业务中须要将一组数据分类后收集总和,本来能够使用Collectors.summingInt()
,可是咱们的数据源是BigDecimal类型的,而Java8原生只提供了summingInt、summingLong、summingDouble三种基础类型的方法。因而就本身动手丰衣足食吧。。app
指望目标:this
Map<String, BigDecimal> result = Arrays.stream(records).parallel().collect(Collectors.groupingBy( Record::getType, CollectorsUtil.summingBigDecimal(Record::getAmount)));
1. 依葫芦
先分析一下Collectors.summingInt()
方法code
public static <T> Collector<T, ?, Integer> summingInt(ToIntFunction<? super T> mapper) { return new CollectorImpl<>( () -> new int[1], (a, t) -> { a[0] += mapper.applyAsInt(t); }, (a, b) -> { a[0] += b[0]; return a; }, a -> a[0], CH_NOID); }
该方法接受ToIntFunction<? super T>
类型的参数,返回CollectorImpl
类型的实例化对象。CollectorImpl
是Collector
接口的惟一实现类对象
CollectorImpl(Supplier<A> supplier, BiConsumer<A, T> accumulator, BinaryOperator<A> combiner, Function<A,R> finisher, Set<Characteristics> characteristics) { this.supplier = supplier; this.accumulator = accumulator; this.combiner = combiner; this.finisher = finisher; this.characteristics = characteristics; }
分析CollectorImpl
的构造器参数,可知summingInt()
方法是这样的接口
arg[0]建立一个计算用的容器: () -> new int[1]
arg[1]为计算逻辑: (a, t) -> { a[0] += mapper.applyAsInt(t); }
arg[2]为合并逻辑: (a, b) -> { a[0] += b[0]; return a; }
arg[3]为返回最终计算值: a -> a[0]
arg[4]为空Set(不知道干什么用。。): Collections.emptySet()
事务
2. 画瓢
很天然的,BigDecimal的就是这样了ip
public static <T> Collector<T, ?, BigDecimal> summingBigDecimal(Function<? super T, BigDecimal> mapper) { return new CollectorImpl<>(() -> new BigDecimal[] { BigDecimal.ZERO }, (a, t) -> { a[0] = a[0].add(mapper.apply(t)); }, (a, b) -> { a[0] = a[0].add(b[0]); return a; }, a -> a[0], CH_NOID); }