search for: sds

..."from=" "to=" [13] "cut=" "na.rm=" Disabling JIT warnings removes the notes, but the call remains to be extremely slow, as the compiler still processes all the exceptions. Thanks, Vitalie. >>>> Sam Steingold <sds at gnu.org> >>>> on Tue, 20 Mar 2012 13:09:07 -0400 wrote: > ess hangs emacs completely. > I just created a brand new Rprofile > options(error = utils::recover) > library(compiler) > compiler::enableJIT(3) > and now and I start R and start typing I...

This is my current work.Now i am trying to use a function to do the normal distribution simulation. rm(list=ls()) t <- u<- mann<- rep(0, 45) Nsimulation<-function(S1,S2,Sds,nSims) { set.seed(1) for (sim in 1:nSims) { matrix_t <-matrix(0,nrow=nSims,ncol=3) matrix_u<-matrix(0,nrow=nSims,ncol=3) matrix_mann <-matrix(0,nrow=nSims,ncol=3) #generate random samples from 2 no...

Hi, Can anyone help me with the following. I have been using R for Monte Carlo simulations and got some results I couldn't explain. Therefor I performed following short test: -------------- mean.sds <- NULL sample.sizes <- 3:30 for(N in sample.sizes){ dum <- NULL for(I in 1:5000){ x <- rnorm(N,0,1) dum <- c(dum,sd(x)) } mean.sds<- c(mean.sds,mean(dum)) } plot(sample.sizes,mean.sds) -------------- My question is why don't I get 1 as a result from my sd() for s...

...0,nrow=nSims,ncol=3) matrix_mann <-matrix(0,nrow=nSims,ncol=3) sample_sizes<- matrix(c(10,10,10,25,25,25,25,50,25,100,50,25,50,100,100,25,100,100), nrow=2) p1<-p2<-p3<-vector() nSims<-10 alpha<-0.05 set.seed(1) simulation<-function(m,n,sds) { for (ss in 1:dim(sample_sizes[2])){ m<-[ss,1] n<[ss,2] for (sim in 1:nSims) { m<-c(10,25,50,100) n<-c(10,25,50,100) sds<-c(1,1.5,2,2.5,3) simulation(m=m,n=n,,sds=sds) p1 <- t.test(x1,y1,var.equal=TRUE)$p.value p2<-t.test(x1,y1,var.equal=FALSE)$p...

...bad form to double post to StackOverflow and R-help? Apologies if so. Here's my task: I've got a matrix of means like so means<-matrix(1:10,nrow=2) colnames(means)<-c("a","b","c","d","e") and a matrix of standard deviations like so sds<-matrix(seq(0.1,1,by=0.1),nrow=2) colnames(sds)<-c("a","b","c","d","e") and I want to output a Latex (or HTML) table where the standard deviations appear in parens just below the corresponding means. Something like this: a & b...

(summary.default): show the vector length in addition to quantiles diff -u -i -p -F '^(def' -b -w -B /home/sds/src/R-3.0.1/src/library/base/R/summary.R.old /home/sds/src/R-3.0.1/src/library/base/R/summary.R --- /home/sds/src/R-3.0.1/src/library/base/R/summary.R.old 2013-03-05 18:02:33.000000000 -0500 +++ /home/sds/src/R-3.0.1/src/library/base/R/summary.R 2013-09-10 10:19:02.682946339 -0400 @@ -39,6 +39,7 @@...

Dear R-community, I'm struggling with a paper that reports only fragmented results of a 2by2by3 experimental design. However, Means and SDs for all cells are given. Does anyone know a package/function that helps computing an ANOVA with only Means and SDs as input (resulting in some sort of effect size and significance test)? The reason why I'm interested is simple: I'm conducting a meta-analysis. If I included only what the a...

...see example below). I can get Figure 1 below to plot the data with error. However, I keep getting a the error message "Error using packet 1 object "y" not found" Can anyone see what I am doing wrong? Thanks! John ------------------------------- a=0.002; b=31.7; c=0.51 sds=rep(c(0,3,5,10,20,50,200), each=3) y1=c(0,0,0.16, 0, 0.33,0.5, 0.16, 0.83, 1.16, 0.67, 0.5, 1.16, 0.83, 2.33, 3.6, 5.5, 4.33, 1.16, 22, 13, 12) lo=y1-0.1*y1 hi=y1+0.1*y1 # Figure 1 xYplot(Cbind(y1, lo, hi)~jitter(sds, amount=1), method="bars",ylim=c(0,max(hi)+1), ylab=&quot...

...g to find sd of my rows in a matrix and i get column sd inspite of extracting rows. I tried to do the sqrt(var(x)) but that did'nt work as well, Here is my data genes 15 24 63 40 25 42 46 35 23 53 37 45 30 37 50 55 40 51 30 48 x<-sd(genes[1:5,]) y<-sqrt(var(genes[1:5,])) I get 4 sds for the 4 columns instead of 5 sds for my 5 rows. Thanks you in advance. Choudary Jagarlamudi Instructor Southwestern Oklahoma State University STF 254 100 campus Drive Weatherford OK 73096 Tel 580-774-7136 ________________________________

Hi all, I'm using the function "fitdistr" (library MASS) to fit a distribution to given data. What I have to do further, is getting the log-Likelihood-Value from this estimation. Is there any simple possibility to realize it? Regards, Carsten

...for each replication and the number of replications. It takes 5-10 minutes to run. ## Put the samples sizes into matrix then use a loop for sample sizes sample_sizes<- matrix(c(10,10,10,25,25,25,25,50,25,100,50,25,50,100,100,25,100,100), nrow=2) #create vector to combine all std deviations sds<-c(4,6,8,10,12,14) # this number is needed below nsds<-length(sds) set.seed(8) #number of simulations nSims<-10000 #set significance level,alpha for the whole simulatio alpha<-0.05 #set empty vector of length no.of _calculations_ to store p-values # Note: you have 54 calculations, not...

