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社区首页 >专栏 >Google Guava-集合类型

Google Guava-集合类型

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用户5166556
发布2019-04-16 12:40:24
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发布2019-04-16 12:40:24
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Guava引入了很多JDK没有的、但我们发现明显有用的新集合类型。 Multiset

统计一个词在文档中出现了多少次,传统的做法是这样的:(这里举个简单的传统方法,下面不在说明)

代码语言:javascript
复制
@Test
	public void test20(){
		List<String> words = new ArrayList<String>();
		Map<String, Integer> counts = new HashMap<String, Integer>();
		for (String word : words) {
		    Integer count = counts.get(word);
		    if (count == null) {
		        counts.put(word, 1);
		    } else {
		        counts.put(word, count + 1);
		    }
		}
	}

这种写法很笨拙,也容易出错,并且不支持同时收集多种统计信息,如总词数。我们可以做的更好。 Guava提供了一个新集合类型 Multiset,它可以多次添加相等的元素。很好使用例如:

代码语言:javascript
复制
@Test
	public void test10() {
		Multiset<String> multiset1 = HashMultiset.create();
		multiset1.add("a", 2);
		multiset1.add("b");
		multiset1.add("b");
		int count = multiset1.count("b");
		System.out.println(count);
		Multiset<String> multiset2 = HashMultiset.create();
		multiset2.add("a", 5);

		multiset1.containsAll(multiset2); // 返回true;因为包含了所有不重复元素,
		// 虽然multiset1实际上包含2个"a",而multiset2包含5个"a"
		Multisets.containsOccurrences(multiset1, multiset2); // returns false
		// multiset2.removeOccurrences(multiset1); // multiset2 现在包含3个"a"
		multiset2.removeAll(multiset1);// multiset2移除所有"a",虽然multiset1只有2个"a"
		multiset2.isEmpty(); // returns true
	}

下面我用测试用例说明一下guava集合类型的新功能

代码语言:javascript
复制
// 在JDK的基础上增加了map的功能
	@Test
	public void test8() {
		Map<String, Integer> left = ImmutableMap.of("a", 1, "b", 2, "c", 3, "d", 4);
		Map<String, Integer> right = ImmutableMap.of("a", 2, "b", 2, "c", 3);
		MapDifference<String, Integer> diff = Maps.difference(left, right);
		System.out.println(diff.entriesDiffering());// 键相同 但是值不相同的映射项
		System.out.println(diff.entriesInCommon());// 键值都相同的映射项
		System.out.println(diff.entriesOnlyOnLeft()); // {"a" => 1}
		System.out.println(diff.entriesOnlyOnRight()); // {"d" => 5}
	}

// 一个非常有用的集合,取得交集,并集,第一个集合中有的数据第二个集合中没有的数据等方法
	@Test
	public void test7() {
		List<String> list = Lists.newArrayList();
		// list.
		Set<String> wordsWithPrimeLength = ImmutableSet.of("one", "two", "three", "six", "seven", "eight");
		Set<String> primes = ImmutableSet.of("two", "three", "five", "seven");
		// primes.add("hello");
		SetView<String> intersection = Sets.intersection(primes, wordsWithPrimeLength);
		// intersection包含"two", "three", "seven"
		ImmutableSet<String> immutableCopy = intersection.immutableCopy();// 可以使用交集,但不可变拷贝的读取效率更高
		// Sets.difference(set1, set2);
		SetView<String> difference = Sets.difference(primes, wordsWithPrimeLength);
		SetView<String> union = Sets.union(primes, wordsWithPrimeLength);
		System.out.println(union);
		System.out.println(difference);
		System.out.println(immutableCopy);
		Set<String> animals = ImmutableSet.of("gerbil", "hamster");
		Set<String> fruits = ImmutableSet.of("apple", "orange", "banana");

		Set<List<String>> product = Sets.cartesianProduct(animals, fruits);
		// {{"gerbil", "apple"}, {"gerbil", "orange"}, {"gerbil", "banana"},
		// {"hamster", "apple"}, {"hamster", "orange"}, {"hamster", "banana"}}

