前言: 基本类型的流没有这个用法

averagingDouble,averagingInt,averagingLong

平均值计算，返回的都是Double类型

        List<Dog> users = Arrays.asList(
                new Dog("a",9),
                new Dog("b",10),
                new Dog("c",10),
                new Dog("d",13),
                new Dog("e",14));

        Double collect = users.stream().collect(Collectors.averagingDouble(x -> x.getAge()));
        System.out.println(collect);//11.2
        Double collect1 = users.stream().collect(Collectors.averagingInt(x -> x.getAge()));
        System.out.println(collect1);//11.2
        Double collect2 = users.stream().collect(Collectors.averagingLong(x -> x.getAge()));
        System.out.println(collect2);//11.2

collectingAndThen

List<Dog> users = Arrays.asList(
                new Dog("a",9),
                new Dog("b",10),
                new Dog("c",10),
                new Dog("d",13),
                new Dog("e",14));

        Map<Integer, Long> collect = users.stream().collect(Collectors.groupingBy(x -> x.getAge(), Collectors.counting()));
        System.out.println(collect);//{9=1, 10=2, 13=1, 14=1}
        Integer collect1 = users.stream().collect(Collectors.collectingAndThen(Collectors.groupingBy(x -> x.getAge()), map -> map.size()));
        System.out.println(collect1);//4

• 第一个参数为要做的操作，第二个参数对第一个参数收集到的结果进行如何的处理

counting

        //创建数组 元素为0~9
        List<Integer> collect = Stream.iterate(0, x -> x + 1).limit(10).collect(Collectors.toList());

        Long collect1 = collect.stream().collect(Collectors.counting());
        System.out.println(collect1);

groupingBy

1. 一个参数表示按照哪个值来分组，这个值为key，其中的value就为符合k的集合
2. 两个参数表示：第一个参数同上，第二个参数可以对value集合再次进行流操作
3. 三个参数表示：第一个参数同上，第二个参数表示要返回什么样的Map，第三个参数可以对value集合再次进行流操作

        List<Dog> users = Arrays.asList(
                new Dog("a",9),
                new Dog("b",10),
                new Dog("c",10),
                new Dog("d",13),
                new Dog("e",14));
        //按照年龄来分类，key为年龄，value为相同年龄的集合
        Map<Integer, List<Dog>> collect = users.stream().collect(Collectors.groupingBy(x -> x.getAge()));
        //{9=[Dog{name='a', age=9}], 10=[Dog{name='b', age=10}, Dog{name='c', age=10}], 13=[Dog{name='d', age=13}], 14=[Dog{name='e', age=14}]}
        System.out.println(collect);
        //{9=[], 10=[Dog{name='c', age=10}], 13=[], 14=[]}
        Map<Integer, Set<Dog>> c = users.stream().collect(Collectors.groupingBy(x -> x.getAge(), Collectors.filtering(x -> x.getName().equals("c"), Collectors.toSet())));
        //
        TreeMap<Object, Set<Dog>> collect1 = users.stream().collect(Collectors.groupingBy(x -> x.getAge(), TreeMap::new, Collectors.toSet()));
        System.out.println(collect1);

groupingByConcurrent

使用同groupingBy，但是返回的是线程安全的集合。

joining

          String[] strings = {"gs","sb","dpz"};
        String collect = Arrays.stream(strings).collect(Collectors.joining());
        String collect1 = Arrays.stream(strings).collect(Collectors.joining(","));
        String collect2 = Arrays.stream(strings).collect(Collectors.joining(",", "[", "]"));
        System.out.println(collect);//gssbdpz
        System.out.println(collect1);//gs,sb,dpz
        System.out.println(collect2);//[gs,sb,dpz]

mapping

        //创建数组 元素为0~9
        List<Integer> collect = Stream.iterate(0, x -> x + 1).limit(10).collect(Collectors.toList());

