Test for equality among all elements of a single numeric vector

I use this method, which compares the min and the max, after dividing by the mean:

# Determine if range of vector is FP 0.
zero_range <- function(x, tol = .Machine$double.eps ^ 0.5) {
  if (length(x) == 1) return(TRUE)
  x <- range(x) / mean(x)
  isTRUE(all.equal(x[1], x[2], tolerance = tol))
}

If you were using this more seriously, you'd probably want to remove missing values before computing the range and mean.

If they're all numeric values then if tol is your tolerance then...

all( abs(y - mean(y)) < tol ) 

is the solution to your problem.

EDIT:

After looking at this, and other answers, and benchmarking a few things the following comes out over twice as fast as the DWin answer.

abs(max(x) - min(x)) < tol

This is a bit surprisingly faster than diff(range(x)) since diff shouldn't be much different than - and abs with two numbers. Requesting the range should optimize getting the minimum and maximum. Both diff and range are primitive functions. But the timing doesn't lie.

And, in addition, as @Waldi pointed out, abs is superfluous here.

Why not simply using the variance:

var(x) == 0

If all the elements of x are equal, you will get a variance of 0. This works only for double and integers though.

Edit based on the comments below:
A more generic option would be to check for the length of unique elements in the vector which must be 1 in this case. This has the advantage that it works with all classes beyond just double and integer from which variance can be calculated from.

length(unique(x)) == 1

You can just check all(v==v[1])