我刚刚测试了上面的方法——它并不完美,但我想就我的目的而言,它应该足够了。(groovy中的代码,粘贴到单元测试中…)
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