浅谈Java之Map 按值排序 (Map sort by value)

发布于 2020-12-29|标签java
复制链接
下面小妖就为大家带来一篇浅谈Java之Map 按值排序 (Map sort by value)。小妖觉得挺不错的,现在就分享给大家,也给大家做个参考。一起跟随小妖过来看看吧
Map是键值对的集合,又叫作字典或关联数组等,是最常见的数据结构之一。在java如何让一个map按value排序呢? 看似简单,但却不容易!比如,Map中key是String类型,表示一个单词,而value是int型,表示该单词出现的次数,现在我们想要按照单词出现的次数来排序: ```java Map map = new TreeMap(); map.put("me", 1000); map.put("and", 4000); map.put("you", 3000); map.put("food", 10000); map.put("hungry", 5000); map.put("later", 6000); ``` 按值排序的结果应该是: ```java key value me 1000 you 3000 and 4000 hungry 5000 later 6000 food 10000 ``` 首先,不能采用SortedMap结构,因为SortedMap是按键排序的Map,而不是按值排序的Map,我们要的是按值排序的Map。 Couldn't you do this with a SortedMap?  No, because the map are being sorted by its keys. 方法一:如下Java代码: ```java import java.util.Iterator; import java.util.Set; import java.util.TreeSet; public class Main { public static void main(String[] args) { Set set = new TreeSet(); set.add(new Pair("me", "1000")); set.add(new Pair("and", "4000")); set.add(new Pair("you", "3000")); set.add(new Pair("food", "10000")); set.add(new Pair("hungry", "5000")); set.add(new Pair("later", "6000")); set.add(new Pair("myself", "1000")); for (Iterator i = set.iterator(); i.hasNext();) System.out.println(i.next()); } } class Pair implements Comparable { private final String name; private final int number; public Pair(String name, int number) { this.name = name; this.number = number; } public Pair(String name, String number) throws NumberFormatException { this.name = name; this.number = Integer.parseInt(number); } public int compareTo(Object o) { if (o instanceof Pair) { int cmp = Double.compare(number, ((Pair) o).number); if (cmp != 0) { return cmp; } return name.compareTo(((Pair) o).name); } throw new ClassCastException("Cannot compare Pair with " + o.getClass().getName()); } public String toString() { return name + ' ' + number; } } ``` 类似的C++代码: ```java typedef pair PAIR; int cmp(const PAIR& x, const PAIR& y) { return x.second > y.second; } map m; vector vec; for (map::iterator curr = m.begin(); curr != m.end(); ++curr) { vec.push_back(make_pair(curr->first, curr->second)); } sort(vec.begin(), vec.end(), cmp); ``` 上面方法的实质意义是:将Map结构中的键值对(Map.Entry)封装成一个自定义的类(结构),或者直接用Map.Entry类。自定义类知道自己应该如何排序,也就是按值排序,具体为自己实现Comparable接口或构造一个Comparator对象,然后不用Map结构而采用有序集合(SortedSet, TreeSet是SortedSet的一种实现),这样就实现了Map中sort by value要达到的目的。就是说,不用Map,而是把Map.Entry当作一个对象,这样问题变为实现一个该对象的有序集合或对该对象的集合做排序。既可以用SortedSet,这样插入完成后自然就是有序的了,又或者用一个List或数组,然后再对其做排序(Collections.sort() or Arrays.sort())。 Encapsulate the information in its own class. Either implement Comparable and write rules for the natural ordering or write a Comparator based on your criteria. Store the information in a sorted collection, or use the Collections.sort() method. 方法二:You can also use the following code to sort by value: ```java public static Map sortByValue(Map map) { List list = new LinkedList(map.entrySet()); Collections.sort(list, new Comparator() { public int compare(Object o1, Object o2) { return ((Comparable) ((Map.Entry) (o1)).getValue()) .compareTo(((Map.Entry) (o2)).getValue()); } }); Map result = new LinkedHashMap(); for (Iterator it = list.iterator(); it.hasNext();) { Map.Entry entry = (Map.Entry) it.next(); result.put(entry.getKey(), entry.getValue()); } return result; } public static Map sortByValue(Map map, final boolean reverse) { List list = new LinkedList(map.entrySet()); Collections.sort(list, new Comparator() { public int compare(Object o1, Object o2) { if (reverse) { return -((Comparable) ((Map.Entry) (o1)).getValue()) .compareTo(((Map.Entry) (o2)).getValue()); } return ((Comparable) ((Map.Entry) (o1)).getValue()) .compareTo(((Map.Entry) (o2)).getValue()); } }); Map result = new LinkedHashMap(); for (Iterator it = list.iterator(); it.hasNext();) { Map.Entry entry = (Map.Entry) it.next(); result.put(entry.getKey(), entry.getValue()); } return result; }         Map map = new HashMap(); map.put("a", 4); map.put("b", 1); map.put("c", 3); map.put("d", 2); Map sorted = sortByValue(map); System.out.println(sorted); // output : {b=1, d=2, c=3, a=4} 或者还可以这样: Map map = new HashMap(); map.put("a", 4); map.put("b", 1); map.put("c", 3); map.put("d", 2); Set> treeSet = new TreeSet>( new Comparator>() { public int compare(Map.Entry o1, Map.Entry o2) { Integer d1 = o1.getValue(); Integer d2 = o2.getValue(); int r = d2.compareTo(d1); if (r != 0) return r; else return o2.getKey().compareTo(o1.getKey()); } }); treeSet.addAll(map.entrySet()); System.out.println(treeSet); // output : [a=4, c=3, d=2, b=1] ``` 另外,Groovy 中实现 sort map by value,当然本质是一样的,但却很简洁 :  用 groovy 中 map 的 sort 方法(需要 groovy 1.6), ```java def result = map.sort(){ a, b -> b.value.compareTo(a.value) } ``` 如:  ["a":3,"b":1,"c":4,"d":2].sort{ a,b -> a.value - b.value } 结果为: [b:1, d:2, a:3, c:4] Python中也类似:  ```py h = {"a":2,"b":1,"c":3} i = h.items() // i = [('a', 2), ('c', 3), ('b', 1)] i.sort(lambda (k1,v1),(k2,v2): cmp(v2,v1) ) // i = [('c', 3), ('a', 2), ('b', 1)] ```
冀ICP备17029012号-4 | 版权所有©鲍亚龙 |免责声明  | GIF图库  | NUXT版