Map是鍵值對的集合,又叫作字典或關聯數組等,是最常見的資料結構之一。在java如何讓一個map按value排序呢? 看似簡單,但卻不容易!
例如,Map中key是String類型,表示一個單詞,而value是int型,表示該單字出現的次數,現在我們想要按照單字出現的次數來排序:
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);
按值排序的結果應該是:
key value me 1000 you 3000 and 4000 hungry 5000 later 6000 food 10000
首先,不能採用SortedMap結構,因為SortedMap是按鍵排序的Map,而不是值排序的Map,我們要的是值排序的Map。
Couldn't you do this with a SortedMap?
No, because the map are being sorted by its keys.
C++程式碼:
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; } }
上面方法的實質意義是:將Map結構中的鍵值對(Map.Entry)封裝成一個自訂的類別(結構),或直接使用Map.Entry類別。自訂類別知道自己應該如何排序,也就是按值排序,具體為自己實作Comparable介面或建構一個Comparator對象,然後不用Map結構而採用有序集合(SortedSet, TreeSet是SortedSet的一種實作),這樣就實現了Map中sort by value要達到的目的。是說,不用Map,而是把Map.Entry當作一個對象,這樣問題變成實現一個該對象的有序集合或對該對象的集合做排序。既可以用SortedSet,這樣插入完成後自然就是有序的了,又或者用一個List或數組,然後再對其做排序(Collections.sort() or Arrays.sort())。
Encapsulate the information in its own class. Either implementComparable and write rules for the natural ordering or write a
Comparator based 是.方法二:
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);
另外,Groovy 實現sort map by value,用groovy 中map 的sort 方法(需groovy 1.6),
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]
如:
["a":3,"b":1,"c":4,"d":2]. { a,b -> a.value - b.value }
Python中也類似:
def result = map.sort(){ a, b -> b.value.compareTo(a.value) }
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