Rounding in R, possible error

Question

I was always taught that for numbers 1,2,3 and 4 you round DOWN and for 5,6,7,8 and 9 you round UP. So can someone please explain to me why when using round or signif in R on 6.5 it rounds it down to 6??

round(6.5)
[1] 6
signif(6.5)
[1] 6

i need my values to round up when it gives a .5 number. Can someone please tell me how to do this?

hmm ok then....but i NEED my numbers to be rounded up when its .5 so how do i do this automatically? — Summer-Jade Gleek'away, Jan 07 '15 at 15:22
Check if you are at a half and use `ceiling` instead of `round` — James, Jan 07 '15 at 15:24
You could add a very small number to your numbers vector too, something like `round(6.5 + 1e-15)`. `+` is vectorised, so you can add `1e-15` to a whole vector within a single line — David Arenburg, Jan 07 '15 at 15:27
To expand on what @DavidArenburg said, you could do `sign(x) * round(abs(x) + 1e-15)` to account for negative numbers as well. — blakeoft, Jan 07 '15 at 15:28
thankyou :) i dont suppose you also might know how to round UP any number? e.g. if it is 4.1 then round it up to 5? — Summer-Jade Gleek'away, Jan 07 '15 at 15:29
@Summer-JadeGleek'away You want to use `ceiling` for rounding up. — blakeoft, Jan 07 '15 at 15:30
Just do `?round` and you'll see all the related functions documentaion — David Arenburg, Jan 07 '15 at 15:31

LyzandeR · Answer 1 · 2015-01-07T18:25:04.170

1

I did this:

round2 <- function(x) {
  index <- gregexpr(pattern='\\.', as.character(x)  ) #find the index of the dot
  if ( substr(as.character(x), index[[1]]+1, index[[1]]+1 ) =='5' ) round(x+0.1) else round(x) #check if the first decimal is a 5
}

> round2(5.5)
[1] 6
> round2(255.5)
[1] 256
> round2(-55.5)
[1] -56

This seems to be working in any case.

edited Jan 07 '15 at 18:25

answered Jan 07 '15 at 15:29

LyzandeR

37,047
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Maybe replace the 3s with nchar(x) in case it has more or less digits? – talat Jan 07 '15 at 15:32
3

This gets harder, because in general you want to look at the character immediately after the decimal place. I think I prefer the answers at the linked/duplicate question. – Ben Bolker Jan 07 '15 at 15:34
@BenBolker aaah yes thanks Ben. I will try something else then. – LyzandeR Jan 07 '15 at 15:36
@docendodiscimus in case I had 4 digits `nchar` wouldn't help though would it? – LyzandeR Jan 07 '15 at 15:36
True, didn't think it through properly.. Just had the feeling that 3,3 might be too static – talat Jan 07 '15 at 15:38
@LyzandeR Would this `sub(".*\\.(.*)", "\\1", as.character(4.5)) == "5"` work as your condition? – blakeoft Jan 07 '15 at 15:41
@blakeoft your way didnt work for x=25.4... – LyzandeR Jan 07 '15 at 15:50
@BenBolker I think this is more generic now. But probably the dup is better – LyzandeR Jan 07 '15 at 15:51
@LyzandeR sorry, I meant `sub(".*\\.(.*)", "\\1", as.character(x)) == "5"`. I had 4.5 in place of x before. – blakeoft Jan 07 '15 at 15:52
1

You should never use text processing functions for mathematical operations. – Roland Jan 07 '15 at 15:55
@ping miss-typed something. This works now. – LyzandeR Jan 07 '15 at 16:05
@Roland I agree 100% but isn't this a simple case? – LyzandeR Jan 07 '15 at 16:05
Why would you do this if you could just use `sign(x) * trunc(abs(x) + 0.5)`, which works also for negative numbers? – Roland Jan 07 '15 at 16:08
@Roland Only because I hadn't thought about it this way. I added to my answer. Thanks – LyzandeR Jan 07 '15 at 16:13
That's the top answer to the duplicate question ... – Roland Jan 07 '15 at 16:22
yep, sorry, my bad... – LyzandeR Jan 07 '15 at 16:24

Rounding in R, possible error

1 Answers1