        //过滤
        Set<Integer> collect1 = collect.stream().collect(Collectors.mapping(x -> x, Collectors.toSet()));
        //等价于
        Set<Integer> collect2 = collect.stream().map(x -> x).collect(Collectors.toSet()); 

maxBy,minBy

1. maxBy:return (a, b) -> comparator.compare(a, b) >= 0 ? a : b 如果返回的小于0，取b
2. minBy:return (a, b) -> comparator.compare(a, b) <= 0 ? a : b 如果返回结果小于0，取a

 //创建数组 元素为0~9
        List<Integer> collect = Stream.iterate(0, x -> x + 1).limit(10).collect(Collectors.toList());

        Optional<Integer> collect1 = collect.stream().collect(Collectors.maxBy((x, y) -> -1));
        Optional<Integer> collect2 = collect.stream().collect(Collectors.minBy((x, y) -> -1));
        System.out.println(collect1); //9
        System.out.println(collect2); //0

partitioningBy

1.一个参数： 根据过滤把结果集分为两组存在Map中，键为true的为一组，键为false的为一组，值为List存放元素
2. 两个参数：同上，但是第二个参数表示再次过滤Map中的Value。

 //创建数组 元素为0~9
        List<Integer> collect = Stream.iterate(0, x -> x + 1).limit(10).collect(Collectors.toList());

        Map<Boolean, List<Integer>> collect2 = collect.stream().collect(Collectors.partitioningBy(x -> x > 2));
        System.out.println(collect2); // {false=[0, 1, 2], true=[3, 4, 5, 6, 7, 8, 9]}
        Map<Boolean, Integer> collect1 = collect.stream().collect(Collectors.partitioningBy(x -> x > 2, Collectors.reducing(0,(x, y) -> Math.min(x, y))));
        System.out.println(collect1); // {false=0, true=0}

reducing

 //创建数组 元素为0~9
        List<Integer> collect = Stream.iterate(0, x -> x + 1).limit(10).collect(Collectors.toList());

        //因为提供了第一个参数，确定了类型，返回值为该类型对象，第二个参数x,y为流中的两个值进行操作
        Integer collect1 = collect.stream().collect(Collectors.reducing(0, (x, y) -> Math.min(x, y)));
        //返回值Optional<T>，第二个参数x,y为流中的两个值进行操作
        Optional<Integer> collect2 = collect.stream().collect(Collectors.reducing((x, y) -> Math.min(x, y)));
        //因为提供了第一个参数，确定了类型，返回值为该类型对象，第二个参数还可以对每个元素进行一次操作，第三个参数x,y为流中的两个值进行操作
        Integer collect3 = collect.stream().collect(Collectors.reducing(0, x -> x+1, (x, y) -> Math.max(x, y)));

summarizingDouble,summarizingInt,summarizingLong

        //创建数组 元素为0~9
        List<Integer> collect = Stream.iterate(0, x -> x + 1).limit(10).collect(Collectors.toList());
        //Collectors.summarizingInt(ToIntFunction<? super T> mapper)
        IntSummaryStatistics collect1 = collect.stream().collect(Collectors.summarizingInt(x -> x));
        System.out.println(collect1);//IntSummaryStatistics{count=10, sum=45, min=0, average=4.500000, max=9}
        DoubleSummaryStatistics collect2 = collect.stream().collect(Collectors.summarizingDouble(x -> x));
        System.out.println(collect2);//DoubleSummaryStatistics{count=10, sum=45.000000, min=0.000000, average=4.500000, max=9.000000}
        LongSummaryStatistics collect3 = collect.stream().collect(Collectors.summarizingLong(x -> x));
        System.out.println(collect3);//LongSummaryStatistics{count=10, sum=45, min=0, average=4.500000, max=9}

summingDouble,summingInt,summingLong

        //创建数组 元素为0~9
        List<Integer> collect = Stream.iterate(0, x -> x + 1).limit(10).collect(Collectors.toList());
        //Collectors.summarizingInt(ToIntFunction<? super T> mapper)
        Double collect1 = collect.stream().collect(Collectors.summingDouble(x -> x));
        Integer collect2 = collect.stream().collect(Collectors.summingInt(x -> x));
        Long collect3 = collect.stream().collect(Collectors.summingLong(x -> x));

