我想在Java中计算任意数量的非空集的笛卡尔积。
我写过那个迭代代码......
public static <T> List<Set<T>> cartesianProduct(List<Set<T>> list) {
List<Iterator<T>> iterators = new ArrayList<Iterator<T>>(list.size());
List<T> elements = new ArrayList<T>(list.size());
List<Set<T>> toRet = new ArrayList<Set<T>>();
for (int i = 0; i < list.size(); i++) {
iterators.add(list.get(i).iterator());
elements.add(iterators.get(i).next());
}
for (int j = 1; j >= 0;) {
toRet.add(Sets.newHashSet(elements));
for (j = iterators.size()-1; j >= 0 && !iterators.get(j).hasNext(); j--) {
iterators.set(j, list.get(j).iterator());
elements.set(j, iterators.get(j).next());
}
elements.set(Math.abs(j), iterators.get(Math.abs(j)).next());
}
return toRet;
}
......但我发现它相当不优雅。 有人有更好的，仍然是迭代的解决方案吗？使用一些奇妙的功能性方法的解决方案？ 否则......关于如何改进它的建议？错误？
答案 0 :(得分：22)
我写了一个解决方案，不要求你在内存中填满大量的集合。不幸的是，所需的代码长达数百行。你可能要等到它出现在番石榴项目（http://guava-libraries.googlecode.com）中，我希望到今年年底。抱歉。 ：（
请注意，如果您使用笛卡儿产品的数量是编译时已知的固定数量，则可能不需要这样的实用程序 - 您可以使用该数量的嵌套for循环。
编辑：代码现已发布。
我认为你会很高兴。它只会在您要求时创建单个列表;没有用它们的所有MxNxPxQ填充内存。
如果您想检查来源，则为here at line 727。
享受！
答案 1 :(得分：2)
使用Google Guava 19和Java 8非常简单：
假设您拥有要关联的所有阵列的列表...
public static void main(String[] args) {
List<String[]> elements = Arrays.asList(
new String[]{"John", "Mary"},
new String[]{"Eats", "Works", "Plays"},
new String[]{"Food", "Computer", "Guitar"}
);
// Create a list of immutableLists of strings
List<ImmutableList<String>> immutableElements = makeListofImmutable(elements);
// Use Guava's Lists.cartesianProduct, since Guava 19
List<List<String>> cartesianProduct = Lists.cartesianProduct(immutableElements);
System.out.println(cartesianProduct);
}
制作不可变列表列表的方法如下：
/**
* @param values the list of all profiles provided by the client in matrix.json
* @return the list of ImmutableList to compute the Cartesian product of values
*/
private static List<ImmutableList<String>> makeListofImmutable(List<String[]> values) {
List<ImmutableList<String>> converted = new LinkedList<>();
values.forEach(array -> {
converted.add(ImmutableList.copyOf(array));
});
return converted;
}
输出如下：
[
[John, Eats, Food], [John, Eats, Computer], [John, Eats, Guitar],
[John, Works, Food], [John, Works, Computer], [John, Works, Guitar],
[John, Plays, Food], [John, Plays, Computer], [John, Plays, Guitar],
[Mary, Eats, Food], [Mary, Eats, Computer], [Mary, Eats, Guitar],
[Mary, Works, Food], [Mary, Works, Computer], [Mary, Works, Guitar],
[Mary, Plays, Food], [Mary, Plays, Computer], [Mary, Plays, Guitar]
]
答案 2 :(得分：1)
以下答案使用迭代而不是递归。它使用我之前回答中的相同Tuple类。
这是一个单独的答案，因为恕我直言都是有效的，不同的方法。
这是新的主要课程：
public class Example {
public static <T> List<Tuple<T>> cartesianProduct(List<Set<T>> sets) {
List<Tuple<T>> tuples = new ArrayList<Tuple<T>>();
for (Set<T> set : sets) {
if (tuples.isEmpty()) {
for (T t : set) {
Tuple<T> tuple = new Tuple<T>();
tuple.add(t);
tuples.add(tuple);
}
} else {
List<Tuple<T>> newTuples = new ArrayList<Tuple<T>>();
for (Tuple<T> subTuple : tuples) {
for (T t : set) {
Tuple<T> tuple = new Tuple<T>();
tuple.addAll(subTuple);
tuple.add(t);
newTuples.add(tuple);
}
}
tuples = newTuples;
}
}
return tuples;
}
}
答案 3 :(得分：1)
这是我写的一个迭代，懒惰的实现。界面与Google的Sets.cartesianProduct非常相似，但它更灵活一点：它处理的是Iterables而不是Sets。此代码及其单元测试位于https://gist.github.com/1911614。
/* Copyright 2012 LinkedIn Corp.
Licensed under the Apache License, Version 2.0 (the "License");
you may not use this file except in compliance with the License.
