Hi Patrick, Spencer,
Thanks for that! Both your solutions are MUCH quicker!
afaik overwork_ratio = 4/pi # what a hard way to calculate this :-)
The real problem I'm actually working on is a little more complicated
:-) - 6 'random' variables, and 18 dependant variables (at last
count)... curre
On Wed, Oct 15, 2003 at 10:06:36AM +0100, Sean O'Riordain wrote:
> n <- 900; # number of valid items required...
>
> x <- numeric(n);
> y <- numeric(n);
> z <- numeric(n);
>
> c <- 1; # current 'array' pointer
> tc <- 0; # total items actually looked at...
>
> while (c <= n) {
> x[c] = ru
For this particular problem you can probably use
polar coordinates.
But something similar to your code could be:
x <- runif(900)
y <- runif(900)
z <- sqrt(x^2 + y^2)
okay <- z < 1
while(any(!okay)) {
n.bad <- sum(!okay)
x[!okay] <- runif(n.bad)
y[!okay] <- runif(n.bad)
z <- sqrt(x^2 +
1. I suggest you avoid using "c" as a loop index, as it conflicts
with the name of a function. R is smart enough to figure out the
difference in some cases but not in all.
2. How about the following:
n <- 900
x <- runif(n)
y <- runif(n)
z <- sqrt(x^2+y^2)
print(list(Overwork.ratio
Hi Folks,
I'm trying to learn R. One of my intentions is to do some Monte-Carlo
type modelling of road "accidents".
Below, to simplify things, I've appended a little program which does a
'monte-carlo' type simulation. However, it is written in a way which
seems a bit un-natural in R. Could
