Java工具库Guava的集合工具类Iterables、Lists、Sets、Maps、Multisets、Multimaps的常用方法示例

本是古典何须时尚 2023-10-11 19:52 71阅读 0赞

## 场景 ##

Java工具库Guava的不可变集合和新集合类型Multiset、Multimap、BiMap、RangeSet、RangeMap等的使用示例：

[Java工具库Guava的不可变集合和新集合类型Multiset、Multimap、BiMap、RangeSet、RangeMap等的使用示例\_霸道流氓气质的博客-CSDN博客][Java_Guava_Multiset_Multimap_BiMap_RangeSet_RangeMap_-CSDN]

Guava提供了一些集合的静态方法。比如下面这样可以很方便的在初始化时指定初始元素。

List<String> list = Lists.newArrayList("badao","de","chengxvyuan");

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### Iterables ###

Iterable,在可能的情况下，Guava提供的工具方法更偏向于接受Iterable而不是Collection类型

1、concat(Iterable<Iterable>) 串联多个iterables的懒视图

Iterable<Integer> concat = Iterables.concat(Ints.asList(1, 2, 3), Ints.asList(4, 5, 6));
            System.out.println(concat);//[1, 2, 3, 4, 5, 6]

2、frequency(Iterable,Object)返回对象在iterable中出现的次数

int frequency = Iterables.frequency(Ints.asList(1, 1, 2, 3, 4, 5), 1);
            System.out.println(frequency);//2

3、partition(Iterable,int) 把iterable按指定大小分割，得到的子集都不能进行修改操作

Iterable<List<Integer>> partition = Iterables.partition(Ints.asList(1, 2, 3, 3, 4, 5, 6), 2);
            System.out.println(partition);//[[1, 2], [3, 3], [4, 5], [6]]

4、getFirst(Iterable,T default)返回iterable的第一个元素，若iterable为空则返回默认值default

Integer first = Iterables.getFirst(Ints.asList(1, 2, 3), 0);
            System.out.println(first);//1
            Integer first1 = Iterables.getFirst(Ints.asList(), 0);
            System.out.println(first1);//0

5、getLast(Iterable)返回iterable的最后一个元素，若iterable为空则抛出NoSuchElementException

Integer last = Iterables.getLast(Ints.asList(1, 2, 3));
            System.out.println(last);//3
            //Object last1 = Iterables.getLast(Ints.asList());//java.util.NoSuchElementException
            //System.out.println(last1);

6、elementsEqual(Iterable,Iterable)如果两个iterable中的所有元素相等且顺序一致，返回true

boolean b = Iterables.elementsEqual(Ints.asList(1, 2, 3), Ints.asList(1, 2, 3));
            System.out.println(b);//true
            boolean b1 = Iterables.elementsEqual(Ints.asList(1, 3, 2), Ints.asList(1, 2, 3));
            System.out.println(b1);//false

7、unmodifiableIterable(Iterable)返回iterable的不可变视图

Iterable<Integer> integers = Iterables.unmodifiableIterable(Ints.asList(1, 2, 3));
            System.out.println(integers);//[1, 2, 3]

8、limit(Iterable.int)限制iterable的元素个数限制给定值

Iterable<Integer> limit = Iterables.limit(Ints.asList(1, 2, 3, 4), 2);
            System.out.println(limit);//[1, 2]

9、getOnlyElement(Iterable)获取iterable中唯一的元素，如果iterable为空或者有多个元素，则快速失败

//getOnlyElement(Iterable)获取iterable中唯一的元素，如果iterable为空或者有多个元素，则快速失败
            //Integer onlyElement = Iterables.getOnlyElement(Ints.asList(1, 2, 3));//java.lang.IllegalArgumentException: expected one element but was: <1, 2, 3>
            //System.out.println(onlyElement);
            Integer onlyElement1 = Iterables.getOnlyElement(Ints.asList(1));
            System.out.println(onlyElement1);//1
    //        Object onlyElement2 = Iterables.getOnlyElement(Ints.asList());//java.util.NoSuchElementException
    //        System.out.println(onlyElement2);

