Hi,
I think you can do this without a loop (well, replicate() is based on
sapply()):
prob<-numeric(1000)
task1 <- replicate(1000,runif(1, min=0.8, max= 0.9))
task2 <- replicate(1000,runif(1, min=0.75, max= 0.85))
task3 <- replicate(1000,runif(1, min=0.81, max= 0.89))
prob <- task1*task2*task3
It might not be faster, but I don't think it can be slower. And I find
the code easier and clearer.
Please correct me if this is not equivalent.
HTH,
Ivan
Le 4/11/2011 01:06, Daniel Malter a écrit :
The loop is correct, you just need to make sure that your result is computed
and stored as the n-th element that is returned by the loop. Pick up any
manual of R, and looping will be explained there. Also, I would recommend
that you draw a random number for every iteration of the loop. Defining the
random vectors outside the loop make sense to me only if they are the same
length as n.
prob<-numeric(1000)
for (n in 1:1000) {
task1<- runif(1, min=0.8, max= 0.9)
task2<- runif(1, min=0.75, max= 0.85)
task3<- runif(1, min=0.81, max= 0.89)
prob[n]<-task1*task2*task3
}
If you wanted to store the individual probabilities (task1..3), you would
proceed accordingly by defining them outside the loop and storing the value
in the loop as the n-th element of that vector just like for prob.
HTH,
Daniel
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