divide a range of values in bins of equal length: cut vs cut2

Question

I\'m using the cut function to split my data in equal bins, it does the job but I\'m not happy with the way it returns the values. What I need is the center of the bin not t

Answer 1

It's not too hard to make the breaks and labels yourself, with something like this. Here since the midpoint is a single number, I don't actually return a factor with labels but instead a numeric vector.

cut2 <- function(x, breaks) {
  r <- range(x)
  b <- seq(r[1], r[2], length=2*breaks+1)
  brk <- b[0:breaks*2+1]
  mid <- b[1:breaks*2]
  brk[1] <- brk[1]-0.01
  k <- cut(x, breaks=brk, labels=FALSE)
  mid[k]
}

There's probably a better way to get the bin breaks and midpoints; I didn't think about it very hard.

Note that this answer is different than Joshua's; his gives the median of the data in each bins while this gives the center of each bin.

> head(cut2(x,3))
[1] 16.666667  3.333333 16.666667  3.333333 16.666667 16.666667
> head(ave(x, cut(x,3), FUN=median))
[1] 18  2 18  2 18 18