currCombin <- c(1,2) #this is the first possible combination 2 out if 5. Since
the vector has length 2, we are doing 2 out of x here.
selectedSet <- NULL
while (!is.na(currCombin)){
if(currCombin passes test) {
selectedSet <- rbind(selectedSet,currCombin)
}
currCombin <- next.combin(currCombin,5) #here we say that we want 2 out of 5
}
On Mar 6, 2010, at 5:54 PM, Herm Walsh wrote:
> Erich-
> This approach would be great for my context. However, in the code below I do
> not see how to restrict the output to the set of combinations I am looking
> for. For example, suppose I am looking for the 10 two element combinations
> of 1:5. Can you give me some psuedocode that shows how to do this?
> After putting the code below in a loop I do not see how to restrict the
> output to containing only numbers from 1:5.
> thanks very much.
>
> From: Erich Neuwirth <erich.neuwi...@univie.ac.at>
> To: Herm Walsh <hermwa...@yahoo.com>
> Cc: David Winsemius <dwinsem...@comcast.net>; r-help@r-project.org
> Sent: Wed, March 3, 2010 2:10:34 PM
> Subject: Re: [R] Working with combinations
>
> The following code takes a combination of type n over k represented by an
> increasing sequence
> as input an produces the lexicographically next combinations.
> So you can single step through all possible combinations and apply your
> filter criteria
> before you produce the next combination.
>
>
> next.combin <- function(oldcomb,n){
> lcomb <- length(oldcomb)
> hole.pos <- last.hole.pos(oldcomb,n)
> if ((hole.pos == lcomb) & oldcomb[lcomb]==n) {
> return(NA)
> }
> newcomb<-oldcomb
> newcomb[hole.pos]<-oldcomb[hole.pos]+1
> return(newcomb)
> }
>
> last.hole.pos <- function(comb,n){
> lcomb <- length(comb)
> diffs <- comb[-1]-comb[-lcomb]
> if (max(diffs)==1) {
> return(lcomb)
> }
> diffpos <- which(diffs>1)
> return(diffpos[length(diffpos)])
>
> }
>
> On Mar 3, 2010, at 7:35 PM, Herm Walsh wrote:
>
>> Thanks David for the thoughts. The challenge I have with this approach is
>> that the criteria I have is defined by a series of tests--which I do not
>> think I could substitute in in place of the logical indexing.
>>
>> In the combinations code I was hoping there is a step where, each new
>> combination is added to the current list of combinations. If this were the
>> case, I could put my series of tests in the code right there and then store
>> the combination if appropriate.
>>
>> However, evalutating the code--which uses recursion--I am not sure if this
>> approach will work. The combinations code is listed below. Is there a
>> simple place(s) where I could insert my tests, operating on the current
>> combination?
>>
>> function (n, r, v = 1:n, set = TRUE, repeats.allowed = FALSE)
>> {
>> if (mode(n) != "numeric" || length(n) != 1 || n < 1 || (n%%1) !=
>> 0)
>> stop("bad value of n")
>> if (mode(r) != "numeric" || length(r) != 1 || r < 1 || (r%%1) !=
>> 0)
>> stop("bad value of r")
>> if (!is.atomic(v) || length(v) < n)
>> stop("v is either non-atomic or too short")
>> if ((r > n) & repeats.allowed == FALSE)
>> stop("r > n and repeats.allowed=FALSE")
>> if (set) {
>> v <- unique(sort(v))
>> if (length(v) < n)
>> stop("too few different elements")
>> }
>> v0 <- vector(mode(v), 0)
>> if (repeats.allowed)
>> sub <- function(n, r, v) {
>> if (r == 0)
>> v0
>> else if (r == 1)
>> matrix(v, n, 1)
>> else if (n == 1)
>> matrix(v, 1, r)
>> else rbind(cbind(v[1], Recall(n, r - 1, v)), Recall(n -
>> 1, r, v[-1]))
>> }
>> else sub <- function(n, r, v) {
>> if (r == 0)
>> v0
>> else if (r == 1)
>> matrix(v, n, 1)
>> else if (r == n)
>> matrix(v, 1, n)
>> else rbind(cbind(v[1], Recall(n - 1, r - 1, v[-1])),
>> Recall(n - 1, r, v[-1]))
>> }
>> sub(n, r, v[1:n])
>> }
>> <environment: namespace:gtools>
>>
>>
>>
>> **************************************************************************************************************************************************************
>>
>>> I am working with the combinations function (available in the gtools
>>> package). However, rather than store all of the possible combinations I
>>> would like to check each combination to see if it meets a certain criteria.
>>> If it does, I will then store it.
>>>
>>> I have looked at the combinations code but am unsure where in the algorithm
>>> I would be able to operate on each combination.
>>
>> Logical indexing:
>>
>>> combinations(3,2,letters[1:3])[,1]=="a"
>> [1] TRUE TRUE FALSE
>>
>>> combinations(3,2,letters[1:3])[ combinations(3,2,letters[1:3])[,1]=="a", ]
>> [,1] [,2]
>> [1,] "a" "b"
>> [2,] "a" "c"
>>
>> --David
>>
>>
>>
>>
>> [[alternative HTML version deleted]]
>>
>> ______________________________________________
>> R-help@r-project.org mailing list
>> 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.
>
> --
> Erich Neuwirth
> Didactic Center for Computer Science and Institute for Scientific Computing
> University of Vienna
>
>
>
>
>
>
--
Erich Neuwirth
Didactic Center for Computer Science and Institute for Scientific Computing
University of Vienna
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