集合：

FluentIterable类：

• 使用FluentIterable.filter过滤, 即利用Predicate实现：

Iterable<Person> adults = FluentIterable.from(ps).filter(
	new Predicate<Person>() {
		@Override
		public boolean apply(Person p) {
			return p.getAge() >= 18; // 年龄>18
		}
	});

• 使用FluentIterable.transform()转换，即利用Function实现：

FluentIterable.from(ps).transform(new Function<Person, String>() {
	@Override
	public String apply(Person p) {
	     return Joiner.on('#').join(p.getName(), p.getAge());
	}
});

Lists类：

• 使用Lists.newArrayList创建列表：

ps = Lists.newArrayList(
	new Person("person1", 22),
	new Person("person2", 23),
	new Person("person3", 17)
);

• 使用Lists.partition()方法分割列表：

//[a, b, c, d, e] --> [[a, b], [c, d, e]] -
List<List<Person>> subList = Lists.partition(ps, 2);

Sets类：

• Sets.difference()求S1-S2

Set<String> s1 = Sets.newHashSet("1", "2", "3");
Set<String> s2 = Sets.newHashSet("2", "3", "4");
Sets.difference(s1, s2); //[1]

• Sets.intersection()求S1,S2交集

Set<String> s1 = Sets.newHashSet("1", "2", "3");
Set<String> s2 = Sets.newHashSet("3", "2", "4");
Sets.SetView<String> sv = Sets.intersection(s1, s2); // [2, 3]

• Sets.union()求合集
  

Set<String> s1 = Sets.newHashSet("1", "2", "3");
Set<String> s2 = Sets.newHashSet("3", "2", "4");
Sets.SetView<String> sv = Sets.union(s1, s2); // [3, 2, 1 ,4]

Maps类：

• Maps.uniqueIndex()将列表转换为map

//iterator各个元素作为Map.values, key为Function.apply返回值
Maps.uniqueIndex(ps.iterator(), new Function<Person, String>() {
	@Override
	public String apply(Person p) {
		return p.getName();
	}
});

• Maps.asMap()，<K, V>和Maps.uniqueIndex()相反

Maps.asMap(ps, new Function<Person, String>() {
	@Override
	public String apply(Person p) {
		return p.getName();
	}
});

• Maps Transform API:

Maps.transformEntries(map, new Maps.EntryTransformer<String, Boolean, String>() {
    @Override
    public String transformEntry(String key, Boolean value) {
        return value ? "yes" : "no";
    }
}); 将Map<String, Boolean> --> Map<String, String>, 其他的还有Maps.transformValues转换值

Multimaps:

• 一个key对应多个value。

ArrayListMultiMap:

ArrayListMultimap<String, String> multiMap = ArrayListMultimap.create();
multiMap.put("Foo", "1");
multiMap.put("Foo", "2");
multiMap.put("Foo", "3");
System.out.println(multiMap); // {Foo=[1,2,3]}

当出现重复值时，依然会被添加，因为ArrayListMultiMap的value时一个ArrayList:

ArrayListMultimap<String, String> multiMap = ArrayListMultimap.create();
multiMap.put("Bar", "1");
multiMap.put("Bar", "2");
multiMap.put("Bar", "3");
multiMap.put("Bar", "3");
multiMap.put("Bar", "3");
System.out.println(multiMap); //{Bar=[1, 2, 3, 3, 3]}，相同的value会重复，value内部是一个List

HashMultiMap:

HashMultimap<String, String> multiMap = HashMultimap.create();
multiMap.put("Bar", "1");
multiMap.put("Bar", "2");
multiMap.put("Bar", "3");
multiMap.put("Bar", "3");
multiMap.put("Bar", "3");
System.out.println(multiMap); //{Bar=[3, 2, 1]}, 相同的value不会重复，value内部是一个Set

其他一些MultiMap:

LinkedHashMultimap //顺序的HashMultimap, 形如LinkedHashMap
TreeMultimap //可排序的MultiMap, 形如TreeMap
//一些不可变的map
ImmutableListMultimap
ImmutableMultimap
ImmutableSetMultimap

