Help with basic loop

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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Hi:

Let's assume the lengths of each vector are the same so that they can 
be multiplied. Here's the timing on my machine:

system.time(replicate(1000, { 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
+ }
+   }))
   user  system elapsed
  16.96    0.01   17.19

system.time(replicate(1000, {

+  task1 = runif(1000, min = 0.8, max = 0.9)
+  task2 <- runif(1000, min = 0.75, max = 0.85)
+  task3 <- runif(1000, min = 0.81, max = 0.89)
+  prob <- task1 * task2 * task3 } ))
   user  system elapsed
   0.37    0.00    0.39

Dennis

On Mon, Apr 11, 2011 at 1:42 AM, Ivan Calandra 
<ivan.calandra at uni-hamburg.de <mailto:ivan.calandra at uni-hamburg.de>> 
wrote:

    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 ?crit :

        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

        --
        View this message in context:
        http://r.789695.n4.nabble.com/Help-with-basic-loop-tp3440190p3440607.html
        Sent from the R help mailing list archive at Nabble.com.