### Lists ###

1、reverse(List)返回给定List的反转视图

List<Integer> integers = Ints.asList(1, 2, 3, 4, 5);
            List<Integer> reverse = Lists.reverse(integers);
            System.out.println(reverse);//[5, 4, 3, 2, 1]

2、partition(List,int) 把List按指定大小分割

List<List<Integer>> partition = Lists.partition(reverse, 2);
            System.out.println(partition);//[[5, 4], [3, 2], [1]]

### Sets ###

1、union(Set,Set)求并集

Sets.SetView<Integer> union = Sets.union(Sets.newHashSet(1, 2, 3), Sets.newHashSet(3, 4));
            System.out.println(union);//[1, 2, 3, 4]

2、intersection(Set,Set)求交集

Sets.SetView<Integer> intersection = Sets.intersection(Sets.newHashSet(1, 2, 3), Sets.newHashSet(2, 3));
            System.out.println(intersection);//[2, 3]

3、difference(Set,Set)属于A但不属于B的差集

Sets.SetView<Integer> difference = Sets.difference(Sets.newHashSet(1, 2, 3), Sets.newHashSet(1, 4, 5));
            System.out.println(difference);//[2, 3]

4、symmetricDifference(Set,Set)对称差集

Sets.SetView<Integer> integers = Sets.symmetricDifference(Sets.newHashSet(1, 2, 3, 7, 8), Sets.newHashSet(1, 2, 3, 5, 6));
            System.out.println(integers);//[8, 7, 5, 6]

5、cartesianProduct(List<Set>)返回所有集合的笛卡尔积

HashSet<Integer> integers1 = Sets.newHashSet(1, 2, 3);
            HashSet<Integer> integers2 = Sets.newHashSet(4, 5);
            Set<List<Integer>> lists = Sets.cartesianProduct(integers1, integers2);
            System.out.println(lists);//[[1, 4], [1, 5], [2, 4], [2, 5], [3, 4], [3, 5]]

6、powerSet(Set)返回给定集合的所有子集

Set<Set<Integer>> sets = Sets.powerSet(Sets.newHashSet(1, 2));
            sets.forEach(System.out::println);
    //        []
    //        [1]
    //        [2]
    //        [1, 2]

### Maps ###

1、Maps.uniqueIndex(Iterable,Function)

针对的场景是，有一组对象，它们在某个属性上分别有独一无二的值，

而我们希望能够按照这个属性值查找对象，比如，有一堆字符串，这些字符串的长度都是独一无二的，希望按照特定长度查找字符串，

注意这里长度必须是独一无二，如果有两个字符串长度一样则不行

ArrayList<String> strings = Lists.newArrayList("a","bd","ddd");
            ImmutableMap<Integer, String> integerStringImmutableMap = Maps.uniqueIndex(strings, new Function<String, Integer>() {
                @Override
                public @Nullable Integer apply(@Nullable String s) {
                    return s.length();
                }
            });
            System.out.println(integerStringImmutableMap);//{1=a, 2=bd, 3=ddd}

2、diffenence(Map,Map)用来比较两个Map以获取所有不同点

ImmutableMap<Integer, String> of = ImmutableMap.of(1, "a", 2, "b",3,"d");
            ImmutableMap<Integer, String> of1 = ImmutableMap.of(1, "c", 2, "b",5,"g");
            MapDifference<Integer, String> difference = Maps.difference(of, of1);
            System.out.println(difference);//not equal: only on left={3=d}: only on right={5=g}: value differences={1=(a, c)}
            //entriesInCommon()两个Map中都有的映射项，包括匹配的键与值
            System.out.println(difference.entriesInCommon());//{2=b}
            //entriesDiffering()键相同但是值不同值映射项
            System.out.println(difference.entriesDiffering());//{1=(a, c)}
            //entriesOnlyOnLeft()键只存在于左边Map的映射项
            System.out.println(difference.entriesOnlyOnLeft());//{3=d}
            //entriesOnlyOnRight()键只存在于右边Map的映射项
            System.out.println(difference.entriesOnlyOnRight());//{5=g}

