55

你知道一些简洁的 Java 库,它们可以让你制作两个(或更多)集合的笛卡尔积吗?

例如:我有三套。一个是 Person 类的对象,第二个是 Gift 类的对象,第三个是 GiftExtension 类的对象。

我想生成一组包含所有可能的三元组 Person-Gift-GiftExtension。

集合的数量可能会有所不同,因此我无法在嵌套的 foreach 循环中执行此操作。在某些情况下,我的应用程序需要制作 Person-Gift pair 的乘积,有时它是三重 Person-Gift-GiftExtension,有时甚至可能会设置 Person-Gift-GiftExtension-GiftSecondExtension-GiftThirdExtension 等。

4

10 回答 10

46

编辑:删除了两组以前的解决方案。有关详细信息,请参阅编辑历史记录。

这是一种对任意数量的集合递归执行的方法:

public static Set<Set<Object>> cartesianProduct(Set<?>... sets) {
    if (sets.length < 2)
        throw new IllegalArgumentException(
                "Can't have a product of fewer than two sets (got " +
                sets.length + ")");

    return _cartesianProduct(0, sets);
}

private static Set<Set<Object>> _cartesianProduct(int index, Set<?>... sets) {
    Set<Set<Object>> ret = new HashSet<Set<Object>>();
    if (index == sets.length) {
        ret.add(new HashSet<Object>());
    } else {
        for (Object obj : sets[index]) {
            for (Set<Object> set : _cartesianProduct(index+1, sets)) {
                set.add(obj);
                ret.add(set);
            }
        }
    }
    return ret;
}

请注意,不可能在返回的集合中保留任何通用类型信息。如果您事先知道要获取多少个集合的乘积,则可以定义一个通用元组来保存那么多元素(例如Triple<A, B, C>),但是在 Java 中没有办法拥有任意数量的通用参数。

于 2009-04-03T14:48:32.997 回答
29

这是一个很老的问题,但为什么不使用Guava 的笛卡尔产品呢?

于 2013-11-15T12:41:50.357 回答
24

下面的方法创建字符串列表的笛卡尔积:

protected <T> List<List<T>> cartesianProduct(List<List<T>> lists) {
    List<List<T>> resultLists = new ArrayList<List<T>>();
    if (lists.size() == 0) {
        resultLists.add(new ArrayList<T>());
        return resultLists;
    } else {
        List<T> firstList = lists.get(0);
        List<List<T>> remainingLists = cartesianProduct(lists.subList(1, lists.size()));
        for (T condition : firstList) {
            for (List<T> remainingList : remainingLists) {
                ArrayList<T> resultList = new ArrayList<T>();
                resultList.add(condition);
                resultList.addAll(remainingList);
                resultLists.add(resultList);
            }
        }
    }
    return resultLists;
}

例子:

System.out.println(cartesianProduct(Arrays.asList(Arrays.asList("Apple", "Banana"), Arrays.asList("Red", "Green", "Blue"))));

会产生这个:

[[Apple, Red], [Apple, Green], [Apple, Blue], [Banana, Red], [Banana, Green], [Banana, Blue]]
于 2012-02-29T09:07:29.140 回答
12

集合的数量可能会有所不同,因此我无法在嵌套的 foreach 循环中执行此操作。

两个提示:

  • A x B x C = A x (B x C)
  • 递归
于 2009-04-03T15:52:38.820 回答
11

基于索引的解决方案

使用索引是一种快速且节省内存的替代方法,并且可以处理任意数量的集合。实现 Iterable 允许在 for-each 循环中轻松使用。有关用法示例,请参见#main 方法。

public class CartesianProduct implements Iterable<int[]>, Iterator<int[]> {

    private final int[] _lengths;
    private final int[] _indices;
    private boolean _hasNext = true;

    public CartesianProduct(int[] lengths) {
        _lengths = lengths;
        _indices = new int[lengths.length];
    }

    public boolean hasNext() {
        return _hasNext;
    }

    public int[] next() {
        int[] result = Arrays.copyOf(_indices, _indices.length);
        for (int i = _indices.length - 1; i >= 0; i--) {
            if (_indices[i] == _lengths[i] - 1) {
                _indices[i] = 0;
                if (i == 0) {
                    _hasNext = false;
                }
            } else {
                _indices[i]++;
                break;
            }
        }
        return result;
    }

