similar to: Permutations with replacement

I have a problem where I need to loop over the total combinations of vectors (combined once chosen via combinatorics). Here is a simplification of the problem: STEP 1: Define three vectors a, b, c. STEP 2: Combine all possible pairwise vectors (i.e., 3 choose 2 = 3 possible pairs of vectors: ab,ac, bc) NOTE: the actual problem has 8 choose 4, 8 choose 5 and 8 choose 6 combinations. STEP

Greetings, fellow R'ers. How can I get this frame in R: ID meas ID.1 meas.1 1 1.1 3 1.2 2 2.1 4 2.2 to look like this (stacking): ID meas 1 1.1 2 2.1 3 1.2 4 2.2 It's not really the reshape function (or is it?) because we can consider the additional columns, viz., ID.1 and meas.1, as independent of ID and meas so it is basically a stacking

My apologies, I forgot to CC: to the list on my previous communication with Daniel. Jesse ----- Forwarded by Jesse Albert Canchola/EMVL/DIAG/US/BAYER on 08/21/2006 10:50 AM ----- Jesse Albert Canchola/EMVL/DIAG/US/BAYER 08/21/2006 09:36 AM To "Daniel Nordlund" <res90sx5 at verizon.net> cc Subject RE: [R] Permutations with replacement Thanks, Daniel. I need to enumerate

I'm using R on Windows XP Professional SP2. I figured out the fix for my buffer overflow problem that did not allow me to use Rgui.exe for about two weeks. I would keep getting "console not found" and even when I uninstalled and re-installed got the same problem. Here's how the problem started. In Rgui.exe, I attempted to get more visible output in the buffer (I wanted

Hello all, I am trying to create a matrix of 1s and -1s without any repetitions for a specified number of columns. e.g. 1s and -1s for 3 columns can be done uniquely in 2^3 ways. -1 -1 -1 -1 -1 1 -1 1 -1 -1 1 1 1 -1 -1 1 -1 1 1 1 -1 1 1 1 and for 4 columns in 2^4 ways and so on. I finally used the function combn([0 1],3) that I found at the following link

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

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 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

Dear useRs! I would like to generate a list of all possible (unique) partitions of N units into k classes. For example, all possible partitions of 4 units into 2 classes are (I hope I have not missed anyone): 1,1,1,2 (this can be read as {1,2,3},{4}) 1,1,2,1 1,2,1,1 2,1,1,1 1,1,2,2 1,2,1,2 1,2,2,1 The partitions 1,1,2,2 and 2,2,1,1 are the same and are therefore not two unique

Dear R users, I want to perform an exact permutation of multisets. I have looked at the coin package, but it doesn't seem to offer what I am looking for. I want to perform permutations (exact - without duplications) on multisets of scalars (e.g., the multiset 0,0,1,2,2). I want to output all the possible permutations into a data frame so that each multiset permutation occupies a row (or

Hi, folks: I need to generate all 0-1 sequences with given length,say,n=3, the ideal result would be the following matrix: 0 0 0 0 0 1 0 1 0 0 1 1 1 0 0 1 0 1 1 1 0 1 1 1 Any help would be appreciated. -- View this message in context: http://www.nabble.com/How-to-generate-all-permutation-of-0-1-sequences-in-R--tf4704377.html#a13446919 Sent from the R help mailing list archive at Nabble.com.

Hi all, I have a vector, e.g., A = c(10, 20, 30, 40). This 4 numbers have 4! = 24 different combination. I want to generate a matrix with one combination in one row, so the output would be B = 10 20 30 40 10 40 20 30 ... Does anyone know how to do this easily in R? Thank you very much. Wendy -- View this message in context:

Hola amigos, muchas gracias por vuestra ayuda. Entonces veo que mi sorpresa era legítima. Por todos vuestros mails la conclusión es que: - En el módulo base de R no incluye combinatoria elemental, ni siquiera el número combinatorio Cm,n hay que cargar el paquete *combinat* - Y para las variaciones con repetición el paquete* gtools* - Y aún así no tenemos ni las combinaciones ni las

Dear R-listers, I am giving part of my R code : ########################################################### n=15 m=1 library("partitions") library("gregmisc") library("combinat") x = t(restrictedparts(n-m,m)) l = length(x[,1]) for(u in 1:l){ A= unique(matrix( unlist(permn(x[u,])), ncol=m, byrow=TRUE )) }

Hola Miguel, Sí se pueden obtener las variaciones con y sin repetición en R. Eso sí están un poco escondidas... Se pueden calcular de esta forma: #---------------------- > #Cargar el paquete gtools > library(gtools) > #Definir el conjunto sobre el que se hará el cálculo > x <- c('rojo', 'azul', 'verde') > #Utilizar la función "permutations()"

Hi all Here is a rather naive implementation of the SEPA algorithm for generating lexical permutations: lexperm3 <- function(x, n=length(x)) { perms <- list() k <- 1 perms[[k]] <- x k <- k + 1 for (y in 1:(factorial(n)-1)) { i <- n-1 while (x[i] > x[i+1] && i > 0) { i <- i - 1 } # i is largest index st x[i] > x[i+1] j <- n #

En relación con lo que comenta Carlos, por ejemplo para el caso de las Variaciones sin Repetición, puede ser instructivoenseñar como se construye como por ejemplo: VsinR <- function(m, n){ return (factorial(m)/factorial(m-n))} VsinR(9,3) ------------------------- Creo que con la función factorial que viene por defecto en R puedes construir siguiendo este modelo rápidadmentecasi cualquier

Sí, también... Para las permutaciones, n=r. Y con el parámetro "repeats.allowed" controlas si son con o sin repetción: #---------------------- > #Permutaciones *con repetición* > permutations(n=3, r=3, v=x, repeats.allowed=TRUE) [,1] [,2] [,3] [1,] "azul" "azul" "azul" [2,] "azul" "azul" "rojo" [3,]

Buena tarde amigos, En días pasados hice algunas consultas y ya pude salir de las dudas que tenia en ese momento, ahora requiero de su colaboración con lo siguiente: Tengo un vector dicotomico (Binario) con la siguiente información que me surgio de algunos procesos anteriores: > MuestraS [1] 1 1 1 1 0 1 1 0 1 1 Ahora necesito hallar todas las posibles combinaciones que se puedan hacer con

Dear R Programmers, I am trying to run a Poisson regression on all pairs of variables in a data set and obtain the permutation distribution. The number of pairs is around 100000. It seems my code will take weeks to run, unless I try something else. Could you give me any suggestions on how to improve the speed of the code below, or any general suggestions on how I may accomplish this task. Thanks