Equal frequency discretization in R

Question

I'm having trouble finding a function in R that performs equal-frequency discretization. I stumbled on the 'infotheo' package, but after some testing I found that the algorithm is broken. 'dprep' seems to no longer be supported on CRAN.

EDIT :

For clarity, I do not need to seperate the values between the bins. I really want equal frequency, it doesn't matter if one value ends up in two bins. Eg :

c(1,3,2,1,2,2) 

should give a bin c(1,1,2) and one c(2,2,3)

Answer 1

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.

Answer 2

This sort of thing is also quite easily solved by using (abusing?) the conditioning plot infrastructure from lattice, in particular function co.intervals():

cutEqual <- function(x, n, include.lowest = TRUE, ...) {
    stopifnot(require(lattice))
    cut(x, co.intervals(x, n, 0)[c(1, (n+1):(n*2))], 
        include.lowest = include.lowest, ...)
}

Which reproduces @Joris' excellent answer:

> set.seed(12345)
> x <- rnorm(50)
> table(cutEqual(x, 5))

 [-2.38,-0.885] (-0.885,-0.115]  (-0.115,0.587]   (0.587,0.938]     (0.938,2.2] 
             10              10              10              10              10
> y <- rpois(50, 5)
> table(cutEqual(y, 5))

 [0.5,3.5]  (3.5,5.5]  (5.5,6.5]  (6.5,7.5] (7.5,11.5] 
        10         13         11          6         10

In the latter, discrete, case the breaks are different although they have the same effect; the same observations are in the same bins.

Answer 3

How about?

a <- rnorm(50)
> table(Hmisc::cut2(a, m = 10))

[-2.2020,-0.7710) [-0.7710,-0.2352) [-0.2352, 0.0997) [ 0.0997, 0.9775) 
               10                10                10                10 
[ 0.9775, 2.5677] 
               10