toCollection

        Integer[] a = {1,2,3,4,5,6,7,8,9,10};
        //Collectors.toCollection 返回指定集合类型，要求：必须为collectors的子类
        Arrays.stream(a).collect(Collectors.toCollection(TreeSet::new));
        Arrays.stream(a).collect(Collectors.toCollection(()->new TreeSet<>()));

toConcurrentMap

 Integer[] a = {1,2,3,4,5,6,7,8,9,10};
        //toConcurrentMap.toMap 使用
        //1.两个参数，一个key，一个value 组成任意Map(key，value的类型由返回值类型决定)
        //{1=1, 2=2, 3=3, 4=4, 5=5, 6=6, 7=7, 8=8, 9=9, 10=10}
        ConcurrentMap<String, String> collect = Arrays.stream(a).collect(Collectors.toConcurrentMap(x -> x.toString(), x -> x.toString()));
        //2.三个参数，第三个参数决定key冲突时怎么处理，(x,y) 代表两个冲突k的value值
        //{1=11} key全为1，value每次都保存最新的那个
        ConcurrentMap<String, String> collect1 = Arrays.stream(a).collect(Collectors.toConcurrentMap((x) -> "1", x -> String.valueOf(x+1),(x, y)->y));
        3.四个参数，第四个代表要返回的Map类型，默认ConcurrentMap,只允许ConcurrentMap的子类
        ConcurrentSkipListMap<String, String> collect2 = Arrays.stream(a).collect(Collectors.toConcurrentMap((x) -> "1", x -> String.valueOf(x + 1), (x, y) -> y, ConcurrentSkipListMap::new));

toList，toSet

 Integer[] a = {1,2,3,4,5,6,7,8,9,10,1};

        //去重，组成set [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
        Set<Integer> set = Arrays.stream(a).collect(Collectors.toSet());
        //直接组成list [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 1]
        List<Integer> list = Arrays.stream(a).collect(Collectors.toList());

toMap

使用例子：

        Integer[] a = {1,2,3,4,5,6,7,8,9,10};
        //Collectors.toMap 使用
        //1.两个参数，一个key，一个value 组成任意Map(key，value的类型由返回值类型决定)
        //{1=1, 2=2, 3=3, 4=4, 5=5, 6=6, 7=7, 8=8, 9=9, 10=10}
        Map<String, String> collect = Arrays.stream(a).collect(Collectors.toMap(x -> x.toString(), x -> x.toString()));
        //2.三个参数，第三个参数决定key冲突时怎么处理，(x,y) 代表两个冲突k的value值
        //{1=11} key全为1，value每次都保存最新的那个
        Map<String, String> collect1 = Arrays.stream(a).collect(Collectors.toMap((x) -> "1", x -> String.valueOf(x+1),(x,y)->y));
        //3.四个参数，第四个代表要返回的Map类型，默认hashMap
        Map<String, String> collect2 = Arrays.stream(a).collect(Collectors.toMap((x) -> "1", x -> String.valueOf(x+1),(x,y)->y,TreeMap::new));

最后

以上就是复杂鞋子为你收集整理的Java8中Collectors的使用averagingDouble,averagingInt,averagingLongcollectingAndThencountinggroupingBygroupingByConcurrentjoiningmappingmaxBy,minBypartitioningByreducingsummarizingDouble,summarizingInt,summarizingLongsummingDouble,summingInt,summingLongtoC的全部内容，希望文章能够帮你解决Java8中Collectors的使用averagingDouble,averagingInt,averagingLongcollectingAndThencountinggroupingBygroupingByConcurrentjoiningmappingmaxBy,minBypartitioningByreducingsummarizingDouble,summarizingInt,summarizingLongsummingDouble,summingInt,summingLongtoC所遇到的程序开发问题。

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