How to round up to the nearest 10 (or 100 or X)?

Question

I am writing a function to plot data. I would like to specify a nice round number for the y-axis max that is greater than the max of the dataset.

Specif

Answer 1

How about:

roundUp <- function(x,to=10)
{
  to*(x%/%to + as.logical(x%%to))
}

Which gives:

> roundUp(c(4,6.1,30.1,100.1))
[1]  10  10  40 110
> roundUp(4,5)
[1] 5
> roundUp(12,7)
[1] 14

Answer 2

Regarding the rounding up to the multiplicity of an arbitrary number, e.g. 10, here is a simple alternative to James's answer.

It works for any real number being rounded up (from) and any real positive number rounded up to (to):

> RoundUp <- function(from,to) ceiling(from/to)*to

Example:

> RoundUp(-11,10)
[1] -10
> RoundUp(-0.1,10)
[1] 0
> RoundUp(0,10)
[1] 0
> RoundUp(8.9,10)
[1] 10
> RoundUp(135,10)
[1] 140

> RoundUp(from=c(1.3,2.4,5.6),to=1.1)  
[1] 2.2 3.3 6.6

Answer 3

You will find an upgraded version of Tommy's answer that takes into account several cases:

• Choosing between lower or higher bound
• Taking into account negative and zero values
• two different nice scale in case you want the function to round differently small and big numbers. Example: 4 would be rounded at 0 while 400 would be rounded at 400.

Below the code :

round.up.nice <- function(x, lower_bound = TRUE, nice_small=c(0,5,10), nice_big=c(1,2,3,4,5,6,7,8,9,10)) {
  if (abs(x) > 100) {
    nice = nice_big
  } else {
    nice = nice_small
  }
  if (lower_bound == TRUE) {
    if (x > 0) {
      return(10^floor(log10(x)) * nice[[max(which(x >= 10^floor(log10(x)) * nice))[[1]]]])
    } else if (x < 0) {
      return(- 10^floor(log10(-x)) * nice[[min(which(-x <= 10^floor(log10(-x)) * nice))[[1]]]])
    } else {
      return(0)
    }
  } else {
    if (x > 0) {
      return(10^floor(log10(x)) * nice[[min(which(x <= 10^floor(log10(x)) * nice))[[1]]]])
    } else if (x < 0) {
      return(- 10^floor(log10(-x)) * nice[[max(which(-x >= 10^floor(log10(-x)) * nice))[[1]]]])
    } else {
      return(0)
    }
  }
}

Answer 4

The plyr library has a function round_any that is pretty generic to do all kinds of rounding. For example

library(plyr)
round_any(132.1, 10)               # returns 130
round_any(132.1, 10, f = ceiling)  # returns 140
round_any(132.1, 5, f = ceiling)   # returns 135

Answer 5

I tried this without using any external library or cryptic features and it works!

Hope it helps someone.

ceil <- function(val, multiple){
  div = val/multiple
  int_div = as.integer(div)
  return (int_div * multiple + ceiling(div - int_div) * multiple)
}

> ceil(2.1, 2.2)
[1] 2.2
> ceil(3, 2.2)
[1] 4.4
> ceil(5, 10)
[1] 10
> ceil(0, 10)
[1] 0

Answer 6

If you add a negative number to the digits-argument of round(), R will round it to the multiples of 10, 100 etc.

    round(9, digits = -1) 
    [1] 10    
    round(89, digits = -1) 
    [1] 90
    round(89, digits = -2) 
    [1] 100