How to check if each element in a vector is intege

Question

Say, I have a vector y, and I want to check if each element in y is integer or not, and if not, stop with an error message. I tried is.integer(y), but it does not work.

Answer 1

The simplest (and fastest!) thing is probably this:

stopifnot( all(y == floor(y)) )

...So trying it out:

y <- c(3,4,9)
stopifnot( all(y == floor(y)) ) # OK

y <- c(3,4.01,9)
stopifnot( all(y == floor(y)) ) # ERROR!

If you want a better error message:

y <- c(3, 9, NaN)
if (!isTRUE(all(y == floor(y)))) stop("'y' must only contain integer values")

Answer 2

I went in a completely different direction then Tim (I like his better though my approach works on a mixed vector that's a character vector with integers etc.):

int.check <- function(vect) {
    vect <- as.character(vect)
    sapply(vect, function(x) all(unlist(strsplit(x, ""))%in% 0:9))
}

x <- c(2.0, 1111,"x", 2.4)
int.check(x)

EDIT: altered the function as it only worked on character vectors.

This works on vectors of the class character as well in case you have a character vector with various number intermixed but that have been coerced to character.

Answer 3

checking the following helps with a crisp if condition which we can use on scripting.

sff <- 5

if(!(is.integer(sff) == is.character(sff))){ 
  sff
} else {
  "hello"
}

gives

hello

sff <- 'a' gives 'a' as the result.

Answer 4

If you have floating-point representation error, try:

round( y, TOLERANCE.DIGITS ) %% 1 == 0

In my application, I had seriously brutal floating-point representation error, such that:

> dictionary$beta[3]
[1] 89
> floor(dictionary$beta[3])
[1] 88
> as.integer( dictionary$beta )[3]
[1] 88
> dictionary$beta[3] %% 1
[1] 1

the remainder divided by one was one. I found that I had to round before I took the integer. I think all of these tests would fail in the case where you wanted the above 89 to count as an integer. The "all.equal" function is meant to be the best way to handle floating-point representation error, but:

all.equal( 88, 89 );

as in my case, would have (and did) given a false negative for an integer value check.

EDIT: In benchmarking, I found that:

(x == as.integer(x)) 

was universally the best performer.

(x == floor(x))
((x - as.integer(x)) == 0)

usually worked well, often just as fast.

(x %% 1 <= tolerance)

works, but not as quickly as the others

!(is.character(all.equal(x, as.integer(x)))) 

when the vector wasn't integers, had terrible performance (certainly because it goes to the trouble of estimating the difference).

identical(x, as.integer(x)) 

when the vector was all integer values, returned the incorrect result (assuming the question was meant to check for integer values, not integer types).

Answer 5

Here's another way (using the same trick as Justin of comparing each number to that number coerced into the 'integer' type):

R> v1 = c(1,2,3)
R> v2 = c(1,2,3.5)
R> sapply(v1, function(i) i == as.integer(i))
[1] TRUE TRUE TRUE
R> sapply(v2, function(i) i == as.integer(i))
[1]  TRUE  TRUE FALSE

To make your test:

R> all(sapply(v2, function(i) i == as.integer(i)))
[1] FALSE

Answer 6

you could do:

   y <- c(3,3.1,1,2.3)
   (y - floor(y)) == 0
    [1]  TRUE FALSE  TRUE FALSE

or

   (y - round(y)) == 0

and if you want a single TRUE or FALSE for the whole thing, put it in all(), e.g.:

   all((y - round(y)) == 0)
    [1] FALSE