2. How about the following:
n <- 900 x <- runif(n) y <- runif(n) z <- sqrt(x^2+y^2) print(list(Overwork.ratio = n/sum(z<1))) plot(x, y) theta <- seq(0, pi/2, length=31) lines(sin(theta), cos(theta)) ######################
This won't give you n numbers z < 1. If you need exactly that, what about first generating, say, 2*n, then either throwing away the excess or generating another 2*n if you don't have enough? You can use a recursive function for this, which would almost never recurse with 2*n but might with 1.1*n.
hope this helps. spencer graves
Sean O'Riordain wrote:
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 someone help me make this a bit more R-ish please?
Or is there a completely different approach I should be taking?
Many thanks in advance, Sean O'Riordain seanpor AT acm.org
-------------------------------------------- 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] = runif(1, 0, 1); y[c] = runif(1, 0, 1);
z[c] = sqrt(x[c]^2 + y[c]^2); if (z[c] < 1) c <- c + 1; tc <- tc + 1; }
print("'overwork' ratio"); print(tc/(c-1)); plot(x,y);
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