similar to: Permutations of configurations on multiple columns

Hi, this is maybe not a real R problem but I want to solve this in R ;-) Consider the set of all permutations of 1:N (=: S, say) and a fixed element a from S. I now need to compute the number of permutations s from S which are elementwise less or equal to a: | { s \in S | s <= a } | Of cource, backtracking using a tree structure is possible. Does anyone know an efficient way? Torsten

Hi, I have a function for doing permutation tests for Manovas, written with the help of folks here on the list. It seems to work ok, and I've found that there is indeed a significant difference among groups in my analysis. I want to follow up on this by testing for which pairs of groups are significantly different. Reasoning that since Wilks Lambda is a generalization of the Hotelling T test,

Hallo All, I would like to apply a function to all permutations of variables in a dataframe (except the first). What is the best way to achieve this? I produce the permutations using: nvar <- ncol(dat) - 1 perms <- as.matrix( expand.grid(rep( list(1:0) , nvar ))[ , nvar:1] ) Thanks in advance Serguei Test-dataframe, comma-delimited: code,wav,w,area,gdp,def,pop,coast,milspend,agr

Question 1. I would calculate all the permutations of numbers (23,46,70,71,89) How can I write correctly the code? (Load gregmisc) > permutations (..?...) ______________________ Question 2. It's possible permute a string? ..for example: EIGER EGIER EREGI etc..etc.. Thanks, E. -- View this message in context: http://www.nabble.com/permutations-tp25834463p25834463.html Sent from the R

Dear All, I'm trying to create this: I've these data (a, b,c, etc. are numbers) : 1 a b 2 c d f 3 g 4 h ... N I'm trying to create the all "time series permutations" (length = N) taking account the possibilities for each value. For example (N=4), 1 a 2 c 3 g 4 h next, 1 b 2 c 3 g 4 h next 1 a 2 d 3 g 4 h and so on. For this example, there are

Dear list, I have a vector of letter strings as follow: > LETTERS[c(1:7,9,15,18:25)]  [1] "A" "B" "C" "D" "E" "F" "G" "I" "O" "R" "S" "T" "U" "V" "W" "X" "Y"   I need to find sequences of length 11, which are made from  all

Dear all, I am using RDA to study plant communities in various land uses (variable LU with values M, U, etc.). For each land use, I sample 3 to 5 fields (M1, M2, U1, U2, etc). I make 5 measurements for both plant communities and environmental variables in each field. I repeat the process every 6 months to study the effect of time and season (D16 for dry season 2017, R16 for rainy season 2016,

Dear R People: Suppose I have the 4 numbers: 1,2,3,4. I would like to create a listing of the permutations of 4 items taken 4 at a time. Is there a built in function for that, please? Thanks in advance! R 2-3-1 for Windows or Linux Sincerely, Erin Hodgess Associate Professor Department of Computer and Mathematical Sciences University of Houston - Downtown mailto: hodgess at gator.uhd.edu

Niels Waller wrote: > Does anyone know of an R (or S-PLUS) function for delineating all possible > combinations and permutations? The following function delivers all permutations of 1:n. all.perm <- function(n) { p <- matrix(1, ncol = 1) for (i in 2:n) { p <- pp <- cbind(p, i) v <- c(1:i, 1:(i - 1)) for (j in 2:i) { v <- v[-1]

Is there a simple function or process that will create a matrix of permutations with replacement? I know that using the combinat package ###### begin R code ###### > library(combinat) > m <- t(array(unlist(permn(3)), dim = c(3, 6))) # we can get the permutations, for example 3!=6 # gives us > m [,1] [,2] [,3] [1,] 1 2 3 [2,] 1 3 2 [3,] 3 1 2 [4,]

Dear R community, I am trying to create a matrix of permutations of a vector: bands <- c("AL", "B", "DB", "DG", "G", "K", "LB", "LG", "MG", "O", "P", "PI", "PK", "PU", "R", "V", "W", "Y") Each permutation must be 4

Hallo List, I'm trying to implemement a restricted permutation scheme in permutest(). More precisely I have dependence in my data that should be allowed for in the permutation - I simulated the problem in the example of the vegan documentation p.24: library(vegan) data(varespec) ## Bray-Curtis distances between samples dis <- vegdist(varespec) ## First 16 sites grazed, remaining 8 sites

Hello R-help Readers I'm currently in the process of trying to write a permutation test procedure for looking at differences between groups with a multivariate data set (something equivalent to ANOSIM - analaysis of similarities for those familiar with this test). As with other permutation tests, for cases where there are large numbers of possible permutations, randomly sampling from these

I have a problem with permutations functions in R I just started using R so maybe it is an easy question I need to obtain all the 9.somthingExp 157 permutations that can be given from the number from 1 to 100 I wrote the following commands: > library(gregmisc) >options(expressions=1e5) cmat <- combinations(300,2) dim(cmat) # 44850 by 2 >permutations(n=100, r=100)

Dear R users, This is a second summary of the permutation problem I previously posted. This summary restates the problem as well as the solution. First of all thanks to everyone including Erich, Robin, Gabor, Christian, Ingmar and others for your suggestions. With the help of an off-list discussion with Gabor I?m going to summarize. THE PROBLEM We have 12 elements in blocks of 3 : 1 2

if i were to have a block size of 4 people and i want to assign a treatment combination to the entire block, there would be 16 different treatment combinations (TTTT, TTTP, TTPP, PTTP, etc.) i am trying to get all 16 permutations and i am able to use this code below. drugs=c('P','T'); comb=expand.grid(drugs,drugs,drugs,drugs) for a block size of 3 the code would be

I'm using functions to return a matrix of all permutations of a specified size from a larger list for predictor selection. For each predictor size I use a seperate function like this: bag2 <- function(n) { rl <- c() for (i1 in seq(n)) { for (i2 in seq(n)) { if (length(unique(c(i1,i2)))==1) {next} rl <- cbind(rl,matrix(c(i1,i2))) } }

Hi Boris, Do a search on "the order of elements of the symmetric group". (This search will also turn up homework questions and solutions.) You will understand why you are seeing this once you understand how a permutation is decomposed into cycles and how the order relates to a partition of n (n=10 in your case). Enjoy! Eric On Fri, Dec 8, 2017 at 6:39 AM, Boris Steipe <boris.steipe

Hi. Does anyone know of a function which will take as input a number n (or a set of n letters) and will give out, one at a time, the permutations of n (or of those n letters) as a vector? So that I can use the permutations one at a time. And such that it will exhaust all the permutations with no repeats. For example if n is 3, I would want a function which I could use in a loop and the first time

Dear R-Users, I'm relativley new to R and have the following problem. I need all permutations of the vectors created by the collumns of a matrix. I will give a small example: p=3 n=2^p-1 #number of obtainable vectors mat=matrix(0,p,p) for(i in 1:p) { mat[i:1,i]=1/i } mat is now a