Here is another solution if you want to count a sequence like 0 as
three overlapping runs of 3 zeros.
> x <- sample(c(rep(0,20), rep(1,37)), 100, TRUE) # vector of 1M
> z <- rle(x) # get the runs
> indx <- which(z$values == 0 & z$lengths > 2) # find the runs
> # now determine the number
Here's a one-liner. Let's see their software do that!
sum(replicate(100, sum(sample(c(rep(0, 20), seq(1:37)), 3, replace =
FALSE)) == 0))
Am 27.06.2011 23:08, schrieb robcinm:
I am taking a basic statistics course this summer, and while the majority of
the class is using a statistical packa
Hi robcinm,
You might also consider:
# data
x <- c(rep(0, 20), 1:37)
# number of simulations
B <- 1000
# result: TRUE/FALSE
out <- replicate(B, {
y <- sample(x, 3, replace = FALSE)
all(y == 0)
})
mean(out)
HTH,
Jorge
On Mon, Jun 27, 2011 at 5:08 PM, robcinm <> wrote:
> I
#sampling elements
x<-c(rep(0, 20), seq(1:37))
#number of simulations to perform
sims<-100
#vector to store results
results<-c()
#using for loop to perform simulations
for(i in 1:sims){
#take your sample
y<-sample(x,3,replace=FALSE)
#check if all elements are zero
I am taking a basic statistics course this summer, and while the majority of
the class is using a statistical package that came with the book, I am doing
everything in R for practical reasons. Forgive me if there is
documentation/instruction easily available on this topic, but Google and
forums sea
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