如果您确定需要来自集合中随机位置的 n 个元素（有点像统计抽样），那么您可能需要考虑只迭代一次集合并按所需的 概率，当您遍历集合时。这种方式效率更高，因为您只需遍历集合一次。
下面的程序演示了这个想法：
import java.util.ArrayList;
import java.util.Iterator;
import java.util.List;
import java.util.Random;
import java.util.Set;
import java.util.TreeSet;
public class SamplingFromSet {
public static void main(String[] args) {
Set<String> population = new TreeSet<>();
/*
* Populate the set
*/
final int popSize = 17;
for (int i=0; i<popSize; i++) {
population.add(getRandomString());
}
List<String> sample
= sampleFromPopulation(population, 3 /*sampleSize */);
System.out.println("population is");
System.out.println(population.toString());
System.out.println("sample is");
System.out.println(sample.toString());
}
/**
* Pick some samples for a population
* @param population
* @param sampleSize - number of samples
* @return
*/
private static <T>
List<T> sampleFromPopulation(Set<T> population
, int sampleSize) {
float sampleProb = ((float) sampleSize) / population.size();
List<T> sample = new ArrayList<>();
Iterator<T> iter = population.iterator();
while (iter.hasNext()) {
T element = iter.next();
if (random.nextFloat()<sampleProb) {
/*
* Lucky Draw!
*/
sample.add(element);
}
}
return sample;
}
private static Random random = new Random();
private static String getRandomString() {
return String.valueOf(random.nextInt());
}
}
这个程序的输出：
population is
[-1488564139, -1510380623, -1980218182, -354029751, -564386445, -57285541, -753388655, -775519772, 1538266464, 2006248253, 287039585, 386398836, 435619764, 48109172, 580324150, 64275438, 860615531]
sample is
[-57285541, -753388655, 386398836]
更新
然而，上述程序有一个警告——因为
样本中的一个遍历集合是通过概率完成的，
返回的sample 可能取决于您当天的运气，
有比指定的更少或更多的样本。
然而，这个问题可以通过稍微改变程序来解决，
它使用了稍微不同的方法签名：
/**
* Pick some samples from a population
* @param population
* @param sampleSize - number of samples
* @param exactSize - a boolean to control whether or not
* the returned sample list must be of the exact size as
* specified.
* @return
*/
private static <T>
List<T> sampleFromPopulation(Set<T> population
, int sampleSize
, boolean exactSize);
防止过采样
在通过总体的一次迭代中，我们过度采样一点，
最后，如果我们确实有太多样本，我们会丢弃一些样本。
防止欠采样
还要注意，即使使用过采样，也存在非零概率
即，在人口的一次迭代结束时，我们仍然
获得比预期更少的样本。如果发生这种情况（不太可能），我们将递归调用
再次尝试相同的方法。 （此递归的概率接近 1
终止，因为它非常不同，在重复递归中
调用该方法，我们总是得到欠采样。）
以下代码实现了新的sampleFromPopulation() 方法：
private static <T>
List<T> sampleFromPopulation(Set<T> population
, int sampleSize
, boolean exactSize) {
int popSize = population.size();
double sampleProb = ((double) sampleSize) / popSize;
final double OVER_SAMPLING_MULIT = 1.2;
if (exactSize) {
/*
* Oversampling to enhance of chance of getting enough
* samples (if we then have too many, we will drop them
* later)
*/
sampleProb = sampleProb * OVER_SAMPLING_MULIT;
}
List<T> sample = new LinkedList<>(); // linked list for fast removal
Iterator<T> iter = population.iterator();
while (iter.hasNext()) {
T element = iter.next();
if (random.nextFloat()<sampleProb) {
/*
* Lucky Draw!
*/
sample.add(element);
}
}
int samplesTooMany = sample.size() - sampleSize;
if (!exactSize || samplesTooMany==0) {
return sample;
} else if (samplesTooMany>0) {
Set<Integer> indexesToRemoveAsSet = new HashSet<>();
for (int i=0; i<samplesTooMany; ) {
int candidate = random.nextInt(sample.size());
if (indexesToRemoveAsSet.add(candidate)) {
/*
* add() returns true if candidate was not
* previously in the set
*/
i++; // proceed to draw next index
}
}
List<Integer> indexesToRemoveAsList
= new ArrayList<>(indexesToRemoveAsSet);
Collections.sort(indexesToRemoveAsList
, (i1, i2) -> i2.intValue() - i1.intValue()); // desc order
/*
* Now we drop from the tail of the list
*/
for (Integer index : indexesToRemoveAsList) {
sample.remove((int) index); // remove by index (not by element)
}
return sample;
} else {
/*
* we were unluckly that we oversampling we still
* get less samples than specified, so here we call
* this very same method again recursively
*/
return sampleFromPopulation(population, sampleSize, exactSize);
}
}