Picking a random element from a set

问题:

How do I pick a random element from a set? I\'m particularly interested in picking a random element from a HashSet or a LinkedHashSet, in Java. Solutions for other languages are also welcome.

回答1:

int size = myHashSet.size();
int item = new Random().nextInt(size); // In real life, the Random object should be rather more shared than this
int i = 0;
for(Object obj : myhashSet)
{
    if (i == item)
        return obj;
    i++;
}

回答2:

A somewhat related Did You Know:

There are useful methods in java.util.Collections for shuffling whole collections: Collections.shuffle(List<?>) and Collections.shuffle(List<?> list, Random rnd).

回答3:

Fast solution for Java using an ArrayList and a HashMap: [element -> index].

Motivation: I needed a set of items with RandomAccess properties, especially to pick a random item from the set (see pollRandom method). Random navigation in a binary tree is not accurate: trees are not perfectly balanced, which would not lead to a uniform distribution.

public class RandomSet<E> extends AbstractSet<E> {

    List<E> dta = new ArrayList<E>();
    Map<E, Integer> idx = new HashMap<E, Integer>();

    public RandomSet() {
    }

    public RandomSet(Collection<E> items) {
        for (E item : items) {
            idx.put(item, dta.size());
            dta.add(item);
        }
    }

    @Override
    public boolean add(E item) {
        if (idx.containsKey(item)) {
            return false;
        }
        idx.put(item, dta.size());
        dta.add(item);
        return true;
    }

    /**
     * Override element at position <code>id</code> with last element.
     * @param id
     */
    public E removeAt(int id) {
        if (id >= dta.size()) {
            return null;
        }
        E res = dta.get(id);
        idx.remove(res);
        E last = dta.remove(dta.size() - 1);
        // skip filling the hole if last is removed
        if (id < dta.size()) {
            idx.put(last, id);
            dta.set(id, last);
        }
        return res;
    }

    @Override
    public boolean remove(Object item) {
        @SuppressWarnings(value = \"element-type-mismatch\")
        Integer id = idx.get(item);
        if (id == null) {
            return false;
        }
        removeAt(id);
        return true;
    }

    public E get(int i) {
        return dta.get(i);
    }

    public E pollRandom(Random rnd) {
        if (dta.isEmpty()) {
            return null;
        }
        int id = rnd.nextInt(dta.size());
        return removeAt(id);
    }

    @Override
    public int size() {
        return dta.size();
    }

    @Override
    public Iterator<E> iterator() {
        return dta.iterator();
    }
}

回答4:

This is faster than the for-each loop in the accepted answer:

int index = rand.nextInt(set.size());
Iterator<Object> iter = set.iterator();
for (int i = 0; i < index; i++) {
    iter.next();
}
return iter.next();

The for-each construct calls Iterator.hasNext() on every loop, but since index < set.size(), that check is unnecessary overhead. I saw a 10-20% boost in speed, but YMMV. (Also, this compiles without having to add an extra return statement.)

Note that this code (and most other answers) can be applied to any Collection, not just Set. In generic method form:

public static <E> E choice(Collection<? extends E> coll, Random rand) {
    if (coll.size() == 0) {
        return null; // or throw IAE, if you prefer
    }

    int index = rand.nextInt(coll.size());
    if (coll instanceof List) { // optimization
        return ((List<? extends E>) coll).get(index);
    } else {
        Iterator<? extends E> iter = coll.iterator();
        for (int i = 0; i < index; i++) {
            iter.next();
        }
        return iter.next();
    }
}

回答5:

If you want to do it in Java, you should consider copying the elements into some kind of random-access collection (such as an ArrayList). Because, unless your set is small, accessing the selected element will be expensive (O(n) instead of O(1)). [ed: list copy is also O(n)]

Alternatively, you could look for another Set implementation that more closely matches your requirements. The ListOrderedSet from Commons Collections looks promising.

