[R] Help on a combinatorial task (lists?)

[jim holtman]

It would appear that if you were extracting the values at the indices
in 'res' that you should get: 2,2,2


> x <- list(c(1,2,3,4), c(1,2,4,3), c(1,4,2,3))
> res <- c(2,2,3)
>
> x
[[1]]
[1] 1 2 3 4

[[2]]
[1] 1 2 4 3

[[3]]
[1] 1 4 2 3

> ?mapply
> mapply(function(a,b)a[b], x, res)
[1] 2 2 2
>


On Tue, Aug 11, 2009 at 5:43 PM, Serguei
Kaniovski<Serguei.Kaniovski at wifo.ac.at> wrote:
> Simple unlist() will not do. In case of repeated weights, unlike
> permutations of indices permn(1:length(w)) some permutations of weights are
> identical.
>
> E.g. w <- c(3,2,2), permutations of indices c(1,2,3) and c(1,3,2) are
> undistinguishable.
>
> I think I have corrected the algorithm, but now I stuck with a rather
> trivial list manipulation at the very end.
>
> library(combinat)
>
> w <- c(5,3,2,1)
> i <- 1:length(w)
> q <- 7
>
> res <- sapply( permn(i), function(x) min(which(cumsum(w[x]) >=q)) )
>
> Now I have the vector 'res' ( of size length(permn(i)) ), and I need to
> extract from each entry of the list produced by permn(i) the element with
> the index stored in 'res'.
>
> E.g. the first three entries:
>
>> permn(i)[1:3]
> [[1]]
> [1] 1 2 3 4
>
> [[2]]
> [1] 1 2 4 3
>
> [[3]]
> [1] 1 4 2 3
>
> ...
>
>> res[1:3]
> [1] 2 2 3
>
> ...
>
> The answer should be 3, 4, 3 ...
>
> Thanks again for you help!
>
> Serguei K
>
>
> jim holtman schrieb:
>> Does 'unlist' do it for you:
>>
>>> w <- c(3,3,2,1)  # vector of weights
>>> q <- 4  # theshold
>>>
>>> # computes which coordinate of w is decisive in each permutation
>>> res <- unlist(sapply( permn(w), function(x) which(w ==
>>> x[min(which(cumsum(x) >=q))]) ))
>>>
>>> # complies the frequencies
>>> prop.table( tabulate( res ))
>> [1] 0.4 0.4 0.1 0.1
>>
>>
>> On Tue, Aug 11, 2009 at 7:03 AM, Serguei
>> Kaniovski<Serguei.Kaniovski at wifo.ac.at> wrote:
>>> Hello!
>>> I have the following combinatorial problem.
>>> Consider the cumulative sums of all permutations of a given weight vector
>>> 'w'. I need to know how often weight in a certain position brings the
>>> cumulative sums equal or above the given threshold 'q'. In other words,
>>> how often each weight is decisive in raising the cumulative sum above
>>> 'q'?
>>>
>>> Here is what I do:
>>>
>>> w <- c(3,2,1)  # vector of weights
>>> q <- 4  # theshold
>>>
>>> # computes which coordinate of w is decisive in each permutation
>>> res <- sapply( permn(w), function(x) which(w == x[min(which(cumsum(x) >=
>>> q))]) )
>>>
>>> # complies the frequencies
>>> prop.table( tabulate( res ))
>>>
>>>
>>> The problem I have is that when the weights are not unique, the which()
>>> function returns a list as opposed to a vector. I don’t know how to
>>> proceed when this happens, as tabulate does not work on lists.
>>>
>>> The answer, of course, should be that equal weights are “decisive”
>>> equally
>>> often.
>>>
>>>
>>> Can you help?
>>> Thanks a lot!
>>>
>>> Serguei Kaniovski
>>>
>>> ______________________________________________
>>> R-help at 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.
>>>
>>
>>
>>
>
> ______________________________________________
> R-help at 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.
>



-- 
Jim Holtman
Cincinnati, OH
+1 513 646 9390

What is the problem that you are trying to solve?