    public Iterator<int[]> iterator() {
        return this;
    }

    public void remove() {
        throw new UnsupportedOperationException();
    }

    /**
     * Usage example. Prints out
     * 
     * <pre>
     * [0, 0, 0] a, NANOSECONDS, 1
     * [0, 0, 1] a, NANOSECONDS, 2
     * [0, 0, 2] a, NANOSECONDS, 3
     * [0, 0, 3] a, NANOSECONDS, 4
     * [0, 1, 0] a, MICROSECONDS, 1
     * [0, 1, 1] a, MICROSECONDS, 2
     * [0, 1, 2] a, MICROSECONDS, 3
     * [0, 1, 3] a, MICROSECONDS, 4
     * [0, 2, 0] a, MILLISECONDS, 1
     * [0, 2, 1] a, MILLISECONDS, 2
     * [0, 2, 2] a, MILLISECONDS, 3
     * [0, 2, 3] a, MILLISECONDS, 4
     * [0, 3, 0] a, SECONDS, 1
     * [0, 3, 1] a, SECONDS, 2
     * [0, 3, 2] a, SECONDS, 3
     * [0, 3, 3] a, SECONDS, 4
     * [0, 4, 0] a, MINUTES, 1
     * [0, 4, 1] a, MINUTES, 2
     * ...
     * </pre>
     */
    public static void main(String[] args) {
        String[] list1 = { "a", "b", "c", };
        TimeUnit[] list2 = TimeUnit.values();
        int[] list3 = new int[] { 1, 2, 3, 4 };

        int[] lengths = new int[] { list1.length, list2.length, list3.length };
        for (int[] indices : new CartesianProduct(lengths)) {
            System.out.println(Arrays.toString(indices) //
                    + " " + list1[indices[0]] //
                    + ", " + list2[indices[1]] //
                    + ", " + list3[indices[2]]);
        }
    }
}
于 2012-06-08T09:12:39.793 回答
3

这是一个 Iterable,它允许您使用简化的 for 循环:

import java.util.*;

// let's begin with the demo. Instead of Person and Gift, 
// I use the well known char and int. 
class CartesianIteratorTest {

    public static void main (String[] args) {
        List <Object> lc = Arrays.asList (new Object [] {'A', 'B', 'C', 'D'});
        List <Object> lC = Arrays.asList (new Object [] {'a', 'b', 'c'});   
        List <Object> li = Arrays.asList (new Object [] {1, 2, 3, 4});
            // sometimes, a generic solution like List <List <String>>
            // might be possible to use - typically, a mixture of types is 
            // the common nominator 
        List <List <Object>> llo = new ArrayList <List <Object>> ();
        llo.add (lc);
        llo.add (lC);
        llo.add (li);

        // Preparing the List of Lists is some work, but then ...    
        CartesianIterable <Object> ci = new CartesianIterable <Object> (llo);

        for (List <Object> lo: ci)
            show (lo);
    }

    public static void show (List <Object> lo) {
        System.out.print ("(");
        for (Object o: lo)
            System.out.print (o + ", ");
        System.out.println (")");
    }
}

它是如何完成的?我们需要一个 Iterable,以使用简化的 for 循环,并且必须从 Iterable 返回一个 Iterator。我们返回一个对象列表——这可以是一个集合而不是列表,但是集合没有索引访问,所以用集合而不是列表来实现它会有点复杂。与通用解决方案不同,Object 可以用于多种用途,但泛型允许更多限制。

class CartesianIterator <T> implements Iterator <List <T>> {

    private final List <List <T>> lilio;    
    private int current = 0;
    private final long last;

    public CartesianIterator (final List <List <T>> llo) {
        lilio = llo;
        long product = 1L;
        for (List <T> lio: lilio)
            product *= lio.size ();
        last = product;
    } 

    public boolean hasNext () {
        return current != last;
    }

    public List <T> next () {
        ++current;
        return get (current - 1, lilio);
    }

    public void remove () {
        ++current;
    }

    private List<T> get (final int n, final List <List <T>> lili) {
        switch (lili.size ())
        {
            case 0: return new ArrayList <T> (); // no break past return;
            default: {
                List <T> inner = lili.get (0);
                List <T> lo = new ArrayList <T> ();
                lo.add (inner.get (n % inner.size ()));
                lo.addAll (get (n / inner.size (), lili.subList (1, lili.size ())));
                return lo;
            }
        }
    }
}

数学工作是在“get”方法中完成的。考虑 2 组 10 个元素。您总共有 100 种组合,从 00、01、02、... 10、... 到 99 枚举。对于 5 X 10 元素 50,对于 2 X 3 元素 6 组合。子列表大小的模有助于为每次迭代选择一个元素。

可迭代我这里最不有趣的事情:

class CartesianIterable <T> implements Iterable <List <T>> {

    private List <List <T>> lilio;  

    public CartesianIterable (List <List <T>> llo) {
        lilio = llo;
    }

    public Iterator <List <T>> iterator () {
        return new CartesianIterator <T> (lilio);
    }
}

为了实现允许for-each循环的Iterable,我们必须实现iterator(),而对于Iterator,我们必须实现hasNext()、next()和remove()。

结果:

(A, a, 1, )
(B, a, 1, )
(C, a, 1, )
(D, a, 1, )
(A, b, 1, )
(B, b, 1, )
(C, b, 1, )
(D, b, 1, )
...
(A, a, 2, )
...
(C, c, 4, )
(D, c, 4, )
于 2012-04-10T05:20:46.387 回答
3

