similar to: Sum of Bernoullis with varying probabilities

Displaying 20 results from an estimated 8000 matches similar to: "Sum of Bernoullis with varying probabilities"

2013 Sep 06
1
Importing function that is previously imported by other package
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
2007 Apr 10
1
R CMD Rdconv drops sections: arguments, seealso, examples (PR#9606)
I've created a .Rd file (below), then converted that to .sgml using R CMD Rdconv --type=Ssgm combn.Rd > combn.sgml The output (shown below) is missing some of the sections: arguments seealso examples If instead I convert to .d (below), the same sections are missing, and the "note" section is included but without the necessary newline.
2006 May 09
1
combn(n, k, ...) and all its re-inventions
It seems people are reinventing the wheel here: The goal is to generate all combinations of 1:n of size k. This (typically) results in a matrix of size k * choose(n,k) i.e. needs O(n ^ k) space, hence is only applicable to relatively small k. Then alternatives have been devised to generate the combinations "one by one", and I think I remember there has been a quiz/challenge about 20
2013 Jan 24
2
Please help R error message "masked from 'package:utils':combn"
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
2006 May 08
3
Non repetitive permutations/combinations of elements
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
2013 Apr 24
2
Distance matrices Combinations
Dear UseRs, MY PROBLEM IS A SMALL PIECE OF A REAL BIG AND A COMPLICATED PROBLEM. IF I DELIBERATE IN A VERY SIMPLE WAY THEN ALL I WANT IS TO PUT ALL THE POSSIBLE COMBINATIONS OF 75 DISTANCE MATRICES (BY TAKING 4 MATRICES, MORE COMMONLY 75C4), in the following equation. t<-as.matrix((MAT1)^2+(MAT2)^2+(MAT3)^2+(MAT4)^2+,upper=T,diag=T)) Then "1215450" values of "t"(one for
2008 Jul 03
1
Problem in applying conditional looping
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
2006 Jan 30
2
yet another vectorization question
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
2005 Aug 17
2
Copying rows from a matrix using a vector of indices
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 ???
2006 Jul 13
1
looping using combinatorics
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
2007 Apr 30
1
R CMD Rdconv drops sections: arguments, seealso, examples (PR#9645)
On Tue, 10 Apr 2007 timh at insightful.com wrote: > I've created a .Rd file (below), then converted that to .sgml using > R CMD Rdconv --type=Ssgm combn.Rd > combn.sgml > The output (shown below) is missing some of the sections: > arguments > seealso > examples > If instead I convert to .d (below), the same sections are missing, > and the "note"
2010 Dec 22
3
A question to get all possible combinations
Let say, I have a matrix with 8 rows and 6 columns:  >  df1  <- matrix(NA, 8, 4)  > df1       [,1] [,2] [,3] [,4]  [1,]   NA   NA   NA   NA  [2,]   NA   NA   NA   NA  [3,]   NA   NA   NA   NA  [4,]   NA   NA   NA   NA  [5,]   NA   NA   NA   NA  [6,]   NA   NA   NA   NA  [7,]   NA   NA   NA   NA  [8,]   NA   NA   NA   NA  Now I want to get **all possible** ways to fetch 6 cells at a
2015 Mar 22
2
Combinatoria
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()"
2010 Mar 26
2
More efficient alternative to combn()?
Hi, i am working on a problem where i need to compute the products of all possible combinations of size m of the elements of a vector. I know that this can be achieved using the function combn(), e.g.: > vector <- 1:6 > combn(x = vector, m = 3, FUN = function(y) prod(y)) In my case the vector has 2000 elements and i need to compute the values specified above for m = 32. Using combn() i
2007 Apr 20
1
simply this loop?
Hi, anyone interested in this: I tried to simply this loop with lapply or something but haven't figured it out: mapt = c("203929_s_at", "203930_s_at", "203928_x_at", "206401_s_at") mapt.combn <- lapply(1:4, function(i) combn(mapt, i)) out = list() k = 1 for (i in 1:length(mapt.combn)){ for (j in 1:ncol(mapt.combn[[i]])){ out[[k]] =
2015 Mar 21
3
Combinatoria
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,
2007 May 29
2
summing up colum values for unique IDs when multiple ID's exist in data frame
I have data.frame's with IDs and multiple columns. B/c some of IDs showed up more than once, I need sum up colum values to creat a new dataframe with unique ids. I hope there are some cheaper ways of doing it... Because the dataframe is huge, it takes almost an hour to do the task. Thanks so much in advance! Young # ------------------------- examples are here and sum.dup.r is at the
2012 Nov 16
1
pairing data using combn with criteria
Dear All, I have a dataframe made up of individual beetles consisting of individual number, family number, mother's family number, father's family number, and sex of the beetle. I would like to pair up the individuals for breeding. I would, however, like to avoid breeding beetles of the same sex (obviously), the same family, and with the same mother's family or father's family,
2010 Nov 17
1
efficient conversion of matrix column rows to list elements
Hi List, I'm hoping to get opinions for enhancing the efficiency of the following code designed to take a vector of probabilities (outcomes) and calculate a union of the probability space. As part of the union calculation, combn() must be used, which returns a matrix, and the parallelized version of lapply() provided in the multicore package requires a list. I've found that
2015 Mar 22
2
Combinatoria
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,]