		Set<Set<String>> animalSets = Sets.powerSet(animals);
		// {{}, {"gerbil"}, {"hamster"}, {"gerbil", "hamster"}}
	}
// 比较器,能够比较数据大小按照自己的要求
	@Test
	public void test6() {
		Ordering<String> order = new Ordering<String>() {
			@Override
			public int compare(String left, String right) {
				return Ints.compare(left.length(), right.length());
			}
		};
		order.compare("a", "b");
		String min = order.min("a", "b1", "c11", "d111");
		System.out.println(min);
	}
	@Test
	public void test11() {
		Multiset<String> multiset = HashMultiset.create();
		multiset.add("a", 3);
		multiset.add("b", 5);
		multiset.add("c", 1);
		multiset.add("a");
		multiset.add("a");
		multiset.add("a");
		multiset.add("a");
		ImmutableMultiset<String> highestCountFirst = Multisets.copyHighestCountFirst(multiset);
		System.out.println(highestCountFirst.elementSet());
		// highestCountFirst,包括它的entrySet和elementSet,按{"b", "a", "c"}排列元素
	}

	// 根据特征进行筛选集合中的数据
	@Test
	public void test12() {
		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);
		System.out.println(digitsByLength);
		/*
		 * digitsByLength maps: 3 => {"one", "two", "six"} 4 => {"zero", "four",
		 * "five", "nine"} 5 => {"three", "seven", "eight"}
		 */
	}

	@Test
	public void test13() {
		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> create = TreeMultimap.create();
		TreeMultimap<Integer, String> inverse = Multimaps.invertFrom(multimap, create);
		System.out.println(inverse);
		// 注意我们选择的实现,因为选了TreeMultimap,得到的反转结果是有序的
		/*
		 * inverse maps: 1 => {"a"} 2 => {"a", "b", "c"} 3 => {"c"} 4 => {"a",
		 * "b"} 5 => {"c"} 6 => {"b"}
		 */
	}
	//map中的key和value值进行反转
	@Test
	public void test14() {
		Map<String, Integer> map = ImmutableMap.of("a", 1, "b", 1, "c", 2);
		SetMultimap<String, Integer> multimap = Multimaps.forMap(map);
		// multimap:["a" => {1}, "b" => {1}, "c" => {2}]
		System.out.println(multimap);
		HashMultimap<Integer, String> dest = HashMultimap.create();
		Multimap<Integer, String> inverse = Multimaps.invertFrom(multimap, dest);
		System.out.println(inverse);
		// inverse:[1 => {"a","b"}, 2 => {"c"}]
		/*
		 * Map<String, Integer> map = ImmutableMap.of("a", 1, "b", 1, "c", 2);
		 * SetMultimap<String, Integer> multimap = Multimaps.forMap(map);
		 */
		// multimap:["a" => {1}, "b" => {1}, "c" => {2}]
		// Multimap<Integer, String> inverse = Multimaps.invertFrom(multimap,
		// Multimap<Integer, String>.create());
		// inverse:[1 => {"a","b"}, 2 => {"c"}]
	}

	// 多功能的map
	@Test
	public void test15() {
		ArrayListMultimap<String, Integer> multimap = ArrayListMultimap.create();
		multimap.put("a", 1);
		multimap.put("a", 2);
		System.out.println(multimap);
		Map<String, Integer> map = new HashMap<String, Integer>();
		map.put("a", 1);
		map.put("a", 2);
		System.out.println(map);
		Map<String, Integer> tree = new TreeMap<String, Integer>();
		tree.put("a", 1);
		tree.put("a", 2);
		System.out.println(tree);
		Map<String, Integer> table = new Hashtable<String, Integer>();
		table.put("a", 1);
		table.put("a", 2);
		System.out.println(table);
		/**
		 * {a=[1, 2]} {a=2} {a=2} {a=2}
		 */
	}
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原始发表:2015年02月06日,如有侵权请联系 cloudcommunity@tencent.com 删除

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