You may obtain a copy of the License at
http://www.apache.org/licenses/LICENSE-2.0
Unless required by applicable law or agreed to in writing, software
distributed under the License is distributed on an "AS IS" BASIS,
WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
See the License for the specific language governing permissions and
limitations under the License.
*/
import com.google.common.base.Function;
import com.google.common.collect.Iterables;
import java.lang.reflect.Array;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.Iterator;
import java.util.List;
import java.util.NoSuchElementException;
/**
* Implements the Cartesian product of ordered collections.
*
* @author <a href="mailto:jmkristian@gmail.com">John Kristian</a>
*/
public class Cartesian {
/**
* Generate the <a href="http://en.wikipedia.org/wiki/Cartesian_product">Cartesian
* product</a> of the given axes. For axes [[a1, a2 ...], [b1, b2 ...], [c1, c2 ...]
* ...] the product is [[a1, b1, c1 ...] ... [a1, b1, c2 ...] ... [a1, b2, c1 ...] ...
* [aN, bN, cN ...]]. In other words, the results are generated in same order as these
* nested loops:
*
* <pre>
* for (T a : [a1, a2 ...])
* for (T b : [b1, b2 ...])
* for (T c : [c1, c2 ...])
* ...
* result = new T[]{ a, b, c ... };
* </pre>
*
* Each result is a new array of T, whose elements refer to the elements of the axes. If
* you prefer a List, you can call asLists(product(axes)).
* <p>
* Don't change the axes while iterating over their product, as a rule. Changes to an
* axis can affect the product or cause iteration to fail (which is usually bad). To
* prevent this, you can pass clones of your axes to this method.
* <p>
* The implementation is lazy. This method iterates over the axes, and returns an
* Iterable that contains a reference to each axis. Iterating over the product causes
* iteration over each axis. Methods of each axis are called as late as practical.
*/
public static <T> Iterable<T[]> product(Class<T> resultType,
Iterable<? extends Iterable<? extends T>> axes) {
return new Product<T>(resultType, newArray(Iterable.class, axes));
}
/** Works like product(resultType, Arrays.asList(axes)), but slightly more efficient. */
public static <T> Iterable<T[]> product(Class<T> resultType, Iterable<? extends T>... axes) {
return new Product<T>(resultType, axes.clone());
}
/**
* Wrap the given arrays in fixed-size lists. Changes to the lists write through to the
* arrays.
*/
public static <T> Iterable<List<T>> asLists(Iterable<? extends T[]> arrays) {
return Iterables.transform(arrays, new AsList<T>());
}
/**
* Arrays.asList, represented as a Function (as used in Google collections).
*/
public static class AsList<T> implements Function<T[], List<T>> {
@Override
public List<T> apply(T[] array) {
return Arrays.asList(array);
}
}
/** Create a generic array containing references to the given objects. */
private static <T> T[] newArray(Class<? super T> elementType, Iterable<? extends T> from) {
List<T> list = new ArrayList<T>();
for (T f : from)
list.add(f);
return list.toArray(newArray(elementType, list.size()));
}
/** Create a generic array. */
@SuppressWarnings("unchecked")
private static <T> T[] newArray(Class<? super T> elementType, int length) {
return (T[]) Array.newInstance(elementType, length);
}
private static class Product<T> implements Iterable<T[]> {
private final Class<T> _resultType;
private final Iterable<? extends T>[] _axes;
/** Caution: the given array of axes is contained by reference, not cloned. */
Product(Class<T> resultType, Iterable<? extends T>[] axes) {
_resultType = resultType;
_axes = axes;
}
@Override
public Iterator<T[]> iterator() {
if (_axes.length <= 0) // an edge case
return Collections.singleton(newArray(_resultType, 0)).iterator();
return new ProductIterator<T>(_resultType, _axes);
}
@Override
public String toString() {
return "Cartesian.product(" + Arrays.toString(_axes) + ")";
}
private static class ProductIterator<T> implements Iterator<T[]> {
private final Iterable<? extends T>[] _axes;
private final Iterator<? extends T>[] _iterators; // one per axis
private final T[] _result; // a copy of the last result
/**
* The minimum index such that this.next() will return an array that contains
* _iterators[index].next(). There are some special sentinel values: NEW means this
* is a freshly constructed iterator, DONE means all combinations have been
* exhausted (so this.hasNext() == false) and _iterators.length means the value is
* unknown (to be determined by this.hasNext).
*/
private int _nextIndex = NEW;
private static final int NEW = -2;
private static final int DONE = -1;
/** Caution: the given array of axes is contained by reference, not cloned. */
ProductIterator(Class<T> resultType, Iterable<? extends T>[] axes) {
_axes = axes;
_iterators = Cartesian.<Iterator<? extends T>> newArray(Iterator.class, _axes.length);
for (int a = 0; a < _axes.length; ++a) {
_iterators[a] = axes[a].iterator();
}
_result = newArray(resultType, _iterators.length);
}
private void close() {
_nextIndex = DONE;
// Release references, to encourage garbage collection:
Arrays.fill(_iterators, null);
Arrays.fill(_result, null);
}
@Override
public boolean hasNext() {
if (_nextIndex == NEW) { // This is the first call to hasNext().