这是一个Iterator给出二维数组的笛卡尔积的结果,其中数组组件表示问题中的集合(总是可以将实际Set的 s 转换为数组):

public class CartesianIterator<T> implements Iterator<T[]> {
    private final T[][] sets;
    private final IntFunction<T[]> arrayConstructor;

    private int count = 0;
    private T[] next = null;

    public CartesianIterator(T[][] sets, IntFunction<T[]> arrayConstructor) {
        Objects.requireNonNull(sets);
        Objects.requireNonNull(arrayConstructor);

        this.sets = copySets(sets);
        this.arrayConstructor = arrayConstructor;
    }

    private static <T> T[][] copySets(T[][] sets) {
        // If any of the arrays are empty, then the entire iterator is empty.
        // This prevents division by zero in `hasNext`.
        for (T[] set : sets) {
            if (set.length == 0) {
                return Arrays.copyOf(sets, 0);
            }
        }
        return sets.clone();
    }

    @Override
    public boolean hasNext() {
        if (next != null) {
            return true;
        }

        int tmp = count;
        T[] value = arrayConstructor.apply(sets.length);
        for (int i = 0; i < value.length; i++) {
            T[] set = sets[i];

            int radix = set.length;
            int index = tmp % radix;

            value[i] = set[index];

            tmp /= radix;
        }

        if (tmp != 0) {
            // Overflow.
            return false;
        }

        next = value;
        count++;

        return true;
    }

    @Override
    public T[] next() {
        if (!hasNext()) {
            throw new NoSuchElementException();
        }

        T[] tmp = next;
        next = null;
        return tmp;
    }
}

基本思想是将count其视为多基数(数字i有自己的基数,等于i“第”组的长度)。每当我们必须解决next时(即,何时hasNext()被调用和nextnull),我们将数字分解为这个多基数中的数字。这些数字现在用作我们从不同集合中提取元素的索引。

示例使用:

String[] a = { "a", "b", "c"};
String[] b = { "X" };
String[] c = { "r", "s" };

String[][] abc = { a, b, c };

Iterable<String[]> it = () -> new CartesianIterator<>(abc, String[]::new);
for (String[] s : it) {
    System.out.println(Arrays.toString(s));
}

输出:

[a, X, r]
[b, X, r]
[c, X, r]
[a, X, s]
[b, X, s]
[c, X, s]

如果不喜欢数组,代码可以很容易地转换为使用集合。

我想这或多或少类似于“用户未知”给出的答案,只是没有递归和集合。

于 2015-04-19T22:24:00.870 回答
0

是的,有函数式 Java

对于一组s

s.bind(P.p2(), s);
于 2009-04-03T21:42:01.523 回答
0

您可以使用Java 9 Streams获取任意数量的不同类型集合的笛卡尔积,并将其存储到一对象中,如下所示:Set<Set<Object>>

在线尝试!

public static Set<Set<Object>> cartesianProduct(Set<?>... sets) {
    // incorrect incoming data
    if (sets == null) return Collections.emptySet();
    return Arrays.stream(sets)
            // non-null and non-empty sets
            .filter(set -> set != null && set.size() > 0)
            // represent each set element as Set<Object>
            .map(set -> set.stream().map(Set::<Object>of)
                    // Stream<Set<Set<Object>>>
                    .collect(Collectors.toSet()))
            // summation of pairs of inner sets
            .reduce((set1, set2) -> set1.stream()
                    // combinations of inner sets
                    .flatMap(inner1 -> set2.stream()
                            // merge two inner sets into one
                            .map(inner2 -> Stream.of(inner1, inner2)
                                    .flatMap(Set::stream)
                                    .collect(Collectors.toCollection(
                                            LinkedHashSet::new))))
                    // set of combinations
                    .collect(Collectors.toCollection(LinkedHashSet::new)))
            // returns Set<Set<Object>>, otherwise an empty set
            .orElse(Collections.emptySet());
}
public static void main(String[] args) {
    Set<Integer> set1 = Set.of(1, 2, 3);
    Set<String> set2 = Set.of("A", "B", "C");
    Set<Object> set3 = Set.of(new Time(0));

    Set<Set<Object>> sets = cartesianProduct(set1, set2, set3);
    // output
    sets.forEach(System.out::println);
}

输出:

[1, A, 03:00:00]
[1, B, 03:00:00]
[1, C, 03:00:00]
[2, A, 03:00:00]
[2, B, 03:00:00]
[2, C, 03:00:00]
[3, A, 03:00:00]
[3, B, 03:00:00]
[3, C, 03:00:00]

另请参阅:如何创建类似于三个不同类型列表的笛卡尔积的数据结构?

于 2021-04-20T07:44:12.323 回答
0

一个简单的解决方案,例如对于整数集应该如下:

void printCombination(List<Set<Integer>> listSet, Set<Integer> combination) {
    if (listSet.isEmpty()) {
        System.out.println("a combination :" + combination);

        return;
    }

    Set<Integer> intSet = listSet.get(0);
    for (Integer it : intSet) {
        Set<Integer> combination1 = new HashSet<Integer>();
        combination1.addAll(combination);
        combination1.add(it);

        List<Set<Integer>> listSet1 = new ArrayList<Set<Integer>>();
        listSet1.addAll(listSet);
        listSet1.remove(0);
        this.printCombination(listSet1, combination1);
    }

} 
于 2021-07-29T10:15:58.017 回答