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浅谈Java之Map 按值排序 (Map sort by value)

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:

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:

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:

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:

typedef pair<string, int> PAIR;

int cmp(const PAIR& x, const PAIR& y)
{
  return x.second > y.second;
}

map<string,int> m;
vector<PAIR> vec;
for (map<wstring,int>::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:

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<Map.Entry<String, Integer>> treeSet = new TreeSet<Map.Entry<String, Integer>>(
        new Comparator<Map.Entry<String, Integer>>() {
          public int compare(Map.Entry<String, Integer> o1,
              Map.Entry<String, Integer> 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),

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:

 

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)]

以上这篇浅谈Java之Map 按值排序 (Map sort by value)就是小编分享给大家的全部内容了,希望能给大家一个参考,也希望大家多多支持。

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