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Java12 Collectors.teeing如何理解,很多新手对此不是很清楚,为了帮助大家解决这个难题,下面小编将为大家详细讲解,有这方面需求的人可以来学习下,希望你能有所收获。
前言
在 Java 12 里面有个非常好用但在官方 JEP 没有公布的功能,因为它只是 Collector 中的一个小改动,它的作用是 merge 两个 collector 的结果,这句话显得很抽象,老规矩,我们先来看个图:
管道改造经常会用这个小东西,通常我们叫它「三通」,它的主要作用就是将 downstream1 和 downstream2 的流入合并,然后从 merger 流出
有了这个形象的说明我们就进入正题吧
Collectors.teeing
上面提到的小功能就是 Collectors.teeing API, 先来看一下 JDK 关于该 API 的说明,看着觉得难受的直接忽略,继续向下看例子就好了:
/** * Returns a {@code Collector} that is a composite of two downstream collectors. * Every element passed to the resulting collector is processed by both downstream * collectors, then their results are merged using the specified merge function * into the final result. * * <p>The resulting collector functions do the following: * * <ul> * <li>supplier: creates a result container that contains result containers * obtained by calling each collector's supplier * <li>accumulator: calls each collector's accumulator with its result container * and the input element * <li>combiner: calls each collector's combiner with two result containers * <li>finisher: calls each collector's finisher with its result container, * then calls the supplied merger and returns its result. * </ul> * * <p>The resulting collector is {@link Collector.Characteristics#UNORDERED} if both downstream * collectors are unordered and {@link Collector.Characteristics#CONCURRENT} if both downstream * collectors are concurrent. * * @param <T> the type of the input elements * @param <R1> the result type of the first collector * @param <R2> the result type of the second collector * @param <R> the final result type * @param downstream1 the first downstream collector * @param downstream2 the second downstream collector * @param merger the function which merges two results into the single one * @return a {@code Collector} which aggregates the results of two supplied collectors. * @since 12 */ public static <T, R1, R2, R> Collector<T, ?, R> teeing(Collector<? super T, ?, R1> downstream1, Collector<? super T, ?, R2> downstream2, BiFunction<? super R1, ? super R2, R> merger) { return teeing0(downstream1, downstream2, merger); }
API 描述重的一句话非常关键:
Every element passed to the resulting collector is processed by both downstream collectors
结合「三通图」来说明就是,集合中每一个要被传入 merger 的元素都会经过 downstream1 和 downstream2 的加工处理
其中 merger 类型是 BiFunction,也就是说接收两个参数,并输出一个值,请看它的 apply 方法
@FunctionalInterface public interface BiFunction<T, U, R> { /** * Applies this function to the given arguments. * * @param t the first function argument * @param u the second function argument * @return the function result */ R apply(T t, U u); }
至于可以如何处理,我们来看一些例子吧
例子
为了更好的说明 teeing 的使用,列举了四个例子,看过这四个例子再回看上面的 API 说明,相信你会柳暗花明了
计数和累加
先来看一个经典的问题,给定的数字集合,需要映射整数流中的元素数量和它们的和
class CountSum { private final Long count; private final Integer sum; public CountSum(Long count, Integer sum) { this.count = count; this.sum = sum; } @Override public String toString() { return "CountSum{" + "count=" + count + ", sum=" + sum + '}'; } }
通过 Collectors.teeing 处理
CountSum countsum = Stream.of(2, 11, 1, 5, 7, 8, 12) .collect(Collectors.teeing( counting(), summingInt(e -> e), CountSum::new)); System.out.println(countsum.toString());
downstream1 通过 Collectors 的静态方法 counting 进行集合计数
downstream2 通过 Collectors 的静态方法 summingInt 进行集合元素值的累加
merger 通过 CountSum 构造器收集结果
运行结果:
CountSum{count=7, sum=46}
我们通过 teeing 一次性得到我们想要的结果,继续向下看其他例子:
最大值与最小值
通过给定的集合, 一次性计算出集合的最大值与最小值,同样新建一个类 MinMax,并创建构造器用于 merger 收集结果
class MinMax { private final Integer min; private final Integer max; public MinMax(Integer min, Integer max) { this.min = min; this.max = max; } @Override public String toString() { return "MinMax{" + "min=" + min + ", max=" + max + '}'; } }
通过 teeing API 计算结果:
MinMax minmax = Stream.of(2, 11, 1, 5, 7, 8, 12) .collect(Collectors.teeing( minBy(Comparator.naturalOrder()), maxBy(Comparator.naturalOrder()), (Optional<Integer> a, Optional<Integer> b) -> new MinMax(a.orElse(Integer.MIN_VALUE), b.orElse(Integer.MAX_VALUE)))); System.out.println(minmax.toString());
