Question

我有一个接受HashMap<String, HashSet<Integer>>的函数。现在我想从HashMap获得一个随机值，但我不知道如何做到这一点。你能给我一个暗示吗？

输出应包含一个包含String和Integer值的元组。

Answer 1

知道地图的大小，您可以选择一个随机的条目编号，然后迭代内容，直到您到达该条目。例如：

final Set<String> keys = allowedInput.keySet();
final int keyNumber = (int)(Math.random() * keys.size());
final Iterator<String> keyIterator = keys.iterator();

String randomKey = null;

for (int i = 0; i < keyNumber && keyIterator.hasNext(); i++) {
    randomKey = keyIterator.next();
}

if (randomKey == null) {
    // This should not happen unless the map was empty, or it was modified
    // externally.  Handle the potential error case accordingly.
}

final HashSet<Integer> value = allowedInput.get(randomKey);

// `value` now contains a random element from the `allowedInput` map.

如果要从结果Integer中检索随机HashSet<Integer>元素，则可以采用相同的技术：只需根据集合的大小选择随机元素编号，然后迭代直到找到它为止。

Answer 2

如果你想重复获取随机值，你可以随机播放，然后按顺序浏览它。

请参阅Picking a random element from a set

Answer 3

我创建了一个利用Mike和SecureRandom的答案的通用解决方案，并包含显式的空值和边界检查，以及单例集合的快速返回（在那里选择不多）。

public static <T> T getRandomElement(Collection<T> collection) {
    if (collection == null || collection.isEmpty()) {
        throw new IllegalArgumentException("Collection should not be null or empty");
    }
    if (collection.size() == 1) {
        return collection.iterator().next();
    }

    // it would be beneficial to make this a field when used a lot
    final Random random = new SecureRandom();
    final int randomIndex  = random.nextInt(collection.size());

    // optimization for list instances, use optimized indexing
    if (collection instanceof List) {
        final List<T> list = (List<T>) collection;
        return list.get(randomIndex);
    }

    int seen = 0;
    for (T e : collection) {
        if (seen++ == randomIndex) {
            return e;
        }
    }
    throw new IllegalStateException("Collection size was altered during operation");
}

现在，您只需先从密钥集中选择一个密钥值，然后从该值中选择一个随机整数，就可以直接检索String和Integer。

String key = getRandomElement(aMap.keySet());
Integer value = getRandomElement(aMap.get(key));

Answer 4

谢谢你的帮助我让我现在工作这是我使用的代码

final Set<String> keys = allowedInput.keySet();
        int keyNumber = (int)(random.nextInt(keys.size()));
        final Iterator<String> keyIterator = keys.iterator();

        String key = null;

        for (int i = 0; i <= keyNumber; i++) {
            key = keyIterator.next();
        }

        if (key == null) {
            // handle empty map
        }

        HashSet<Integer> field = allowedInput.get(key);

        final int fieldNumber = (int)(random.nextInt(field.size()));
        int fieldID = 0;
        int i = 0;
        for(Object obj : field)
        {

            if (i == fieldNumber)
            {
                //fieldID =  Integer.parseInt(obj.toString());
                fieldID =  (int)obj;
                break;
            }
            i = i + 1;

        }

        // Start constructing the userinput
        // Note: we need an instance of the UserInputSystem to create the UserInput instance!
        UserInputSystem userInputSystem = new UserInputSystem();
        UserInput input = userInputSystem.new UserInput(fieldID, key);
        System.err.println("ai fieldID = "+fieldID+" key = "+key);

从HashMap中获取随机值<string，hashset <integer =“”>＆gt; </string，>

4 个答案: