Check if the number is integer

Question

Another alternative is to check the fractional part:

x%%1==0

or, if you want to check within a certain tolerance:

min(abs(c(x%%1, x%%1-1))) < tol

Here's a solution using simpler functions and no hacks:

all.equal(a, as.integer(a))

What's more, you can test a whole vector at once, if you wish. Here's a function:

testInteger <- function(x){
  test <- all.equal(x, as.integer(x), check.attributes = FALSE)
  if(test == TRUE){ return(TRUE) }
  else { return(FALSE) }
}

You can change it to use *apply in the case of vectors, matrices, etc.

Here is one, apparently reliable way:

check.integer <- function(N){
    !grepl("[^[:digit:]]", format(N,  digits = 20, scientific = FALSE))
}

check.integer(3243)
#TRUE
check.integer(3243.34)
#FALSE
check.integer("sdfds")
#FALSE

This solution also allows for integers in scientific notation:

> check.integer(222e3)
[1] TRUE

Reading the R language documentation, as.integer has more to do with how the number is stored than if it is practically equivalent to an integer. is.integer tests if the number is declared as an integer. You can declare an integer by putting a L after it.

> is.integer(66L)
[1] TRUE
> is.integer(66)
[1] FALSE

Also functions like round will return a declared integer, which is what you are doing with x==round(x). The problem with this approach is what you consider to be practically an integer. The example uses less precision for testing equivalence.

> is.wholenumber(1+2^-50)
[1] TRUE
> check.integer(1+2^-50)
[1] FALSE

So depending on your application you could get into trouble that way.