downstream1 通过 Collectors 的静态方法 minBy,通过 Comparator 比较器按照自然排序找到最小值
downstream2 通过 Collectors 的静态方法 maxBy,通过 Comparator 比较器按照自然排序找到最大值
merger 通过 MinMax 构造器收集结果,只不过为了应对 NPE,将 BiFunction 的两个入参经过 Optional 处理
运行结果:
MinMax{min=1, max=12}
为了验证一下 Optional,我们将集合中添加一个 null 元素,并修改一下排序规则来看一下排序结果:
MinMax minmax = Stream.of(null, 2, 11, 1, 5, 7, 8, 12) .collect(Collectors.teeing( minBy(Comparator.nullsFirst(Comparator.naturalOrder())), maxBy(Comparator.nullsLast(Comparator.naturalOrder())), (Optional<Integer> a, Optional<Integer> b) -> new MinMax(a.orElse(Integer.MIN_VALUE), b.orElse(Integer.MAX_VALUE))));
downstream1 处理规则是将 null 放在排序的最前面
downstream2 处理规则是将 null 放在排序的最后面
merger 处理时,都会执行 optional.orElse 方法,分别输出最小值与最大值
运行结果:
MinMax{min=-2147483648, max=2147483647}
瓜的总重和单个重量
接下来举一个更贴合实际的操作对象的例子
// 定义瓜的类型和重量 class Melon { private final String type; private final int weight; public Melon(String type, int weight) { this.type = type; this.weight = weight; } public String getType() { return type; } public int getWeight() { return weight; } } // 总重和单个重量列表 class WeightsAndTotal { private final int totalWeight; private final List<Integer> weights; public WeightsAndTotal(int totalWeight, List<Integer> weights) { this.totalWeight = totalWeight; this.weights = weights; } @Override public String toString() { return "WeightsAndTotal{" + "totalWeight=" + totalWeight + ", weights=" + weights + '}'; } }
通过 teeing API 计算总重量和单个列表重量
List<Melon> melons = Arrays.asList(new Melon("Crenshaw", 1200), new Melon("Gac", 3000), new Melon("Hemi", 2600), new Melon("Hemi", 1600), new Melon("Gac", 1200), new Melon("Apollo", 2600), new Melon("Horned", 1700), new Melon("Gac", 3000), new Melon("Hemi", 2600) ); WeightsAndTotal weightsAndTotal = melons.stream() .collect(Collectors.teeing( summingInt(Melon::getWeight), mapping(m -> m.getWeight(), toList()), WeightsAndTotal::new)); System.out.println(weightsAndTotal.toString());
downstream1 通过 Collectors 的静态方法 summingInt 做重量累加
downstream2 通过 Collectors 的静态方法 mapping 提取出瓜的重量,并通过流的终结操作 toList() 获取结果
merger 通过 WeightsAndTotal 构造器获取结果
运行结果:
WeightsAndTotal{totalWeight=19500, weights=[1200, 3000, 2600, 1600, 1200, 2600, 1700, 3000, 2600]}
继续一个更贴合实际的例子吧:
预约人员列表和预约人数
class Guest { private String name; private boolean participating; private Integer participantsNumber; public Guest(String name, boolean participating, Integer participantsNumber) { this.name = name; this.participating = participating; this.participantsNumber = participantsNumber; } public boolean isParticipating() { return participating; } public Integer getParticipantsNumber() { return participantsNumber; } public String getName() { return name; } } class EventParticipation { private List<String> guestNameList; private Integer totalNumberOfParticipants; public EventParticipation(List<String> guestNameList, Integer totalNumberOfParticipants) { this.guestNameList = guestNameList; this.totalNumberOfParticipants = totalNumberOfParticipants; } @Override public String toString() { return "EventParticipation { " + "guests = " + guestNameList + ", total number of participants = " + totalNumberOfParticipants + " }"; } }
通过 teeing API 处理
var result = Stream.of( new Guest("Marco", true, 3), new Guest("David", false, 2), new Guest("Roger",true, 6)) .collect(Collectors.teeing( Collectors.filtering(Guest::isParticipating, Collectors.mapping(Guest::getName, Collectors.toList())), Collectors.summingInt(Guest::getParticipantsNumber), EventParticipation::new )); System.out.println(result);
downstream1 通过 filtering 方法过滤出确定参加的人,并 mapping 出他们的姓名,最终放到 toList 集合中
downstream2 通过 summingInt 方法计数累加
merger 通过 EventParticipation 构造器收集结果
其中我们定义了 var result 来收集结果,并没有指定类型,这个语法糖也加速了我们编程的效率
运行结果:
EventParticipation { guests = [Marco, Roger], total number of participants = 11 }
其实 teeing API 就是灵活应用 Collectors 里面定义的静态方法,将集合元素通过 downstream1 和 downstream2 进行处理,最终通过 merger 收集起来,当项目中有同时获取两个收集结果时,是时候应用我们的 teeing API 了。
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