On Monday, 3 June 2013 at 10:10:15 UTC, Yann wrote:
Thanks a lot for the suggestions!
Cheers,
Yann
On Monday, 3 June 2013 at 10:06:30 UTC, bearophile wrote:
Yann:
Is there a better way to accomplish this? Naively, I would
expect something like
"return iota(1, 1000).randomShuffle.take(10).sort;"
Two ways, the first gives items in random order, the second
ordered as you seem to desire:
import std.stdio, std.random, std.range, std.array;
void main() {
iota(1, 1001).randomCover(rndGen).take(10).writeln;
iota(1, 1001).randomSample(10).writeln;
}
But be careful with randomCover, because maybe it takes the
rnd generator by value.
Bye,
bearophile
For small samples from very large ranges an efficient algorithm
would be:
int[] randomGen(int N, int M) {
if (N == 0)
return [];
int[] result = randomGen(N-1, M-1);
int num = rand(M);
foreach (ref int i; result)
if (i >= num)
++i;
result ~= num;
return result;
}
