rank <- trunc(rank(x,na.last = NA))/sum(!is.na(x))
   x = na.omit(x)
   x[which.min(x = abs(perc.rank(x-(p.ile/100))))]
}

