# [R] Variation of bubble sort (based on divisors)

Hervé Pagès hpages at fredhutch.org
Tue Apr 4 00:27:14 CEST 2017

Hi Piotr,

On 03/31/2017 11:16 AM, Piotr Koller wrote:
> Hi, I'd like to create a function that will sort values of a vector on a
> given basis:
>
> -zeros
>
> -ones
>
> -numbers divisible by 2
>
> -numbers divisible by 3 (but not by 2)
>
> -numbers divisible by 5 (but not by 2 and 3)

In other words, you want to sort your values by smallest divisor
(not regarding 1 as a divisor). The sorting part is easy if you can
map each value to its smaller divisor (mapping 0 to 0 and 1 to 1):

1) Map the values in 'x' to their smallest divisor:

x <- as.integer(x)
smallest_divisor <- sapply(x, smallestDivisor)
smallest_divisor[x == 0L] <- 0L
smallest_divisor[x == 1L] <- 1L

2) Then sort 'x' based on the values in 'smallest_divisor':

x[order(smallest_divisor)]

So the real difficulty here is not the sorting, it's to find the
smallest divisor. Here is a function that does this:

smallestDivisor <- function(x)
{
if (!is.integer(x) || length(x) != 1L || is.na(x))
stop("'x' must be a single integer")

## All prime numbers <= 2*3*5*7 = ND
pm210 <- as.integer(c(2, 3, 5, 7, 11, 13, 17, 19,
23, 29, 31, 37, 41, 43, 47,
53, 59, 61, 67, 71, 73, 79,
83, 89, 97, 101, 103, 107, 109,
113, 127, 131, 137, 139, 149,
151, 157, 163, 167, 173, 179,
181, 191, 193, 197, 199))
ans <- which(x %% pm210 == 0L)[1L]
if (!is.na(ans))
return(pm210[ans])

## Use Sieve of Eratosthenes to prepare the divisors that
## are > ND and <= 2*ND.
pm0 <- c(3L, 5L, 7L)  # must start with 3, not 2
prod0 <- as.integer(cumprod(pm0)[length(pm0)])
ND <- 2L * prod0
div <- 1L + 2L*(0L:(prod0-1L))
for (p in pm0)
div <- setdiff(div, p*(1L:(ND%/%p-1L)))
div <- div + ND
sqrtx <- sqrt(x)
while (div[1L] <= sqrtx) {
ans <- which(x %% div == 0L)[1L]
if (!is.na(ans))
return(div[ans])
div <- div + ND
}
x
}

I'm sure there are faster ways to do this.

Cheers,
H.

>
> etc.
>
> I also want to omit zeros in those turns. So when I have a given vector of
> c(0:10), I want to receive 0 1 2 4 6 8 10 3 9 5 7 I think it'd be the best
> to use some variation of bubble sort, so it'd look like that
>
> sort <- function(x) {
>  for (j in (length(x)-1):1) {
>    for (i in j:(length(x)-1)) {
>      if (x[i+1]%%divisor==0 && x[i]%%divisor!=0) {
>       temp <- x[i]
>       x[i] <- x[i+1]
>       x[i+1] <- temp
>       }
>     }
>   }
>  return(x)}
>
> This function works out well on a given divisor and incresing sequences.
>
> sort <- function(x) {
>   for (j in (length(x)-1):1) {
>      for (i in j:(length(x)-1)) {
>        if (x[i+1]%%5==0 && x[i]%%5!=0) {
>         temp <- x[i]
>         x[i] <- x[i+1]
>         x[i+1] <- temp
>        }
>       }
>      }
>   return(x)
>  }
>
> x <- c(1:10)
> print(x)
> print(bubblesort(x))
>
> This function does its job. It moves values divisible by 5 on the
> beginning. The question is how to increase divisor every "round" ?
>
> Thanks for any kind of help
>
> 	[[alternative HTML version deleted]]
>
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>

--
Hervé Pagès

Program in Computational Biology
Division of Public Health Sciences
Fred Hutchinson Cancer Research Center
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