...roblem I am having is that R only replaces the last value from the loops, and I am not experienced enough with R to know how it stores data when looping and how to overcome this problem. By having it print the avg, I was expecting to see 24 numbers, but it only gives me one. Here is my code. cts.sds<-array(1:24,c(3,8)) for(i in 3){ for(j in 8){ avg<-sum(exprdata[3*i-2:3*i,j]/3) cts.sds[i,j]<-avg print(avg) } } print(cts.sds) Any help with this pesky matter will be greatly appreciated. -- View this message in context: http://www.nabble.com/Help-with-loops-and-how-R-...

...eans in brackets. Can anyone check this code to see how this can be adjusted? library(xtable) dataset1 = matrix( c(1,2,3,4, 5, 6 ), 2 , 3) dataset2 = matrix( c(4,3,5,10, 1, 0), 2, 3) dataset <- rbind(dataset1,dataset2) #combine dataset means <- apply(dataset,1,mean) #calculate row means sds <- apply(dataset,1,sd) #calculate row standard deviation msd <- paste(round(means,2)," (",round(sds,2),")",sep="") #mean and standard deviation rn <- c("Var1","Var2") #rownames cn <- c("Dataset1","Dataset2") #...

Hi all, Very surprising (to me!) and mystifying result from predict.glm(): the predictions vary depending on whether or not I use ns() or splines::ns(). Reprex follows: library(splines) set.seed(12345) dat <- data.frame(claim = rbinom(1000, 1, 0.5)) mns <- c(3.4, 3.6) sds <- c(0.24, 0.35) dat$wind <- exp(rnorm(nrow(dat), mean = mns[dat$claim + 1], sd = sds[dat$claim + 1])) dat <- dat[order(dat$wind), ] m1 <- glm(claim ~ ns(wind, df = 5), data = dat, family = binomial) m2 <- glm(claim ~ splines::ns(wind, df = 5), data = dat, family = binomial) # The...

...ount.0 <- length(which(x == 0)) --8<---------------cut here---------------end--------------->8--- however, this approach allocates and discards 2 vectors: a logical vector of length=length(x) and an integer vector in which. is there a cheaper alternative? Thanks! -- Sam Steingold (http://sds.podval.org/) on Ubuntu 12.04 (precise) X 11.0.11103000 http://www.childpsy.net/ http://iris.org.il http://honestreporting.com http://jihadwatch.org http://pmw.org.il http://www.PetitionOnline.com/tap12009/ War doesn't determine who's right, just who's left.

...). That is, I have a list/vector of say 100 means and another of the corresponding 100 SD's, and I want a matrix of 100 rows (one for each mean and SD pair) each having 1000 random samples. Something like: sample_matrix = rnorm(n=1000, mean=vector_of_100_means, sd=vector_of_100_corresponding_SDs) Is this possible? Thanks for any help in advance.

...is because that i have to consider the outer loop and inner loop also... so the iterative for i need to minus one because it have ran one times simulation already ,then times the number of sizes of inner loop, then plus the iterative of j.... then for the simulation, i think there will be ((ss-1)*nsds +sim) for the index in the body of for loop... Is it correct? Sent from my phone On Jim Lemon <drjimlemon at gmail.com>, Apr 6, 2016 6:52 PM wrote: You are running through two loops and putting the output into a vector. Without this calculation you will overwrite the same elements of the ou...

...19), c(5e+05, 16)) as.matrix(a) [,1] [1,] Numeric,2 [2,] Numeric,2 [3,] Numeric,2 [4,] Numeric,2 [5,] Numeric,2 [6,] Numeric,2 [7,] Numeric,2 [8,] Numeric,2 [9,] Numeric,2 --8<---------------cut here---------------end--------------->8--- thanks! -- Sam Steingold (http://sds.podval.org/) on Ubuntu 12.04 (precise) X 11.0.11103000 http://www.childpsy.net/ http://palestinefacts.org http://dhimmi.com http://jihadwatch.org http://www.PetitionOnline.com/tap12009/ http://memri.org Rhinoceros has poor vision, but, due to his size, it's not his problem.

How I can generate a random sample from a mixed norml distribution? (i.e. a mixed normal distribution in the form f(x)=.6*N(1,5)+.4*N(2,3) in which N(A,B) is a normal distribution with mean A and variance B.) -- Free e-mail accounts at http://zworg.com

Is there an analogue of common lisp "*" variable which contains the value of the last expression? E.g., in lisp: > (+ 1 2) 3 > * 3 I wish I could recover the value of the last expression without re-evaluating it. thanks -- Sam Steingold (http://sds.podval.org/) on Ubuntu 11.10 (oneiric) X 11.0.11004000 http://www.childpsy.net/ http://camera.org http://ffii.org http://truepeace.org http://memri.org http://americancensorship.org The early bird may get the worm, but the second mouse gets the cheese.