回答6:

In Java:

Set<Integer> set = new LinkedHashSet<Integer>(3);
set.add(1);
set.add(2);
set.add(3);

Random rand = new Random(System.currentTimeMillis());
int[] setArray = (int[]) set.toArray();
for (int i = 0; i < 10; ++i) {
    System.out.println(setArray[rand.nextInt(set.size())]);
}

回答7:

List asList = new ArrayList(mySet);
Collections.shuffle(asList);
return asList.get(0);

回答8:

Clojure solution:

(defn pick-random [set] (let [sq (seq set)] (nth sq (rand-int (count sq)))))

回答9:

Perl 5

@hash_keys = (keys %hash);
$rand = int(rand(@hash_keys));
print $hash{$hash_keys[$rand]};

Here is one way to do it.

回答10:

C++. This should be reasonably quick, as it doesn\'t require iterating over the whole set, or sorting it. This should work out of the box with most modern compilers, assuming they support tr1. If not, you may need to use Boost.

The Boost docs are helpful here to explain this, even if you don\'t use Boost.

The trick is to make use of the fact that the data has been divided into buckets, and to quickly identify a randomly chosen bucket (with the appropriate probability).

//#include <boost/unordered_set.hpp>  
//using namespace boost;
#include <tr1/unordered_set>
using namespace std::tr1;
#include <iostream>
#include <stdlib.h>
#include <assert.h>
using namespace std;

int main() {
  unordered_set<int> u;
  u.max_load_factor(40);
  for (int i=0; i<40; i++) {
    u.insert(i);
    cout << \' \' << i;
  }
  cout << endl;
  cout << \"Number of buckets: \" << u.bucket_count() << endl;

  for(size_t b=0; b<u.bucket_count(); b++)
    cout << \"Bucket \" << b << \" has \" << u.bucket_size(b) << \" elements. \" << endl;

  for(size_t i=0; i<20; i++) {
    size_t x = rand() % u.size();
    cout << \"we\'ll quickly get the \" << x << \"th item in the unordered set. \";
    size_t b;
    for(b=0; b<u.bucket_count(); b++) {
      if(x < u.bucket_size(b)) {
        break;
      } else
        x -= u.bucket_size(b);
    }
    cout << \"it\'ll be in the \" << b << \"th bucket at offset \" << x << \". \";
    unordered_set<int>::const_local_iterator l = u.begin(b);
    while(x>0) {
      l++;
      assert(l!=u.end(b));
      x--;
    }
    cout << \"random item is \" << *l << \". \";
    cout << endl;
  }
}

回答11:

Solution above speak in terms of latency but doesn\'t guarantee equal probability of each index being selected.
If that needs to be considered, try reservoir sampling. http://en.wikipedia.org/wiki/Reservoir_sampling.
Collections.shuffle() (as suggested by few) uses one such algorithm.

回答12:

Since you said \"Solutions for other languages are also welcome\", here\'s the version for Python:

>>> import random
>>> random.choice([1,2,3,4,5,6])
3
>>> random.choice([1,2,3,4,5,6])
4

回答13:

Can\'t you just get the size/length of the set/array, generate a random number between 0 and the size/length, then call the element whose index matches that number? HashSet has a .size() method, I\'m pretty sure.

In psuedocode -

function randFromSet(target){
 var targetLength:uint = target.length()
 var randomIndex:uint = random(0,targetLength);
 return target[randomIndex];
}

回答14:

PHP, assuming \"set\" is an array:

$foo = array(\"alpha\", \"bravo\", \"charlie\");
$index = array_rand($foo);
$val = $foo[$index];

The Mersenne Twister functions are better but there\'s no MT equivalent of array_rand in PHP.

回答15:

Icon has a set type and a random-element operator, unary \"?\", so the expression

? set( [1, 2, 3, 4, 5] )

will produce a random number between 1 and 5.

