similar to: matching a given sd range

folks, if i have a matrix of means: a <- matrix(1:4, 2) and a matrix of std deviations: b <- matrix(5:8, 2) and i want to create a matrix X of random variates such that X[i, j] is a draw from normal distribution with mean = a[i, j] and std dev = b[i, j], i think i can do this? X <- matrix(rnorm(4, mean = a, sd = b), ncol = 2) now if I want to create 10 such matrices, I could possibly do

Thank you very much for your reply. Your code work well with this example. I modified a little to fit my real data, I got an error massage. Error in split.default(x = seq_len(nrow(x)), f = f, drop = drop, ...) : Group length is 0 but data length > 0 On Thu, Jan 31, 2013 at 12:21 PM, arun kirshna [via R] < ml-node+s789695n4657196h87@n4.nabble.com> wrote: > Hi, > Try this: >

Hi there, I have some data which are convenient to enter as lists. For example: t1<-list(fname="animal1",testname="hyla",dspkr="left",res1=39.7,res2=15.0) t2<-list(fname="animal1",testname="bufo",dspkr="left",res1=14.4,res2=56.1)

Hello R community, I am trying to combine two CSV files that look like this: File A Row_ID_CR, Data1, Data2, Data3 1, aa, bb, cc 2, dd, ee, ff File B Row_ID_N, Src_Row_ID, DataN1 1a, 1, This is comment 1 2a, 1, This is comment 2 3a,

Dear R users, I have a data structure as follows: id two res1 res2 c1 c2 inter 1 -0.786093166 1 0 1 2 6 3 -0.308495749 1 0 0 1 2 5 -0.738033048 1 0 0 0 1 7 -0.52176252 1 0

Hi R helpers! I have a question. I'm trying to create a function for an exercise. Here are the arguments I should include: x and y are numeric z is a name ("plus","minus","multiply","divide") and swap is logical. Here is what the function should do: When z="plus", then x+y is performed and so on for the other z names. It should give a NA

Hi, I use lme to fit models like R> res1 <- lme(y~A+B, data=mydata, random=~1|subject) R> res2 <- lme(y~B+A, data=mydata, random=~1|subject) (only difference between these two models are the sequence in which the indep variables are written in formula) where y is continuous and A, B, and subject are factors. To get ANOVA table I used R> anova(res1) R> anova(res2) and found

Dear R-users, I would like to maximize the function g above which depends on 4 parameters (2 vectors, 1 real number, and 1 matrix) using optim() and BFGS method. Here is my code: # fonction to maximize g=function(x) { x1 = x[1:ncol(X)] x2 = x[(ncol(X)+1)] x3 = matrix(x[(ncol(X)+2):(ncol(X)+1+ncol(X)*ncol(Y))],nrow=ncol(X),ncol=ncol(Y)) x4 = x[(ncol(X)+1+ncol(X)*ncol(Y)+1):length(x)]

Hi everybody, I am beginer in R and I need your precious help. I want to create a small function in R as in sas to retrieve date. I have a file with data that import in R. DATE PAYS nb_pays.ILI. 1 24/04/2009 usa 0 2 24/04/2009 usa 0 3 24/04/2009 Mexique 0 4 24/04/2009

Thanks. Yes. Your approach can identify: Glaxy ace S 5830 and S 5830 Glaxy ace But you can not identify using same program: Iphone 4S 16 G Iphone 4S 16G How should I solve both in same time. Kind regards,Tammy [[alternative HTML version deleted]]

I am using R 1.1 with Redhat 6.2 and RW 1.001 with Win98 (the upkey doesn't work on my IBM either as has been previously reported by others). The function aov doesn't return either the residuals or the residual degrees of freedom for split-plot designs. If you use the following code from Baron and Li's "Notes on the use of R for psycology experiments and questionnaires"

HI Elisa, You could use ?cut() vec1<-c(33,18,13,47,30,10,6,21,39,25,40,29,14,16,44,1,41,4,15,20,46,32,38,5,31,12,48,27,36,24,34,2,35,11,42,9,8,7,26,22,43,17,19,28,23,3,49,37,50,45) label1<-unlist(lapply(mapply(c,lapply(seq(0,45,5),function(x) x),lapply(seq(5,50,5),function(x) x),SIMPLIFY=FALSE),function(i) paste(i[1],"<x<=",i[2],sep="")))

I initially thought, this should better be posted to r-devel but alas! no response. so I try it here. sory for the lengthy explanation but it seems unavoidable. to quickly see the problem simply copy the litte example below and execute f(n=5) which crashes. called with n != 5 (and of course n>3 since there are 3 parameters in the model...) everything is as it should be. in detail:

Dear all! I have a data frame composed by 13 columns (year, and 12 months). I want to transform this data base in another like this year month values 1901 1 1901 2 1901 3 ..... 1901 12 1902 1 1902 2 .... 1902 12 Is there a possibility to succeed that in R? Thank you! best regards! CR -- --- Catalin-Constantin ROIBU Lecturer PhD, Forestry engineer Forestry Faculty of Suceava Str.

Dear all...Merry Christmas I would like to split a long dataframe. The dataframe looks like this x<-c('0:00:00', '0:30:00', '1:00:00', '1:30:00', '2:00:00', '2:30:00', '3:00:00', '0:00:00', '0:30:00', '1:00:00', '1:30:00', '2:00:00', '2:30:00', '3:00:00', '3:30:00',

Hi Elisa, I hope this is what you wanted. dat1<-read.csv("peaks.csv",sep=",") #Subset dat2<-dat1[1:5,] res1<-do.call(cbind,lapply(seq_len(nrow(dat2)),function(i) do.call(rbind,lapply(split(rbind(dat2[i,],dat2[-i,]),1:nrow(rbind(dat2[i,],dat2[-i,]))), function(x) {x1<-rbind(dat2[i,],x);

I am trying to run separate regressions for different groups of observations using the lapply function. It works fine when I write the formula inside the lm() function. But I would like to pass formulae into lm(), so I can do multiple models more easily. I got an error message when I tried to do that. Here is my sample code: #generating data x1 <- rnorm(100,1) x2 <- rnorm(100,1) y <-

Hi, You could try: A<- matrix(unlist(read.table(text=" 1 2 3 4 5 6 7 8 9 9 8 7 6 5 4 3 2 1 ",sep="",header=FALSE)),ncol=3,byrow=FALSE,dimnames=NULL) library(matrixStats) ?res1<-t(sapply(split(as.data.frame(A),as.numeric(gl(nrow(A),2,6))),colProds)) ?res1 #? [,1] [,2] [,3] #1??? 4?? 10?? 18 #2?? 63?? 64?? 63 #3?? 18?? 10??? 4

Dear R users. I'm trying to avoid using nested loops in the following code but I'm not sure how to proceed. Any help would be greatly appreciated. With regards,Phil X = matrix(rnorm(100), 10, 10) ## Version with nested loopsresult = 0 for(m in 1:nrow(X)){ for(n in 1:ncol(X)){ if(X[m,n] != 0){ result = result + (X[m,n] / (1 + abs(m - n))) } }} ## No loop-sum(ifelse(M

Deal all, as MCMClogit does not allow for the specification of several chains, I have run my model 3 times with different random number seeds and differently dispersed multivariate normal priors. For example: res1 = MCMClogit(y~x,b0=0,B0=0.001,data=mydat, burnin=500, mcmc=5500, seed=1234, thin=5) res2 = MCMClogit(y~x,b0=1,B0=0.01,data=mydat, burnin=500, mcmc=5500, seed=5678, thin=5) res3 =