_nextIndex = 0; // start here
for (Iterator<? extends T> iter : _iterators) {
if (!iter.hasNext()) {
close(); // no combinations
break;
}
}
} else if (_nextIndex >= _iterators.length) {
// This is the first call to hasNext() after next() returned a result.
// Determine the _nextIndex to be used by next():
for (_nextIndex = _iterators.length - 1; _nextIndex >= 0; --_nextIndex) {
Iterator<? extends T> iter = _iterators[_nextIndex];
if (iter.hasNext()) {
break; // start here
}
if (_nextIndex == 0) { // All combinations have been generated.
close();
break;
}
// Repeat this axis, with the next value from the previous axis.
iter = _axes[_nextIndex].iterator();
_iterators[_nextIndex] = iter;
if (!iter.hasNext()) { // Oops; this axis can't be repeated.
close(); // no more combinations
break;
}
}
}
return _nextIndex >= 0;
}
@Override
public T[] next() {
if (!hasNext())
throw new NoSuchElementException("!hasNext");
for (; _nextIndex < _iterators.length; ++_nextIndex) {
_result[_nextIndex] = _iterators[_nextIndex].next();
}
return _result.clone();
}
@Override
public void remove() {
for (Iterator<? extends T> iter : _iterators) {
iter.remove();
}
}
@Override
public String toString() {
return "Cartesian.product(" + Arrays.toString(_axes) + ").iterator()";
}
}
}
}
答案 4 :(得分：1)
基于索引的解决方案
使用索引是一种简单的替代方法，可以快速且节省内存，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，请参阅#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]]);
}
}
}
答案 5 :(得分：0)
您可能对另一个关于笛卡儿产品的问题感兴趣（编辑：删除以保存超链接，搜索标签笛卡儿产品）。这个答案有一个很好的递归解决方案，我很难改进。您是否特别需要迭代解决方案而不是递归解决方案？
修改
在查看perl和a clean explanation中关于堆栈溢出的另一个迭代解决方案之后，这是另一个解决方案：
public static <T> List<Set<T>> uglyCartesianProduct(List<Set<T>> list) {
List<Iterator<T>> iterators = new ArrayList<Iterator<T>>(list.size());
List<T> elements = new ArrayList<T>(list.size());
List<Set<T>> toRet = new ArrayList<Set<T>>();
for (int i = 0; i < list.size(); i++) {
iterators.add(list.get(i).iterator());
elements.add(iterators.get(i).next());
}
for(int i = 0; i < numberOfTuples(list); i++)
{
toRet.add(new HashSet<T>());
}
int setIndex = 0;
for (Set<T> set : list) {
int index = 0;
for (int i = 0; i < numberOfTuples(list); i++) {
toRet.get(index).add((T) set.toArray()[index % set.size()]);
index++;
}
setIndex++;
}
return toRet;
}
private static <T> int numberOfTuples(List<Set<T>> list) {
int product = 1;
for (Set<T> set : list) {
product *= set.size();
}
return product;
}
答案 6 :(得分：0)
我相信这是正确的。它不是追求效率，而是通过递归和抽象来实现清晰的风格。
关键的抽象是引入一个简单的Tuple类。这有助于后来的泛型：
class Tuple<T> {
private List<T> list = new ArrayList<T>();
public void add(T t) { list.add(t); }
public void addAll(Tuple<T> subT) {
for (T t : subT.list) {
list.add(t);
}
}
public String toString() {
String result = "(";
for (T t : list) { result += t + ", "; }
result = result.substring(0, result.length() - 2);
result += " )";
return result;
}
}
使用这个类，我们可以编写一个类似的类：
public class Example {
public static <T> List<Tuple<T>> cartesianProduct(List<Set<T>> sets) {
List<Tuple<T>> tuples = new ArrayList<Tuple<T>>();
if (sets.size() == 1) {
Set<T> set = sets.get(0);
for (T t : set) {
Tuple<T> tuple = new Tuple<T>();
tuple.add(t);
tuples.add(tuple);
}
} else {
Set<T> set = sets.remove(0);
List<Tuple<T>> subTuples = cartesianProduct(sets);
System.out.println("TRACER size = " + tuples.size());
for (Tuple<T> subTuple : subTuples) {
for (T t : set) {
Tuple<T> tuple = new Tuple<T>();
tuple.addAll(subTuple);
tuple.add(t);
tuples.add(tuple);
}
}
}
return tuples;
}
}
我有一个很好的例子，但为了简洁起见省略了它。
答案 7 :(得分：0)
这是一个惰性迭代器方法，它使用函数生成适当的输出类型。
public static <T> Iterable<T> cartesianProduct(
final Function<Object[], T> fn, Object[]... options) {
final Object[][] opts = new Object[options.length][];
for (int i = opts.length; --i >= 0;) {
// NPE on null input collections, and handle the empty output case here
// since the iterator code below assumes that it is not exhausted the
// first time through fetch.
