从Java中的Map中选择随机键和值集-SelectingrandomkeyandvaluesetsfromaMapinJava

47  

HashMap x;

Random       random    = new Random();
List keys      = new ArrayList(x.keySet());
String       randomKey = keys.get( random.nextInt(keys.size()) );
String       value     = x.get(randomKey);

4  

This question should be of help to you Is there a way to get the value of a HashMap randomly in Java? and this one also Picking a random element from a set because HashMap is backed by a HashSet. It would be either O(n) time and constant space or it would be O(n) extra space and constant time.

这个问题应该对你有所帮助有没有办法在Java中随机获取HashMap的值?这个也从一个集中挑选一个随机元素,因为HashMap由一个HashSet支持。它可以是O(n)时间和恒定空间,也可以是O(n)额外空间和恒定时间。

3  

If you don't mind the wasted space, one approach would be to separately keep a List of all keys that are in the Map. For best performance, you'll want a List that has good random-access performance (like an ArrayList). Then, just get a random number between 0 (inclusive) and list.size() (exclusive), pull out the key at that index, and look that key up.

如果您不介意浪费的空间,一种方法是单独保留Map中所有键的List。为了获得最佳性能,您需要一个具有良好随机访问性能的List(如ArrayList)。然后,只需获得0(含)和list.size()(不包括)之间的随机数,拉出该索引处的键,然后查看该键。

Random rand = something
int randIndex = rand.nextInt(list.size());
K key = list.get(randIndex);
V value = map.get(key);

This approach also means that adding a key-value pair is a good deal cheaper than removing one. To add the key-value pair, you would test to see if the key is already in the map (if your values can be null, you'll have to separately call map.containsKey; if not, you can just add the key-value pair and see if the "old value" it returns is null). If the key is already in the map, the list is unchanged, but if not, you add the key to the list (an O(1) operation for most lists). Removing a key-value pair, though, involves an O(N) operation to remove the key from the list.

这种方法也意味着添加键值对比删除键值便宜。要添加键值对,您将测试键是否已经在地图中(如果您的值可以为null,则必须单独调用map.containsKey;如果不是,您只需添加键 - 值对并查看它返回的“旧值”是否为空)。如果密钥已经在映射中,则列表将保持不变,但如果不是,则将密钥添加到列表中(对于大多数列表,为O(1)操作)。但是,删除键值对涉及O(N)操作以从列表中删除键。

If space is a big concern, but performance is less so, you could also get an Iterator over the map's entry set (Map.entrySet()), and skip randIndex entries before returning the one you want. But that would be an O(N) operation, which kinda defeats the whole point of a map.

如果空间是一个很大的问题,但性能不那么重要,你也可以在地图的入口集(Map.entrySet())上获得一个迭代器,并在返回你想要的那个之前跳过randIndex条目。但这将是一个O(N)操作,它有点击败了地图的整个点。

Finally, you can just get the entry set's toArray() and randomly index into that. That's simpler, though less efficient.

最后,您可以获取条目集的toArray()并随机索引到该条目。这更简单,但效率更低。

2  

if your keys are integer, or something comparable, you can use TreeMap to do that.

如果你的键是整数或类似的东西,你可以使用TreeMap来做到这一点。

TreeMap treeMap = new TreeMap<>();
int key = RandomUtils.ranInt(treeMap.lastKey());
int value = treeMap.ceilingKey(key);

1  

I would copy the Map into an array and select the entry you want at random. This avoid the need to also lookup the value from the key.

我会将Map复制到一个数组中并随机选择你想要的条目。这样就无需从密钥中查找值。

Map x = new HashMap();
Map.Entry[] entries = x.entrySet().toArray(new Map.Entry[0]);
Random rand = new Random();

// call repeatedly
Map.Entry keyValue = entries[rand.nextInt(entries.length)];

If you want to avoid duplication, you can randomize the order of the entries

如果要避免重复,可以随机化条目的顺序

Map x = new HashMap();
List> entries = new ArrayList> (x.entrySet());
Collections.shuffle(entries);
for (Map.Entry entry : entries) {
    System.out.println(entry);
}

1  

Use reservoir sampling to select a list of random keys, then insert them into a map (along with their corresponding values in the source map.)

使用油藏采样选择随机键列表,然后将它们插入地图(以及源地图中的相应值)。

This way you do not need to copy the whole keySet into an array, only the selected keys.

这样您就不需要将整个keySet复制到一个数组中,只需复制选定的键。

public static Map sampleFromMap(Map source, int n, Random rnd) {
    List chosenKeys = new ArrayList();
    int count = 0;
    for (K k: source.keySet()) {
        if (count++  result = new HashMap();
    for (K k: chosenKeys) {
        result.put(k, source.get(k));
    }
    return Collections.unmodifiableMap(result);
}

0  

In some cases you might want to preserve an order you put the elements in the Set,
In such scenario you can use, This 

Set alldocsId = new HashSet<>();
            for (int i=0;i alldocIDlst = new ArrayList<>();
        Iterator it = alldocsId.iterator();
        while (it.hasNext())
        {
            alldocIDlst.add(Integer.valueOf(it.next().toString()));
        }

-1  

Been a while since a played with java, but doesn't keySet() give you a list that you can select from using a numerical index? I think you could pick a random number and select that from the keySet of myMap, then select the corresponding value from myMap. Can't test this right now, but it seems to strike me as possible!

自从玩过java以来​​已经有一段时间了,但是keySet()没有给你一个可以使用数字索引选择的列表吗?我想你可以选择一个随机数并从myMap的keySet中选择,然后从myMap中选择相应的值。现在无法测试,但它似乎尽可能地打击我!