similar to: sample

Hello If I have this vector x=c(5,1,2,9) and n=length(x) and I want to sample one value from x , and each value of x has equal probability to appear (1/n). Are the following codes equivalent?? sample(x,1,replace=TRUE)  and   sample(x,1,replace=TRUE,prob=rep(1/n , n))   Regards [[alternative HTML version deleted]]

Hello I have a loop to draw few samples and I want to but this samples in one list, how I can make this? when I use it , the list show just the last sample. I want to know the correct way to put all samples in one list.   Regards Sulafah [[alternative HTML version deleted]]

Hello I have a loop to sample 20 samples and I want to put them in one list, how I can make this??   Regards Sulafah [[alternative HTML version deleted]]

Hello, R friends... I am very new to R, and I need some help. I am trying to construct a simulation for my dissertation. I need to create 1000 datasets of 1000 subjects with the following variables... Treatment variable - Drawn from a binomial distribution (1 run, prob=.13) Covariate 1 - Drawn from a normal distribution (mean=100, sd=16) Covariate 2 - Drawn from a normal distribution

Hello If i want to resampl from the tails of normal distribution , are these commans equivelant??   upper tail:qnorm(runif(n,pnorm(b),1))  if b is an upper tail boundary   or   upper tail:qnorm((1-p)+p(runif(n))  if p is the probability of each interval (the observatins are divided to intervals)   Regards [[alternative HTML version deleted]]

Greetings.? My wife is teaching an introductory stat class at UC Davis.? The class emphasizes the use of simulations, rather than mathematics, to get insight into statistics, and R is the mandated tool.?? A student in the class recently inquired about different approaches to sampling from a binomial distribution.? I've appended some code that exhibits the idea, the gist of which is that using

When `length( skewed.probs ) > 200' uniform samples are generated in R-devel. R-3.5.1 behaves as expected. `epsilon` can be a lot bigger than illustrated and still the uniform distribution is produced. Chuck > set.seed(123) > > epsilon <- 1e-10 > > ## uniform to 200 then small > p200 <- prop.table( rep( c(1, epsilon), c(200, 999-200))) > ## uniform to 201

Hello I used SIGN.test function in R 2.12.2 to apply one sample sign test and it is worked well ,but I want to put statistic in a variable and I could not get this value, I used : x=rnorm(100) t=SIGN.test(x,md=0,alt="less") t$rval[1]$statistic   the last command work in the old version of R but it does not work in R 2.12.2. what can I do to get value of statistic??   Regards Sulafah

Hello all, Consider the following problem: There is a matrix of probabilities: > set.seed(1) > probs <- array(abs(rnorm(25, sd = 0.33)), dim = c(5,5), dimnames = list(1:5, letters[1:5])) > probs a b c d e 1 0.21 0.27 0.50 0.0148 0.303 2 0.06 0.16 0.13 0.0053 0.258 3 0.28 0.24 0.21 0.3115 0.025 4 0.53 0.19 0.73 0.2710 0.656 5 0.11 0.10 0.37 0.1960

Hi all, For some odd reason when running na?ve bayes, k-NN, etc., I get slightly different results (e.g., error rates, classification probabilities) from run to run even though I am using the same random seed. Nothing else (input-wise) is changing, but my results are somewhat different from run to run. The only randomness should be in the partitioning, and I have set the seed before this

In case you don't get an answer from someone more knowledgeable: 1. I don't know. 2. But it is possible that other packages that are loaded after set.seed() fool with the RNG. 3. So I would call set.seed just before you invoke each random number generation to be safe. Cheers, Bert Bert Gunter "The trouble with having an open mind is that people keep coming along and sticking

On Mon, Feb 26, 2018 at 3:25 PM, Gary Black <gwblack001 at sbcglobal.net> wrote: (Sorry to be a bit slow responding.) You have not supplied a complete example, which would be good in this case because what you are suggesting could be a serious bug in R or a package. Serious journals require reproducibility these days. For example, JSS is very clear on this point. To your question >

Thank you, everybody, who replied! I appreciate your valuable advise! I will move the location of the set.seed() command to after all packages have been installed and loaded. Best regards, Gary Sent from my iPad > On Mar 4, 2018, at 12:18 PM, Paul Gilbert <pgilbert902 at gmail.com> wrote: > > On Mon, Feb 26, 2018 at 3:25 PM, Gary Black <gwblack001 at sbcglobal.net> >

Dear all, Suppose I have a dataset like below, then I take for example, 100 random sample "class" variable where contains "yes" and "no" respectively, 70% and 30%. I need a new 100 random sample from mydat dataset, but I can't get the result. Thanks you very much for any helps. Best, Muhammad Subianto mydat <- data.frame(size=c(30,12,15,10,12,12,25,30,20,14),

Hello How to compare  two statistical histograms? How i can know if these histograms are equivalent or not??   Regards [[alternative HTML version deleted]]

The following helps identify when .GlobalEnv$.Random.seed has changed: rng_tracker <- local({ last <- .GlobalEnv$.Random.seed function(...) { curr <- .GlobalEnv$.Random.seed if (!identical(curr, last)) { warning(".Random.seed changed") last <<- curr } TRUE } }) addTaskCallback(rng_tracker, name = "RNG tracker") EXAMPLE: >

On 04/03/2018 5:54 PM, Henrik Bengtsson wrote: > The following helps identify when .GlobalEnv$.Random.seed has changed: > > rng_tracker <- local({ > last <- .GlobalEnv$.Random.seed > function(...) { > curr <- .GlobalEnv$.Random.seed > if (!identical(curr, last)) { > warning(".Random.seed changed") > last <<- curr

Hi All, I'm generating 10 different data sets with 1 and 0 in a matrix form and writing the output in separate files. Now I need to stack all these data sets in one vector and I know that stack only operates on list or data frame however I got these data sets by converting list to a matrix so can't go backwards now. Is there a way i can still use Stack? Please see the program:

R 2.7.2 Windows XP I am using apply to compute a series of Student's t-test from two matrices, sample1 and sample2. boo<-apply(sample1,1,t.test,sample2) I want to pick of the p-values from the tests, but can't seem to get it to work. I have tried several methods to get the values including: boo<-apply(sample1,1,t.test$t.test,sample2) boo<-apply(sample1,1,t.test,sample2)$t.test

Under platform x86_64-pc-linux-gnu arch x86_64 os linux-gnu system x86_64, linux-gnu status major 2 minor 13.1 year 2011 month 07