similar to: Two packages and one method

Hola Jorge, gracias por ayudarme antes de empezar. Ayer me fui a cama con tan solo escribir el título "Ecuaciones no lineales" Básicamente lo que voy a buscar es lo clásico, es decir, resolver cualquier ecuación no lineal, ya sea polinómica o trascendente, con una o varias incógnitas. Ejemplo típico: x^2 + y^2 = 1 y = sin x Si me van surgiendo dudas, aquí estaré, no lo dudéis. Por

I would appreciate any suggestions on which function to use to write subsequent functions analysing combinations of treatments. This refers to experimental trials of medical treatments. I want to write routines to analyse various comparisons (combinations) So .... if 5 treatments are available (t01, t02, t03, t04 and t05) I want to write a general routine that works out all possible

Error: could not find function "hier.part" things I have tried: 1. reinstall R (lastest version, on windows XP) 2. install package "gtools" 3. include: library(gtools) require(gtools) 4. how I call this function: hier.part(c$Y, xdata, fam = "gaussian", gof = "Rsqu") 5.when I try to check what's in the package "gtools", I get (hier.part is

Ya me extrañaba a mi! Mañana a primera hora miro todo lo que me comentas. Ahora estoy con ecuaciones no lineales y ya veo que hay que cargar librerias Muchas gracias. Un saludo El 23 de marzo de 2015, 0:08, Carlos Ortega <cof en qualityexcellence.es> escribió: > Hola, > > Por precisar un par de detalles: > > > - Con el paquete gtools se pueden generar: > - las

When I want to install "dprep" and I always get information: > install.packages("dprep") Warning in install.packages("dprep") : argument 'lib' is missing: using 'C:\Users\Documents/R/win-library/2.8' --- Please select a CRAN mirror for use in this session --- Warning message: package ?dprep? is not available I have tried a lot of mirror...

I am working with the combinations function (available in the gtools package).  However, rather than store all of the possible combinations I would like to check each combination to see if it meets a certain criteria.  If it does, I will then store it. I have looked at the combinations code but am unsure where in the algorithm I would be able to operate on each combination. Thanks!

Dear R-users, Let me ask about the 'stack overflow' error which I got when I used the function 'combinations' in "gtools". The following is what I did: --------- library(gtools) options(expressions=1e5) combinations(500, 3, 1:500) # or combinations(400, 2, 1:400) Error: protect(): stack overflow --------- How can I overcome this error? Is there perhaps any other

Jeff, I wanted to let you know that your function is faster than generating the directional circular permutations and weeding. Here is the time for n = 10. I compared with just doing the permutations, there is no point in proceeding further with the weeding since it is slower at the start itself. system.time(directionless_circular_permutations(10)) user system elapsed 1.576 0.000

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

I don't know if this is more efficient than enumerating with distinct directions and weeding... it seems kind of heavyweight to me: ####### library(gtools) directionless_circular_permutations <- function( n ) { v <- seq.int( n-1 ) ix <- combinations( n-1, 2 ) jx <- permutations( n-3, n-3 ) x <- lapply( seq.int( nrow( ix ) ) , function( i ) {

New function below is a bit faster due to more efficent memory handling. for-loop FTW! directionless_circular_permutations2 <- function( n ) { n1 <- n - 1L v <- seq.int( n1 ) ix <- combinations( n1, 2L ) jx <- permutations( n-3L, n-3L ) jxrows <- nrow( jx ) jxoffsets <- seq.int( jxrows ) result <- matrix( n, nrow = factorial( n1 )/2L, ncol = n ) k

Hi All, I'm trying to come up with a clear and concise (and fast?) solution to the following problem. I would like to take a vector 'v' and enumerate all of the ways in which it can be broken into n sets of length 2 (if the length of the vector is odd, and an additional set of length 1). An element of 'v' can only appear in one set. Order within sets is not important.

Hello, I am attempting to install the gtools package, so that I can compute the permutations of a set. However, I having difficulties. My problems are: 1) When I use the GUI "Install package(s)" command, the package "gtools" does not appear in my list. 2) I dowloaded the package from http://cran.r-project.org/web/packages/gtools/ . I Then used the GUI command

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,]

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()"

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

Thanks! Yes, however, this seems a bit wasteful. Just wondering if there are other, more efficient options possible. Best wishes, Ranjan On Thu, 29 Mar 2018 22:20:19 -0400 Boris Steipe <boris.steipe at utoronto.ca> wrote: > If one is equal to the reverse of another, keep only one of the pair. > > B. > > > > > On Mar 29, 2018, at 9:48 PM, Ranjan Maitra

Hello All, I am trying to carry out variable reduction. I do not have information about the dependent variable, and have only the X variables as it were. In selecting variables I wish to keep, I have considered the following criteria. 1) Percentage of missing value in each column/variable 2) Variance of each variable, with a cut-off value. I recently came across Weka and found that there is an

Hello, I've recently transitioned to using R in Linux. My OS/installation versions are: $ Red Hat Enterprise Linux AS release 4 (Nahant Update 6) $ Linux lx-chgmqsd05 2.6.9-67.0.1.ELsmp #1 SMP Fri Nov 30 11:57:43 EST 2007 x86_64 x86_64 x86_64 GNU/Linux I am trying to create a repository of packages under ~/R/library which I've created since I do not have admin access on my system. From my

Hello, I've recently transitioned to using R in Linux. My OS/installation versions are: $ Red Hat Enterprise Linux AS release 4 (Nahant Update 6) $ Linux lx-chgmqsd05 2.6.9-67.0.1.ELsmp #1 SMP Fri Nov 30 11:57:43 EST 2007 x86_64 x86_64 x86_64 GNU/Linux I am trying to create a repository of packages under ~/R/library which I've created since I do not have admin access on my system. From my