我刚刚测试了上面的方法 - 它不完美,但我想我的目的,它应该是足够的。 (代码在常规,粘贴到一个单元测试...)
void test() {
for (int i = 0; i < 10; i++) {
once()
}
}
private def once() {
def double[] probs = [1/11, 2/11, 3/11, 1/11, 2/11, 2/11]
def int[] whoCounts = new int[probs.length]
def Random r = new Random()
def int who
int TIMES = 1000000
for (int i = 0; i < TIMES; i++) {
who = selectPerson(probs, r.nextDouble())
whoCounts[who]++
}
for (int j = 0; j < probs.length; j++) {
System.out.printf(" %10f ", (probs[j] - (whoCounts[j]/TIMES)))
}
println ""
}
public int selectPerson(double[] probabilies, double r) {
double t = r
double p = 0.0f;
for (int i = 0; i < probabilies.length; i++) {
p += probabilies[i];
if (t < p) {
return i;
}
}
return probabilies.length - 1;
}
outputs: the difference betweenn the probability, and the actual count/total
obtained over ten 1,000,000 runs:
-0.000009 0.000027 0.000149 -0.000125 0.000371 -0.000414
-0.000212 -0.000346 -0.000396 0.000013 0.000808 0.000132
0.000326 0.000231 -0.000113 0.000040 -0.000071 -0.000414
0.000236 0.000390 -0.000733 -0.000368 0.000086 0.000388
-0.000202 -0.000473 -0.000250 0.000101 -0.000140 0.000963
0.000076 0.000487 -0.000106 -0.000044 0.000095 -0.000509
0.000295 0.000117 -0.000545 -0.000112 -0.000062 0.000306
-0.000584 0.000651 0.000191 0.000280 -0.000358 -0.000181
-0.000334 -0.000043 0.000484 -0.000156 0.000420 -0.000372
对于记录,这不是别名方法。有关可以轻松移植到Java的C#实现,请参阅http://stackoverflow.com/a/9958717/1913277。 – Carl 2015-10-23 17:34:36