round but .5 should be floored - r

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

Answer 4

This function works by finding elements that have decimal part equal to 0.5, and adding a small negative number to to them before rounding, ensuring that they'll be rounded downwards. (It relies -- harmlessly but in slightly obfuscated manner --- on the fact that a Boolean vector in R will be converted to a vector of 0's and 1's when multiplied by a numeric vector.)
f <- function(x) {
round(x - .1*(x%%1 == .5))
}
x <- c(0.5,1,1.5,2,2.5,2.01,2.99)
f(x)
[1] 0 1 1 2 2 2 3

Answer 5

The function (not golfed) is very simple and checks whether the decimals that are left are .5 or less. In effect you could easily make it more useful and take 0.5 as an argument:
nice.round <- function(x, myLimit = 0.5) {
bX <- x
intX <- as.integer(x)
decimals <- x%%intX
if(is.na(decimals)) {
decimals <- 0
}
if(decimals <= myLimit) {
x <- floor(x)
} else {
x <- round(x)
}
if (bX > 0.5 & bX < 1) {
x <- 1
}
return(x)
}
Tests
Currently, this function does not work properly with values between 0.5 and 1.0.
> nice.round(1.5)
[1] 1
> nice.round(1.6)
[1] 2
> nice.round(10000.624541)
[1] 10001
> nice.round(0.4)
[1] 0
> nice.round(0.6)
[1] 1