按照value排序（倒序），取前N位
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public class RelationSpanSorter {
private RelationSpanSorter() {
}
// test
public static void main(String[] args) {
Map<Integer, Integer> map = new HashMap<Integer, Integer>();
map.put(1, 4);
map.put(5, 1);
map.put(2, 3);
map.put(4, 2);
map.put(12, 12);
map.put(21, 21);
map.put(11, 11);
map.put(41, 41);
map.put(31, 31);
map.put(15, 15);
map.put(62, 62);
map.put(14, 14);
map.put(111, 111);
map.put(523, 523);
map.put(92, 92);
map.put(40, 40);
Map<Integer, Integer> sorted = sortByValue(map, 8);
System.out.println(sorted);
}
@SuppressWarnings("unchecked")
public static Map<Integer, Integer> sortByValue(Map<Integer, Integer> map, int topN) {
List<Object> list = new LinkedList<Object>(map.entrySet());
Collections.sort(list, new Comparator<Object>() {
public int compare(Object o1, Object o2) {
return -((Comparable<Integer>) ((Map.Entry<Integer, Integer>) (o1)).getValue()).compareTo(((Map.Entry<Integer, Integer>) (o2)).getValue());
}
});
Map<Integer, Integer> result = new LinkedHashMap<Integer, Integer>();
int i = 0;
for (Object it : list) {
Map.Entry<Integer, Integer> entry = (Map.Entry<Integer, Integer>) it;
if (i >= topN) {
break;
}
result.put(entry.getKey(), entry.getValue());
i++;
}
return result;
}
}
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以上代码改编自：http://blog.csdn.net/srjklssj/article/details/6324880
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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();或者还可以这样：
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}
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)]