The random seed is initialized to 0 when a program is run, so to produce different results on each run use randomize()

回答16:

In C#

        Random random = new Random((int)DateTime.Now.Ticks);

        OrderedDictionary od = new OrderedDictionary();

        od.Add(\"abc\", 1);
        od.Add(\"def\", 2);
        od.Add(\"ghi\", 3);
        od.Add(\"jkl\", 4);


        int randomIndex = random.Next(od.Count);

        Console.WriteLine(od[randomIndex]);

        // Can access via index or key value:
        Console.WriteLine(od[1]);
        Console.WriteLine(od[\"def\"]);

回答17:

Javascript solution ;)

function choose (set) {
    return set[Math.floor(Math.random() * set.length)];
}

var set  = [1, 2, 3, 4], rand = choose (set);

Or alternatively:

Array.prototype.choose = function () {
    return this[Math.floor(Math.random() * this.length)];
};

[1, 2, 3, 4].choose();

回答18:

In lisp

(defun pick-random (set)
       (nth (random (length set)) set))

回答19:

In Mathematica:

a = {1, 2, 3, 4, 5}

a[[ ⌈ Length[a] Random[] ⌉ ]]

Or, in recent versions, simply:

RandomChoice[a]

---

This received a down-vote, perhaps because it lacks explanation, so here one is:

Random[] generates a pseudorandom float between 0 and 1. This is multiplied by the length of the list and then the ceiling function is used to round up to the next integer. This index is then extracted from a.

Since hash table functionality is frequently done with rules in Mathematica, and rules are stored in lists, one might use:

a = {\"Badger\" -> 5, \"Bird\" -> 1, \"Fox\" -> 3, \"Frog\" -> 2, \"Wolf\" -> 4};

回答20:

How about just

public static <A> A getRandomElement(Collection<A> c, Random r) {
  return new ArrayList<A>(c).get(r.nextInt(c.size()));
}

回答21:

This is identical to accepted answer (Khoth), but with the unnecessary size and i variables removed.

    int random = new Random().nextInt(myhashSet.size());
    for(Object obj : myhashSet) {
        if (random-- == 0) {
            return obj;
        }
    }

Though doing away with the two aforementioned variables, the above solution still remains random because we are relying upon random (starting at a randomly selected index) to decrement itself toward 0 over each iteration.

回答22:

Unfortunately, this cannot be done efficiently (better than O(n)) in any of the Standard Library set containers.

This is odd, since it is very easy to add a randomized pick function to hash sets as well as binary sets. In a not to sparse hash set, you can try random entries, until you get a hit. For a binary tree, you can choose randomly between the left or right subtree, with a maximum of O(log2) steps. I\'ve implemented a demo of the later below:

import random

class Node:
    def __init__(self, object):
        self.object = object
        self.value = hash(object)
        self.size = 1
        self.a = self.b = None

class RandomSet:
    def __init__(self):
        self.top = None

    def add(self, object):
        \"\"\" Add any hashable object to the set.
            Notice: In this simple implementation you shouldn\'t add two
                    identical items. \"\"\"
        new = Node(object)
        if not self.top: self.top = new
        else: self._recursiveAdd(self.top, new)
    def _recursiveAdd(self, top, new):
        top.size += 1
        if new.value < top.value:
            if not top.a: top.a = new
            else: self._recursiveAdd(top.a, new)
        else:
            if not top.b: top.b = new
            else: self._recursiveAdd(top.b, new)

    def pickRandom(self):
        \"\"\" Pick a random item in O(log2) time.
            Does a maximum of O(log2) calls to random as well. \"\"\"
        return self._recursivePickRandom(self.top)
    def _recursivePickRandom(self, top):
        r = random.randrange(top.size)
        if r == 0: return top.object
        elif top.a and r <= top.a.size: return self._recursivePickRandom(top.a)
        return self._recursivePickRandom(top.b)

if __name__ == \'__main__\':
    s = RandomSet()
    for i in [5,3,7,1,4,6,9,2,8,0]:
        s.add(i)

    dists = [0]*10
    for i in xrange(10000):
        dists[s.pickRandom()] += 1
    print dists

I got [995, 975, 971, 995, 1057, 1004, 966, 1052, 984, 1001] as output, so the distribution seams good.

