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]]
______________________________________________
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
https://urldefense.proofpoint.com/v2/url?u=https-3A__stat.ethz.ch_mailman_listinfo_r-2Dhelp&d=DwICAg&c=eRAMFD45gAfqt84VtBcfhQ&r=BK7q3XeAvimeWdGbWY_wJYbW0WYiZvSXAJJKaaPhzWA&m=AeHAcF7RnhWvQIqG5c2ucgFS0WIOmMFeRheLIeSwu0U&s=xJMmDOJLaQZ0QMmI7rkkNd2T5-zrh843rlJ-R1LQ9G8&e=
PLEASE do read the posting guide
https://urldefense.proofpoint.com/v2/url?u=http-3A__www.R-2Dproject.org_posting-2Dguide.html&d=DwICAg&c=eRAMFD45gAfqt84VtBcfhQ&r=BK7q3XeAvimeWdGbWY_wJYbW0WYiZvSXAJJKaaPhzWA&m=AeHAcF7RnhWvQIqG5c2ucgFS0WIOmMFeRheLIeSwu0U&s=c9IcZitWvur2grg8C54Jnt5LmajX0ODDANY-BGRzMbk&e=
and provide commented, minimal, self-contained, reproducible code.
--
Hervé Pagès
Program in Computational Biology
Division of Public Health Sciences
Fred Hutchinson Cancer Research Center
1100 Fairview Ave. N, M1-B514
P.O. Box 19024
Seattle, WA 98109-1024
E-mail: hpa...@fredhutch.org
Phone: (206) 667-5791
Fax: (206) 667-1319
______________________________________________
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
