一、Maps

Maps有许多很酷的实用程序，值得单独解释。

1、uniqueIndex

Maps.uniqueIndex（Iterable，Function）解决了一个常见的情况，即有一堆对象，每个对象都有一些唯一的属性，并希望能够根据该属性查找这些对象。
假设我们有一堆字符串，我们知道它们有唯一的长度，我们希望能够查找具有特定长度的字符串。

ImmutableMap<Integer, String> stringsByIndex = Maps.uniqueIndex(strings, new Function<String, Integer> () {public Integer apply(String string) {return string.length();}});

如果索引不是唯一的，请参阅下面的Multimaps.index。

2、difference

Maps.difference（Map，Map）允许您比较两张地图之间的所有差异。它返回一个MapDifference对象，该对象将Venn图分解为：

Method	Description
entriesInCommon()	The entries which are in both maps, with both matching keys and values.
entriesDiffering()	具有相同键但不同值的条目。此映射中的值属于MapDifference.ValueDifference类型，可以查看左右值。
entriesOnlyOnLeft()	Returns the entries whose keys are in the left but not in the right map.
entriesOnlyOnRight()	Returns the entries whose keys are in the right but not in the left map.

Map<String, Integer> left = ImmutableMap.of("a", 1, "b", 2, "c", 3);
Map<String, Integer> right = ImmutableMap.of("b", 2, "c", 4, "d", 5);
MapDifference<String, Integer> diff = Maps.difference(left, right);diff.entriesInCommon(); // {"b" => 2}
diff.entriesDiffering(); // {"c" => (3, 4)}
diff.entriesOnlyOnLeft(); // {"a" => 1}
diff.entriesOnlyOnRight(); // {"d" => 5}

3、BiMap 

BiMap上的Guava实用程序位于Maps类中，因为BiMap也是Map。

BiMap utility	Corresponding Map utility
synchronizedBiMap(BiMap)	Collections.synchronizedMap(Map)
unmodifiableBiMap(BiMap)	Collections.unmodifiableMap(Map)

4、Static Factories

映射提供以下静态工厂方法。

Implementation	Factories
HashMap	basic, from Map, with expected size
LinkedHashMap	basic, from Map
TreeMap	basic, from Comparator, from SortedMap
EnumMap	from Class, from Map
ConcurrentMap	basic
IdentityHashMap	basic

二、Multisets

标准集合操作（如containsAll）忽略多集合中元素的计数，只关心元素是否在多集合中。多集提供了许多考虑多集中元素多重性的操作。

Method	Explanation	Difference from Collection method
containsOccurrences(Multiset sup, Multiset sub)	Returns true if sub.count(o) <= super.count(o) for all o.	Collection.containsAll忽略计数，只测试是否包含元素。
removeOccurrences(Multiset removeFrom, Multiset toRemove)	Removes one occurrence in removeFrom for each occurrence of an element in toRemove.	Collection.removeAll将删除在to Remove中出现一次的任何元素的所有实例。
retainOccurrences(Multiset removeFrom, Multiset toRetain)	Guarantees that removeFrom.count(o) <= toRetain.count(o) for all o.	Collection.retail将所有出现的元素保留为Retain。
intersection(Multiset, Multiset)	返回两个多集的交集的视图；一种非破坏性的替代方案。	Has no analogue.

Multiset<String> multiset1 = HashMultiset.create();
multiset1.add("a", 2);Multiset<String> multiset2 = HashMultiset.create();
multiset2.add("a", 5);multiset1.containsAll(multiset2); // returns true: all unique elements are contained,// even though multiset1.count("a") == 2 < multiset2.count("a") == 5
Multisets.containsOccurrences(multiset1, multiset2); // returns falseMultisets.removeOccurrences(multiset2, multiset1); // multiset2 now contains 3 occurrences of "a"multiset2.removeAll(multiset1); // removes all occurrences of "a" from multiset2, even though multiset1.count("a") == 2
multiset2.isEmpty(); // returns true

Multiset中的其他实用程序包括：

Method	Description
copyHighestCountFirst(Multiset)	返回multiset的不可变副本，该副本按频率降序迭代元素。
unmodifiableMultiset(Multiset)	Returns an unmodifiable view of the multiset.
unmodifiableSortedMultiset(SortedMultiset)	Returns an unmodifiable view of the sorted multiset.

