similar to: Replace values on seq

Hi there, I have a multiple for loop over a list of data frames for ( i in 1:(N-1) ) { for ( j in (i+1):N ) { for ( p in 1:M ) { v_i[p] = alist[[p]][i,"v"] v_j[p] = alist[[p]][j,"v"] } rho_s = cor(v_i, v_j, method = "spearman") rho_p = cor(v_i, v_j, method = "pearson"

Dear R-Users, I have the following problem: I would like to create a postscript file containing an r-plot with the string "\\vartheta" in it (reason: this is later converted to the TeX-string "\vartheta" and a vartheta is printed in the figure). In the minimal example below, the problem is that the created postscript file does _not_ contain the string "\\vartheta

I am trying to use lme to fit a mixed effects model to get the same results as when using the following SAS code: proc mixed; class refseqid probeid probeno end; model expression=end logpgc / ddfm=satterth; random probeno probeid / subject=refseqid type=cs; lsmeans end / diff cl; run; There are 3 genes (refseqid) which is the large grouping factor, with 2 probeids nested within each refseqid,

Hello all, Is there a way in R to compute the multivariate normal density of every pair of entries in a vector efficiently instead of using for loop? For example Suppose I have a vector a=c(v_1,...,v_p)=c(0.5343909, -0.7784353, -0.0568370, 1.8772838, -1.3183407, 0.8227418,...) I want to compute density(v_i, v_j) for every pair of entries (i,j) (i!=j) in a. The joint bivariate distribution

Hi. If I have a vector, v_1, and another vector of positive integers, i_1, the same length as v_1, then rep(v_1,i_1) will repeat v_i[j] exactly i_1[j] times, like so: >rep(c(1,2,3),c(3,2,1)) [1] 1 1 1 2 2 3 > I'd like to do the same sort of thing where I replace v_1 with a matrix, and the jth row of the matrix is repeated i_1 times. Obviously, I could do this with for loops, like

It is frustrating to see the labels I want in the dimensions of a list but not be able to extract those labels into titles for plots generated from component objects. If someone could set me straight, I would appreciate it. For your amusement, I have provided an example of the Byzantine code I am currently using to avoid loops: # Simulate ANOVA type test data sex<-c(rep(1,8),rep(0,8))

Dear All, Thank you for the replies to my first thread here: http://r.789695.n4.nabble.com/global-optimisation-with-inequality-constraints-td3799258.html. So far the best result is achieved via a penalised objective function. This was suggested by someone on this list privately. I am still looking into some of the options mentioned in the original thread, but I have been advised that there may

Hello. Slope estimates in lme are shrinkage estimates which pull the OLS slope estimates towards the population estimates, the degree of which depends on the group sample size and the distance between the group-based estimate and the overall population estimate. Although these shrinkage estimates as said to be more precise with respect to the true values, they are also biased. So there is a

I am trying to use lme to fit a mixed effects model to get the same results as when using the following SAS code: proc mixed; class refseqid probeid probeno end; model expression=end logpgc / ddfm=satterth; random probeno probeid / subject=refseqid type=cs; lsmeans end / diff cl; run; There are 3 genes (refseqid) which is the large grouping factor, with 2 probeids nested within each refseqid,

R Experts: I am conducting a meta-analysis where the effect measures to be pooled are simple proportions. For example, consider this data from Fleiss/Levin/Paik's Statistical methods for rates and proportions (2003, p189) on smokers: Study N Event P(Event) 1 86 83 0.965 2 93 90 0.968 3 136 129 0.949 4 82 70 0.854 Total

I buy flowers at a local market on a fairly regular basis. The flower vendors post their prices and if I want to buy only one or two flowers I will generally get the posted price. From time to time I want to buy large quantities of flowers, and sometimes a vendor will give me a better price than their posted price for the bulk order, but more often I have to offer them a higher price than the

Hello! I have noticed that mapply() drops names in R 2.3.1 as well as in r-devel. Here is a simple example: l <- list(a=1, b=2) k <- list(1) mapply(FUN="+", l, k) [1] 2 3 mapply(FUN="+", l, k, SIMPLIFY=FALSE) [[1]] [1] 2 [[2]] [1] 3 Help page does not indicate that this should happen. Argument USE.NAMES does not have any effect here as it used only in a bit special

On Thu, 12 Apr 2007, Fernando Magno Quintao Pereira wrote: >> I'm definitely interested in improving coalescing and it sounds like >> this would fall under that work. Do you have references to papers >> that talk about the various algorithms? > > Some suggestions: > > @InProceedings{Budimlic02, > AUTHOR = {Zoran Budimlic and Keith D. Cooper and Timothy

Hello all I would like to ask your help use mapply. I have a function called findCell that takes two arguments(x,sr) where x is a vector of size two (e.g x<-c(2,3) and sr is a matrix. I would like to call many times the findCell function (thus I need mapply) for different x inputs but always for the same sr. as x is a vector of size two (two cells) I want to pass inside inside the following

I can run one-sample t-test on an array, for example a matrix myData1, with the following apply(myData1, 2, t.test) Is there a similar fashion using apply() or something else to run 2-sample t-test with datasets from two groups, myData1 and myData2, without looping? TIA, Gang

I have a data frame whose first colum contains the names of the variables and whose second colum contains the values to assign to them: : kkk <- data.frame(vars=c("var1", "var2", "var3"), vals=c(10, 20, 30), stringsAsFactors=F) If I do : assign(kkk$vars[1], kkk$vals[1]) it works : var1 [1] 10 However, if I try with mapply

List: Thank you for the replies to my post yesterday. Gabor and Phil also gave useful replies on how to improve the function by relying on mapply rather than the explicit for loop. In general, I try and use the family of apply functions rather than the looping constructs such as for, while etc as a matter of practice. However, it seems the mapply function in this case is slower (in terms of CPU

Dear all, #Last week, I asked about merge x and y as list. #Now I have a dataset with list of list like: x <- list(list(matrix(1:20, 5, 4),matrix(1:20, 5, 4)), list(matrix(1:20, 5, 4),matrix(1:20, 5, 4))) y <- list(list(c(1, -1, -1, 1, 1),c(1, 1, -1, -1, -1)), list(c(1, 1, 1, 1, 1),c(1, 1, -1, 1, -1))) x y #I need merge x and y, I have tried with list.uni <-

Hi, Starting with ordinary vectors, so we know what to expect: > mapply(function(x, y) {x * y}, 101:106, rep(1:3, 2)) [1] 101 204 309 104 210 318 > mapply(function(x, y) {x * y}, 101:106, 1:3) [1] 101 204 309 104 210 318 Now with an S4 object: setClass("A", representation(aa="integer")) a <- new("A", aa=101:106) > length(a)

Hi, I have a problem that I'm finding a bit tricky. I'm trying to use mapply and assign to generate curried functions. For example, if I have the function divide divide <- function(x, y) { x / y } And I want the end result to be functionally equivalent to: half <- function(x) divide(x, 2) third <- function(x) divide(x, 3) quarter <- function(x) divide(x, 4) But I want