Hi Carrie, Use the first approach:
n <- 5 p <- c(0.2, 0.9, 0.15, 0.8, 0.75) rbinom(n, 1, p) # [1] 0 0 0 1 1 rbinom(n, 1, p) # [1] 1 1 0 1 1 To check, replicate the analysis 5000 times and then estimate the probability for each subject: rowMeans(replicate(5000, rbinom(n, 1, p))) # 0.2002 0.9026 0.1550 0.8020 0.7562 Using Erik's solution you will get the "same" results: > rowMeans(replicate(5000, sapply(p, function(x) sample(0:1, 1, prob = c(1-x, x))))) # [1] 0.1958 0.9006 0.1492 0.8066 0.7446 Note that in both cases the result is close to the original values of p. HTH, Jorge On Sat, May 22, 2010 at 10:52 PM, Carrie Li <> wrote: > Dear R-helpers, > > I would like to generate a variable that takes 0 or 1, and each subject has > different probabilities of taking the draw. > > So, which of the following code I should use ? > > suppose there are 5 subjects, and their probabilities of this Bernoulli > variable is p=c(0.2, 0.9, 0.15, 0.8, 0.75) > n<-5 > Ber.var <- rbimon(n,1,p) ## I doubt if this will take the first > probability, > which is 0.2, but won't change for each subject ?? > > ## or should I use > Ber.var <- sample(c(0,1), n, prob=p, replace=TRUE) > > Or any suggestions on how I can check if the probability of drawing this > binary variable is based on the subject's probabilities ? > > Thanks for help!! > > Carrie > > [[alternative HTML version deleted]] > > ______________________________________________ > R-help@r-project.org mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide > http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. > [[alternative HTML version deleted]] ______________________________________________ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
