java中的接口是类吗
221
2023-07-09
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
int cmp(const PAIR& x, const PAIR& y)
{
return x.second > y.second;
}
map
vector
for (map
{
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());
http:// 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.kIipxnext();
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
new Comparator
public int compare(Map.Entry
Map.Entry
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)]
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