### Multisets ###

1、containsOccurrences(Multiset sup,Multiset sub) 如果sub.count(o)<=sup.conunt(o)返回true

HashMultiset<Object> multiset = HashMultiset.create();
            multiset.add("a",2);
    
            HashMultiset<Object> multiset1 = HashMultiset.create();
            multiset1.add("a",5);
    
            //multiset包含2个a,multiset1包含5个a,返回true
            System.out.println(multiset1.containsAll(multiset));//true
    
            //containsOccurrences(Multiset sup,Multiset sub) 如果sub.count(o)<=sup.conunt(o)返回true
            boolean b = Multisets.containsOccurrences(multiset, multiset1);
            System.out.println(b);//false
    
            Multisets.removeOccurrences(multiset1,multiset);
            System.out.println(multiset1);//[a x 3]

2、retainOccurrences(Multiset removeFrom,Multiset toRetain)修改removeFrom,以保证任意o都符合removeFrom.count(o)<=toRetain.count(o)

System.out.println(multiset);//[a x 2]
            System.out.println(multiset1);//[a x 3]
            Multisets.retainOccurrences(multiset1,multiset);
            System.out.println(multiset);//[a x 2]
            System.out.println(multiset1);//[a x 2]

3、intersection(Multiset,Multiset)返回两个multiset的交集

Multiset<Object> intersection = Multisets.intersection(multiset1, multiset);
            System.out.println(intersection);//[a x 2]

### Multimaps ###

1、index 通常针对的场景是:有一组对象，它们有共同的特定的属性，我们希望能按照这个属性的值查询对象

ImmutableSet<String> of = ImmutableSet.of("a", "bb", "cc", "ddd");
            ImmutableListMultimap<Integer, String> index = Multimaps.index(of, new Function<String, Integer>() {
                @Override
                public @Nullable Integer apply(@Nullable String s) {
                    return s.length();
                }
            });
            System.out.println(index);//{1=[a], 2=[bb, cc], 3=[ddd]}

2、invertFrom,Multimap可以把多个键映射到同一个值，也可以把一个键映射到多个值，反转Multimap时可以供invertFrom,并且可以自由选择反转后的Multimap的实现

ArrayListMultimap<String, Integer> multimap = ArrayListMultimap.create();
            multimap.putAll("a",Ints.asList(2,4,6));
            multimap.putAll("b",Ints.asList(1,2,6));
            multimap.putAll("c",Ints.asList(3,4,6));
    
            TreeMultimap<Integer, String> invertFrom = Multimaps.invertFrom(multimap, TreeMultimap.create());
            System.out.println(invertFrom);//{1=[b], 2=[a, b], 3=[c], 4=[a, c], 6=[a, b, c]}

3、想在Map对象上使用Multimap的方法吗，forMap(Map)把Map包装成SetMultimap

例如，与Multimaps.invertFrom结合使用，可以把多对一的Map反转成一对多的Multimap

Map<String, Integer> map = ImmutableMap.of("a", 1, "b", 1, "c", 2);
            SetMultimap<String, Integer> multimap1 = Multimaps.forMap(map);
            System.out.println(multimap1);//{a=[1], b=[1], c=[2]}
            HashMultimap<Integer, String> invertFrom1 = Multimaps.invertFrom(multimap1, HashMultimap.create());
            System.out.println(invertFrom1);//{1=[a, b], 2=[c]}

[Java_Guava_Multiset_Multimap_BiMap_RangeSet_RangeMap_-CSDN]: https://blog.csdn.net/BADAO_LIUMANG_QIZHI/article/details/127900306
[CSDN_-C_SpringBoot]: https://blog.csdn.net/badao_liumang_qizhi