EDIT : given your real goal, why don't you just do (corrected) :
EqualFreq2 <- function(x,n){
nx <- length(x)
nrepl <- floor(nx/n)
nplus <- sample(1:n,nx - nrepl*n)
nrep <- rep(nrepl,n)
nrep[nplus] <- nrepl+1
x[order(x)] <- rep(seq.int(n),nrep)
x
}
This returns a vector with indicators for which bin they are. But as some values might be present in both bins, you can't possibly define the bin limits. But you can do :
x <- rpois(50,5)
y <- EqualFreq2(x,15)
table(y)
split(x,y)
Original answer:
You can easily just use cut()
for this :
EqualFreq <-function(x,n,include.lowest=TRUE,...){
nx <- length(x)
id <- round(c(1,(1:(n-1))*(nx/n),nx))
breaks <- sort(x)[id]
if( sum(duplicated(breaks))>0 stop("n is too large.")
cut(x,breaks,include.lowest=include.lowest,...)
}
Which gives :
set.seed(12345)
x <- rnorm(50)
table(EqualFreq(x,5))
[-2.38,-0.886] (-0.886,-0.116] (-0.116,0.586] (0.586,0.937] (0.937,2.2]
10 10 10 10 10
x <- rpois(50,5)
table(EqualFreq(x,5))
[1,3] (3,5] (5,6] (6,7] (7,11]
10 13 11 6 10
As you see, for discrete data an optimal equal binning is rather impossible in most cases, but this method gives you the best possible binning available.