Collectors.groupingBy

转自https://www.jianshu.com/p/21b20c375599

1）计数

List fruitList = Lists.newArrayList(new Fruit("apple", 6),

       new Fruit("apple", 6),

       new Fruit("banana", 7), new Fruit("banana", 7),

       new Fruit("banana", 7), new Fruit("grape",8));Map map = fruitList.stream().

 collect(Collectors.groupingBy(Fruit::getName,Collectors.counting()));

输出结果是：{banana=3, apple=2, grape=1}

换一种啰嗦、复杂的写法，但能加深理解。

Map map = fruitList.stream().map(Fruit::getName).

 collect(Collectors.groupingBy(Function.identity(), Collectors.counting()));

2）排序

现在要按照水果map中value的数量逆序打印每个entry

map.entrySet().stream().sorted(Map.Entry.comparingByValue().reversed())

       .forEachOrdered(System.out::println);

3）累加求和

Map sumMap = fruitList.stream().collect.

(Collectors.groupingBy(Fruit::getName, Collectors.summingInt(Fruit::getPrice))); 

输出结果是：{banana=21, apple=12, grape=8}

4）分组

Map> groupMap =

fruitList.stream().collect(Collectors.groupingBy(Fruit::getName));

上述代码根据name将list分组，如果name是唯一的，那么上述代码就会显得啰嗦。我们需要知道，Guava补JDK之不足，现在改Guava一显身手了。

Map map = Maps.uniqueIndex(fruitList, Fruit::getName); 

生成的Map是ImmutableMap，不可更改里面的值。比如map.remove("apple")会抛出异常：java.lang.UnsupportedOperationException

根据不同的名字分为若干组

// group by price, uses 'mapping' to convert List to ListMap> groupMap =

fruitList.stream().collect(Collectors.groupingBy(Fruit::getName,

Collectors.mapping(Fruit::getPrice, Collectors.toList()))); 

上面一段代码可以用Guava代替

Multimap multiMap = ArrayListMultimap.create();

fruitList.forEach(fruit -> multiMap.put(fruit.getName(), fruit.getPrice()));