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<[email protected]> 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<[email protected]> 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 >>> >>> ______________________________________________ >>> [email protected] 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. >>> >> >> >> > > ______________________________________________ > [email protected] 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? ______________________________________________ [email protected] 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.