Multiset<String> multiset = HashMultiset.create();
multiset.add("a", 3);
multiset.add("b", 5);
multiset.add("c", 1);ImmutableMultiset<String> highestCountFirst = Multisets.copyHighestCountFirst(multiset);// highestCountFirst, like its entrySet and elementSet, iterates over the elements in order {"b", "a", "c"}

三、Multimaps

Multimaps提供了许多通用的实用程序操作，值得单独解释。

1、index

Maps.uniqueIndex，Multimaps.index（Iterable，Function）的表亲回答了这样一种情况，即当您希望能够查找具有某些特定共同属性的所有对象时，这些属性不一定是唯一的。
假设我们希望根据字符串的长度对其进行分组。

ImmutableSet<String> digits = ImmutableSet.of("zero", "one", "two", "three", "four","five", "six", "seven", "eight", "nine");
Function<String, Integer> lengthFunction = new Function<String, Integer>() {public Integer apply(String string) {return string.length();}
};
ImmutableListMultimap<Integer, String> digitsByLength = Multimaps.index(digits, lengthFunction);
/** digitsByLength maps:*  3 => {"one", "two", "six"}*  4 => {"zero", "four", "five", "nine"}*  5 => {"three", "seven", "eight"}*/

2、invertFrom

由于多映射可以将多个关键点映射到一个值，也可以将一个关键点映像到多个值，因此反转多映射非常有用。Guava提供inverteFrom（Multimap to Invert，Multimap dest），让您无需选择实现即可完成此操作。
注意：如果您使用的是ImmutableMultimap，请考虑使用ImmutableMultimap.inverse（）。

ArrayListMultimap<String, Integer> multimap = ArrayListMultimap.create();
multimap.putAll("b", Ints.asList(2, 4, 6));
multimap.putAll("a", Ints.asList(4, 2, 1));
multimap.putAll("c", Ints.asList(2, 5, 3));TreeMultimap<Integer, String> inverse = Multimaps.invertFrom(multimap, TreeMultimap.<Integer, String>create());
// note that we choose the implementation, so if we use a TreeMultimap, we get results in order
/** inverse maps:*  1 => {"a"}*  2 => {"a", "b", "c"}*  3 => {"c"}*  4 => {"a", "b"}*  5 => {"c"}*  6 => {"b"}*/

3、forMap

需要在地图上使用Multimap方法吗？forMap（Map）将Map视为SetMultimap。例如，与Multimaps.inverteFrom组合使用时，这一功能特别有用。

Map<String, Integer> map = ImmutableMap.of("a", 1, "b", 1, "c", 2);
SetMultimap<String, Integer> multimap = Multimaps.forMap(map);
// multimap maps ["a" => {1}, "b" => {1}, "c" => {2}]
Multimap<Integer, String> inverse = Multimaps.invertFrom(multimap, HashMultimap.<Integer, String> create());
// inverse maps [1 => {"a", "b"}, 2 => {"c"}]

四、Tables

Tables类提供了一些方便的实用程序。

1、customTable

与Multimaps.newXXXMultimap（Map，Supplier）实用程序类似，Tables.newCustomTable（Map，供应商＜Map＞）允许您使用任何喜欢的行或列映射指定表实现。

// use LinkedHashMaps instead of HashMaps
Table<String, Character, Integer> table = Tables.newCustomTable(Maps.<String, Map<Character, Integer>>newLinkedHashMap(),new Supplier<Map<Character, Integer>> () {public Map<Character, Integer> get() {return Maps.newLinkedHashMap();}});