我希望这个问题对这个论坛来说不是太基本的,但我们会看到的。我想知道如何重构一些代码以获得更好的性能,这是运行了很多次。

假设我正在使用Map(可能是HashMap)创建一个词频列表,其中每个键都是一个包含要统计的单词的String,值是一个Integer,该Integer在每次找到单词的标记时递增。

在Perl中,增加这样一个值非常简单:

$map{$word}++;

但在Java中,这要复杂得多。下面是我目前的做法:

int count = map.containsKey(word) ? map.get(word) : 0;
map.put(word, count + 1);

这当然依赖于新Java版本中的自动装箱特性。我想知道您是否可以建议一种更有效的方法来增加这个值。是否有更好的性能理由避开Collections框架而使用其他框架呢?

更新:我已经对几个答案做了测试。见下文。


当前回答

在java 8中,简单易行的方法如下:

final ConcurrentMap<String, AtomicLong> map = new ConcurrentHashMap<String, AtomicLong>();
    map.computeIfAbsent("foo", key -> new AtomicLong(0)).incrementAndGet();

其他回答

作为我自己评论的后续:Trove看起来是可行的。如果出于某种原因,您希望坚持使用标准JDK,那么ConcurrentMap和AtomicLong可以使代码稍微好一点,不过使用YMMV。

    final ConcurrentMap<String, AtomicLong> map = new ConcurrentHashMap<String, AtomicLong>();
    map.putIfAbsent("foo", new AtomicLong(0));
    map.get("foo").incrementAndGet();

将在foo的map中保留1作为值。实际上,增加线程的友好性是这种方法所推荐的。

你应该意识到你最初的尝试

int count = map.containsKey(word) ? map.get(word) : 0;

包含映射上两个可能代价高昂的操作,即containsKey和get。前者执行的操作可能与后者非常相似,因此您要做两次相同的工作!

如果查看Map的API,当Map不包含所请求的元素时,get操作通常返回null。

注意,这将得到一个像

map.put( key, map.get(key) + 1 );

dangerous, since it might yield NullPointerExceptions. You should check for a null first. Also note, and this is very important, that HashMaps can contain nulls by definition. So not every returned null says "there is no such element". In this respect, containsKey behaves differently from get in actually telling you whether there is such an element. Refer to the API for details. For your case, however, you might not want to distinguish between a stored null and "noSuchElement". If you don't want to permit nulls you might prefer a Hashtable. Using a wrapper library as was already proposed in other answers might be a better solution to manual treatment, depending on the complexity of your application. To complete the answer (and I forgot to put that in at first, thanks to the edit function!), the best way of doing it natively, is to get into a final variable, check for null and put it back in with a 1. The variable should be final because it's immutable anyway. The compiler might not need this hint, but its clearer that way. final HashMap map = generateRandomHashMap(); final Object key = fetchSomeKey(); final Integer i = map.get(key); if (i != null) { map.put(i + 1); } else { // do something } If you do not want to rely on autoboxing, you should say something like map.put(new Integer(1 + i.getValue())); instead.

由于很多人在Java主题中搜索Groovy的答案,下面是如何在Groovy中做到这一点:

dev map = new HashMap<String, Integer>()
map.put("key1", 3)

map.merge("key1", 1) {a, b -> a + b}
map.merge("key2", 1) {a, b -> a + b}

查看谷歌Collections Library总是一个好主意。在这种情况下,Multiset将做的伎俩:

Multiset bag = Multisets.newHashMultiset();
String word = "foo";
bag.add(word);
bag.add(word);
System.out.println(bag.count(word)); // Prints 2

有类似map的方法用于遍历键/条目等。在内部实现目前使用HashMap<E, AtomicInteger>,所以您不会产生装箱成本。

希望我正确理解了你的问题,我从Python来到Java,所以我可以同情你的挣扎。

如果你有

map.put(key, 1)

你会这么做

map.put(key, map.get(key) + 1)

希望这能有所帮助!