if (options[i].length == 0) { return Collections.emptySet(); }
opts[i] = options[i].clone();
}
return new Iterable<T>() {
public Iterator<T> iterator() {
return new Iterator<T>() {
final int[] pos = new int[opts.length];
boolean hasPending;
T pending;
boolean exhausted;
public boolean hasNext() {
fetch();
return hasPending;
}
public T next() {
fetch();
if (!hasPending) { throw new NoSuchElementException(); }
T out = pending;
pending = null; // release for GC
hasPending = false;
return out;
}
public void remove() { throw new UnsupportedOperationException(); }
private void fetch() {
if (hasPending || exhausted) { return; }
// Produce a result.
int n = pos.length;
Object[] args = new Object[n];
for (int j = n; --j >= 0;) { args[j] = opts[j][pos[j]]; }
pending = fn.apply(args);
hasPending = true;
// Increment to next.
for (int i = n; --i >= 0;) {
if (++pos[i] < opts[i].length) {
for (int j = n; --j > i;) { pos[j] = 0; }
return;
}
}
exhausted = true;
}
};
}
};
}
答案 8 :(得分：0)
我为字符串表写了一个递归的笛卡尔积算法。您可以将其修改为具有集合。以下是算法。我的article
也解释了这一点public class Main {
public static void main(String[] args) {
String[] A = new String[]{ "a1", "a2", "a3" };
String[] B = new String[]{ "b1", "b2", "b3" };
String[] C = new String[]{ "c1" };
String[] cp = CartesianProduct(0, A, B, C);
for(String s : cp) {
System.out.println(s);
}
}
public static String[] CartesianProduct(int prodLevel, String[] res, String[] ...s) {
if(prodLevel < s.length) {
int cProdLen = res.length * s[prodLevel].length;
String[] tmpRes = new String[cProdLen];
for (int i = 0; i < res.length; i++) {
for (int j = 0; j < s[prodLevel].length; j++) {
tmpRes[i * res.length + j] = res[i] + s[prodLevel][j];
}
}
res = Main.CartesianProduct(prodLevel + 1, tmpRes, s);
}
return res;
}}
答案 9 :(得分：0)
这里我只有一个简单的实现，只需少量代码，运行时所需空间更少 https://github.com/treesong/cartesian-set/blob/master/src/main/java/net/ruiqi/collection/CartesianSet.java
/**
* cartesian
*
* @author ruiqi.zss@alibaba-inc.com
* @date 2019/3/28
*/
public class CartesianSet<T> {
private final T[][] source;
private final long count;
public CartesianSet(T[][] source) {
this.source = source;
int total = 1;
for (T[] array : source) {
total *= array.length;
}
this.count = total;
}
public long getCount() {
return count;
}
public List<T> get(int index) {
if (index < 0 || count <= index) { return null; }
List<T> result = new ArrayList<T>(this.source.length);
int weight = 1;
for (T[] row : this.source) {
int times = index / weight;
int column = times % row.length;
result.add(row[column]);
weight *= row.length;
}
return result;
}
}
答案 10 :(得分：0)
您可以使用 Stream.reduce 方法。
Java 9 没有额外的库。
public static <U> List<Set<U>> cartesianProduct(List<Set<? extends U>> sets) {
// incorrect incoming data
if (sets == null) return Collections.emptyList();
return sets.stream()
// non-null and non-empty sets
.filter(set -> set != null && set.size() > 0)
// represent each set element as Set<U>
.map(set -> set.stream().map(Set::<U>of)
// Stream<List<Set<U>>>
.collect(Collectors.toList()))
// 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.toSet())))
// list of combinations
.collect(Collectors.toList()))
// List<Set<U>>
.orElse(Collections.emptyList());
}
public static void main(String[] args) {
Set<Integer> set1 = Set.of(1, 2);
Set<Double> set2 = Set.of(3.0, 4.0);
Set<Long> set3 = Set.of(5L, 6L);
List<Set<Number>> sets = cartesianProduct(List.of(set1, set2, set3));
// output
sets.forEach(System.out::println);
}
输出（元素的顺序可能不同）：
[1, 3.0, 5]
[1, 3.0, 6]
[1, 4.0, 5]
[1, 4.0, 6]
[2, 3.0, 5]
[2, 3.0, 6]
[2, 4.0, 5]
[2, 4.0, 6]