round but .5 should be floored

Question

From R help function: Note that for rounding off a 5, the IEC 60559 standard is expected to be used, ‘go to the even digit’. Therefore round(0.5) is 0 and round(-1.5) is -2.

> round(0.5)
[1] 0
> round(1.5)
[1] 2
> round(2.5)
[1] 2
> round(3.5)
[1] 4
> round(4.5)
[1] 4

But I need all values ending with .5 to be rounded down. All other values should be rounded as it they are done by round() function. Example:

round(3.5) = 3
round(8.6) = 9
round(8.1) = 8
round(4.5) = 4

Is there a fast way to do it?

Answer 1

Per Dietrich Epp's comment, you can use the ceiling() function with an offset to get a fast, vectorized, correct solution:

round_down <- function(x) ceiling(x - 0.5) round_down(seq(-2, 3, by = 0.5)) ## [1] -2 -2 -1 -1  0  0  1  1  2  2  3 

I think this is faster and much simpler than many of the other solutions shown here.

As noted by Carl Witthoft, this adds much more bias to your data than simple rounding. Compare:

mean(round_down(seq(-2, 3, by = 0.5))) ## [1] 0.2727273 mean(round(seq(-2, 3, by = 0.5))) ## [1] 0.4545455 mean(seq(-2, 3, by = 0.5)) ## [1] 0.5 

What is the application for such a rounding procedure?

Answer 2

Check if the remainder of x %% 1 is equal to .5 and then floor or round the numbers:

x <- seq(1, 3, 0.1) ifelse(x %% 1 == 0.5, floor(x), round(x)) > 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2 2 3 3 3 3 3 

Answer 3

I'll join the circus too:

rndflr <- function(x) {
  sel <- vapply(x - floor(x), function(y) isTRUE(all.equal(y, 0.5)), FUN.VALUE=logical(1))
  x[sel] <- floor(x[sel])
  x[!sel] <- round(x[!sel])  
  x
}

rndflr(c(3.5,8.6,8.1,4.5))
#[1] 3 9 8 4