Re: [R] Help on a combinatorial task (lists?)
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 Kaniovskiserguei.kaniov...@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 Kaniovskiserguei.kaniov...@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@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@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? __ 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.
[R] Help on a combinatorial task (lists?)
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@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.
Re: [R] Help on a combinatorial task (lists?)
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 Kaniovskiserguei.kaniov...@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@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? __ 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.
Re: [R] Help on a combinatorial task (lists?)
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 Kaniovskiserguei.kaniov...@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@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@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.
