similar to: Algorithm needed

Dear R users, I want to aggregate all *d *data from all combination of n *plots* taken by k. Thank very much! My data is like that: plot d 1 14 1 13 1 12 1 14 1 18 1 20 1 21 1 43 1 108 1 43 2 41 2 61 2 83 2 61 2 84 2 45 2 21 2 12 2 11 ... 100 10 100 12 -- --- Catalin-Constantin ROIBU Forestry engineer, PhD Forestry Faculty of Suceava

Hi The message occurred from R, when I was selected of "optimization > block diagonal Fhiser matrix" and used the attached file on PFIM. Could you please advise me about the following message? ***************************** Loading required pakage: combinat Attaching package:'combinat' The following object(s) are masked from 'package:utils':combn

Dear developeRs, I encounter the following problem: in the current version of my package FrF2, certain calls to a functioni do not work when package combinat is loaded, because function combn from combinat masks the function from utils that my package uses. I tried to solve this issue by importing function combn into the namespace of FrF2; I don't need to export it, I just want to use it

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

Hi, I am trying to use a vector of indices to select some rows from a matrix. But before I can do that I somehow need to convert 'combinations' into a list, since 'mode(combinations)' says it's 'numerical'. Any idea how I can do that? library("combinat") combinations <- t(combn(8,2)) indices <- c(sample(1:length(combinations),10)) # convert ???

 Hi R-listers, Savings regression results after a loop is straightforward. But what about when you have nested loops? I am running a regression of the form  lm(y~1+x+M+ D[,i] + D[,j] + D[,k]) where x is the variable of interest. M and D are vectors with other covariates. Vectors "M" and "x" are included in every regression. Then i loop over the columns of D to use all

Dear useRs! I would like to list all possible samples of size n form a population of size N. Obviously, N must be small (up to 20??) for this to be possible. For example, let say that N = 3 and n = 2. Therefore, we can say we have units 1, 2 and 3. I believe all possible samples are : {1,2},{2,3} and {1,3}. I would like to emphasize that I am not looking for the number of different

Hi dear all, Can aynone help me about delete-d jackknife usually normal jackknife code for my data is: n <- nrow(data) y <- data$y z <- data$z theta.hat <- mean(y) / mean(z) print (theta.hat) theta.jack <- numeric(n) for (i in 1:n) theta.jack[i] <- mean(y[-i]) / mean(z[-i]) bias <- (n - 1) * (mean(theta.jack) - theta.hat) print(bias) but how i can apply delete-d jackknife

The prob package has been archived because it depends upon some other packages which have issues. However, such projects as Introduction to Probability and Statistics in R depend upon it for learning. There are a few other resources that also use it. Does anyone know of any workarounds? Someone at stack exchange mentioned using R 2.9. However, that broke my RStudio (WSOD) and the dependent

Hi everyone! I have a data frame with 1112 time series and I am going to randomly sampling r samples for z times to compose different portfolio size(r securities portfolio). As for r=2 and z=10000,that's: z=10000 A=seq(1:1112) x1=sample(A,z,replace =TRUE) x2=sample(A,z,replace =TRUE) M=cbind(x1,x2) # combination of 2 series Because in a portfolio with x1[i]=x2[i],(i=1,2,...,10000) means a 1

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

Hola buenos días, me presento, me llamo Miguel y 'soy de' y 'vivo en' Galicia. Soy profesor de secundaria (Bachillerato Adultos) y llevo 15 días estudiando R a un buen ritmo, pero todavía me faltan miles de cosas. He visto que R facilita, no solo el análisis de datos y que posee una potencia en cálculos estadísticos a cualquier nivel, sino gran caudal de recursos para Data Mining,

Hi Folks, Given a series of n independent Bernoulli trials with outcomes Yi (i=1...n) and Prob[Yi = 1] = Pi, I want P = Prob[sum(Yi) = r] (r = 0,1,...,n) I can certainly find a way to do it: Let p be the vector c(P1,P2,...,Pn). The cases r=0 and r=n are trivial (and also are exceptions for the following routine). For a given value of r in (1:(n-1)), library(combinat) Set <- (1: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

Respected All, I hope you are enjoying good health, I am tring to write a program in R but could not be very sucessful. My program draws random sample form bivariate normal distribution and then compute a variable PIJ. For certian samples some entries of variable PIJ is apearing as negative, which result in negative variance estimator. I want to introduce a loop in my program that verify the each

The issue is fAsianOptions. Is there a version that works with the latest version of R? If not, which version of it works with which version of R and where can it be found? I tried several at the archive already. Alternatively, is there another package that behaves similarly to prob? On Wed, Nov 1, 2017 at 6:17 PM, David Winsemius <dwinsemius at comcast.net> wrote: > > > On Nov

Dear R-helpers, I'm trying to develop a function which specifies all possible expressions that can be formed using a certain number of variables. For example, with three variables A, B and C we can have - presence/absence of A; B and C - presence/absence of combinations of two of them - presence/absence of all three A B C 1 0 2 1 3 0 4 1 5 0 6 1

The formula attribute of the builtin CO2 dataset seems a bit strange: > formula(CO2) Plant ~ Type + Treatment + conc + uptake What is one supposed to do with that? Certainly its not suitable for input to lm and none of the examples in ?CO2 use the above.

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

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()"