Test for equality among all elements of a single numeric vector

I'm trying to test whether all elements of a vector are equal to one another. The solutions I have come up with seem somewhat roundabout, both involving checking length().

x <- c(1, 2, 3, 4, 5, 6, 1)  # FALSE
y <- rep(2, times = 7)       # TRUE

With unique():

length(unique(x)) == 1
length(unique(y)) == 1

With rle():

length(rle(x)$values) == 1
length(rle(y)$values) == 1

A solution that would let me include a tolerance value for assessing 'equality' among elements would be ideal to avoid FAQ 7.31 issues.

Is there a built-in function for type of test that I have completely overlooked? identical() and all.equal() compare two R objects, so they won't work here.

Edit 1

Here are some benchmarking results. Using the code:

library(rbenchmark)

John <- function() all( abs(x - mean(x)) < .Machine$double.eps ^ 0.5 )
DWin <- function() {diff(range(x)) < .Machine$double.eps ^ 0.5}
zero_range <- function() {
  if (length(x) == 1) return(TRUE)
  x <- range(x) / mean(x)
  isTRUE(all.equal(x[1], x[2], tolerance = .Machine$double.eps ^ 0.5))
}

x <- runif(500000);

benchmark(John(), DWin(), zero_range(),
  columns=c("test", "replications", "elapsed", "relative"),
  order="relative", replications = 10000)

With the results:

          test replications elapsed relative
2       DWin()        10000 109.415 1.000000
3 zero_range()        10000 126.912 1.159914
1       John()        10000 208.463 1.905251

So it looks like diff(range(x)) < .Machine$double.eps ^ 0.5 is fastest.

Why not simply using the variance:

var(x) == 0

If all the elements of x are equal, you will get a variance of 0.

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.