Test for equality among all elements of a single v

问题:

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.

回答1:

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.

回答2:

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.

回答3:

Why not simply using the variance:

var(x) == 0

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

回答4:

> isTRUE(all.equal( max(y) ,min(y)) )
[1] TRUE
> isTRUE(all.equal( max(x) ,min(x)) )
[1] FALSE

Another along the same lines:

> diff(range(x)) < .Machine$double.eps ^ 0.5
[1] FALSE
> diff(range(y)) < .Machine$double.eps ^ 0.5
[1] TRUE

回答5:

You can use identical() and all.equal() by comparing the first element to all others, effectively sweeping the comparison across:

R> compare <- function(v) all(sapply( as.list(v[-1]), 
+                         FUN=function(z) {identical(z, v[1])}))
R> compare(x)
[1] FALSE
R> compare(y)
[1] TRUE
R> 

That way you can add any epsilon to identical() as needed.

回答6:

Since I keep coming back to this question over and over, here's an Rcpp solution that will generally be much much faster than any of the R solutions if the answer is actually FALSE (because it will stop the moment it encounters a mismatch) and will have the same speed as the fastest R solution if the answer is TRUE. For example for the OP benchmark, system.time clocks in at exactly 0 using this function.

library(inline)
library(Rcpp)

fast_equal = cxxfunction(signature(x = 'numeric', y = 'numeric'), '
  NumericVector var(x);
  double precision = as<double>(y);

  for (int i = 0, size = var.size(); i < size; ++i) {
    if (var[i] - var[0] > precision || var[0] - var[i] > precision)
      return Rcpp::wrap(false);
  }

  return Rcpp::wrap(true);
', plugin = 'Rcpp')

fast_equal(c(1,2,3), 0.1)
#[1] FALSE
fast_equal(c(1,2,3), 2)
#[2] TRUE

回答7:

I wrote a function specifically for this, which can check not only elements in a vector, but also capable of checking if all elements in a list are identical. Of course it as well handle character vectors and all other types of vector well. It also has appropriate error handling.

all_identical <- function(x) {
  if (length(x) == 1L) {
    warning("'x' has a length of only 1")
    return(TRUE)
  } else if (length(x) == 0L) {
    warning("'x' has a length of 0")
    return(logical(0))
  } else {
    TF <- vapply(1:(length(x)-1),
                 function(n) identical(x[[n]], x[[n+1]]),
                 logical(1))
    if (all(TF)) TRUE else FALSE
  }
}

Now try some examples.

x <- c(1, 1, 1, NA, 1, 1, 1)
all_identical(x)       ## Return FALSE
all_identical(x[-4])   ## Return TRUE
y <- list(fac1 = factor(c("A", "B")),
          fac2 = factor(c("A", "B"), levels = c("B", "A"))
          )
all_identical(y)     ## Return FALSE as fac1 and fac2 have different level order

回答8:

You do not actually need to use min, mean, or max. Based on John's answer:

all(abs(x - x[[1]]) < tolerance)

回答9:

Here an alternative using the min, max trick but for a data frame. In the example I am comparing columns but the margin parameter from apply can be changed to 1 for rows.

valid = sum(!apply(your_dataframe, 2, function(x) diff(c(min(x), max(x)))) == 0)

If valid == 0 then all the elements are the same