I\'ve struggled with the same problem for myself, and I haven\'t yet decided weather the performance gain of this more efficient pick is worth the overhead of using a python based collection. I could of course refine it and translate it to C, but that is too much work for me today :)

回答23:

In Java 8:

static <E> E getRandomSetElement(Set<E> set) {
    return set.stream().skip(new Random().nextInt(set.size())).findFirst().orElse(null);
}

回答24:

PHP, using MT:

$items_array = array(\"alpha\", \"bravo\", \"charlie\");
$last_pos = count($items_array) - 1;
$random_pos = mt_rand(0, $last_pos);
$random_item = $items_array[$random_pos];

回答25:

For fun I wrote a RandomHashSet based on rejection sampling. It\'s a bit hacky, since HashMap doesn\'t let us access it\'s table directly, but it should work just fine.

It doesn\'t use any extra memory, and lookup time is O(1) amortized. (Because java HashTable is dense).

class RandomHashSet<V> extends AbstractSet<V> {
    private Map<Object,V> map = new HashMap<>();
    public boolean add(V v) {
        return map.put(new WrapKey<V>(v),v) == null;
    }
    @Override
    public Iterator<V> iterator() {
        return new Iterator<V>() {
            RandKey key = new RandKey();
            @Override public boolean hasNext() {
                return true;
            }
            @Override public V next() {
                while (true) {
                    key.next();
                    V v = map.get(key);
                    if (v != null)
                        return v;
                }
            }
            @Override public void remove() {
                throw new NotImplementedException();
            }
        };
    }
    @Override
    public int size() {
        return map.size();
    }
    static class WrapKey<V> {
        private V v;
        WrapKey(V v) {
            this.v = v;
        }
        @Override public int hashCode() {
            return v.hashCode();
        }
        @Override public boolean equals(Object o) {
            if (o instanceof RandKey)
                return true;
            return v.equals(o);
        }
    }
    static class RandKey {
        private Random rand = new Random();
        int key = rand.nextInt();
        public void next() {
            key = rand.nextInt();
        }
        @Override public int hashCode() {
            return key;
        }
        @Override public boolean equals(Object o) {
            return true;
        }
    }
}

回答26:

you can also transfer the set to array use array it will probably work on small scale i see the for loop in the most voted answer is O(n) anyway

Object[] arr = set.toArray();

int v = (int) arr[rnd.nextInt(arr.length)];

回答27:

If you really just want to pick \"any\" object from the Set, without any guarantees on the randomness, the easiest is taking the first returned by the iterator.

    Set<Integer> s = ...
    Iterator<Integer> it = s.iterator();
    if(it.hasNext()){
        Integer i = it.next();
        // i is a \"random\" object from set
    }

回答28:

The easiest with Java 8 is:

outbound.stream().skip(n % outbound.size()).findFirst().get()

where n is a random integer. Of course it is of less performance than that with the for(elem: Col)

回答29:

A generic solution using Khoth\'s answer as a starting point.

/**
 * @param set a Set in which to look for a random element
 * @param <T> generic type of the Set elements
 * @return a random element in the Set or null if the set is empty
 */
public <T> T randomElement(Set<T> set) {
    int size = set.size();
    int item = random.nextInt(size);
    int i = 0;
    for (T obj : set) {
        if (i == item) {
            return obj;
        }
        i++;
    }
    return null;
}

回答30:

If set size is not large then by using Arrays this can be done.

int random;
HashSet someSet;
<Type>[] randData;
random = new Random(System.currentTimeMillis).nextInt(someSet.size());
randData = someSet.toArray();
<Type> sResult = randData[random];