BiMap:

• 其限制value是唯一的，且通过value可以找到key

BiMap<String,String> biMap = HashBiMap.create();
biMap.put("1","Tom");         
biMap.put("2","Tom"); //抛出异常

• BiMap.forcePut()强制放入value相等的entry:

BiMap<String, String> biMap = HashBiMap.create();
biMap.put("1", "Tom");
biMap.forcePut("2", "Tom");
System.out.println(biMap); //{2=Tom}

• BiMap.inverse()反转key-value

BiMap<String, String> biMap = HashBiMap.create();
biMap.put("1", "Tom");
biMap.put("2", "Harry");
BiMap<String, String> inverseMap = biMap.inverse();
System.out.println(biMap); //{2=Harry, 1=Tom}
System.out.println(inverseMap); //{Harry=2, Tom=1

Table:

• 它具有2个key[行, 列]，对应一个值。

HashBasedTable:

• 通用操作

HashBasedTable<Integer, Integer, String> table = HashBasedTable.create();
table.put(1, 1, "Rook");
table.put(1, 2, "Knight");
table.put(1, 3, "Bishop");
System.out.println(table.contains(1, 1)); //true
System.out.println(table.containsColumn(2)); //true
System.out.println(table.containsRow(1)); //true
System.out.println(table.containsValue("Rook")); //true
System.out.println(table.remove(1, 3)); //Bishop
System.out.println(table.get(3, 4)); //null

• Table views，表行列视图

Map<Integer,String> columnMap = table.column(1);
Map<Integer,String> rowMap = table.row(2);

• 其他table

ArrayTable //二维数组实现
ImmutableTable //不可变table，创建后不能改变
TreeBasedTable //对行列排序的table

Range:

• 代表一种范围的类。

Range<Integer> numberRange = Range.closed(1, 10); //包括首尾
System.out.println(numberRange.contains(10)); //true
System.out.println(numberRange.contains(1)); //true
		
Range<Integer> numberRange1 = Range.open(1,10); //除去首尾
System.out.println(numberRange1.contains(10)); //false
System.out.println(numberRange1.contains(1)); //false

• Range和Function组合成Predicate的过滤条件

Range<Integer> ageRange = Range.closed(35, 50);
//Person到age转换的function
Function<Person, Integer> ageFunction = new Function<Person, Integer>() {
	@Override
	public Integer apply(Person person) {
		return person.getAge();
	}
};
//这样得到年龄再[35, 50]之间的人
Predicate<Person> predicate = Predicates.compose(ageRange,ageFunction);

• 创建不可变的集合

MultiMap<Integer,String> map = new ImmutableListMultimap.Builder<Integer,String>()
                                   .put(1,"Foo").putAll(2,"Foo","Bar","Baz")
                                   .putAll(4,"Huey","Duey","Luey")
                                   .put(3,"Single").build();

Ordering:

• Ordering提供了一些简单强大的排序功能。

/**
 * 城市人口比较器
 */
public class CityByPopluation implements Comparator<City> {
	@Override
	public int compare(City city1, City city2) {
		return Ints.compare(city1.getPopulation(), city2.getPopulation());
	}
}

• 反序

Ordering.from(cityByPopluation).reverse();

• 处理Null

Ordering.from(comparator).nullsFirst();//null值最小放在前面

• 二次排序

/**
 * 雨量比较器
 */
public class CityByRainfall implements Comparator<City> {
	@Override
	public int compare(City city1, City city2) {
		return Doubles.compare(city1.getAverageRainfall(),
				city2.getAverageRainfall());
	}
}

Ordering<City> secondaryOrdering = Ordering.from(cityByPopulation).compound(cityByRainfall);//组合比较器
Collections.sort(cities,secondaryOrdering); //排序

• 获取最大最小

Ordering<City> ordering = Ordering.from(cityByPopluation);
List<City> topFive = ordering.greatestOf(cityList,5); //前5个
List<City> bottomThree = ordering.leastOf(cityList,3); //最后3个

看来Guava在集合方面还是很给力的。

来源：http://my.oschina.net/indestiny/blog/215839