如果我有一个用Java实现Map接口的对象,并且我希望对其中包含的每一对进行迭代,那么最有效的方法是什么?

元素的顺序是否取决于我对接口的特定映射实现?


当前回答

在Java 8中,您可以使用新的lambdas功能快速、干净地完成任务:

 Map<String,String> map = new HashMap<>();
 map.put("SomeKey", "SomeValue");
 map.forEach( (k,v) -> [do something with key and value] );

 // such as
 map.forEach( (k,v) -> System.out.println("Key: " + k + ": Value: " + v));

k和v的类型将由编译器推断,不再需要使用Map.Entry。

小菜一碟

其他回答

这里有一个泛型类型安全方法,可以调用它来转储任何给定的Map。

import java.util.Iterator;
import java.util.Map;

public class MapUtils {
    static interface ItemCallback<K, V> {
        void handler(K key, V value, Map<K, V> map);
    }

    public static <K, V> void forEach(Map<K, V> map, ItemCallback<K, V> callback) {
        Iterator<Map.Entry<K, V>> it = map.entrySet().iterator();

        while (it.hasNext()) {
            Map.Entry<K, V> entry = it.next();

            callback.handler(entry.getKey(), entry.getValue(), map);
        }
    }

    public static <K, V> void printMap(Map<K, V> map) {
        forEach(map, new ItemCallback<K, V>() {
            @Override
            public void handler(K key, V value, Map<K, V> map) {
                System.out.println(key + " = " + value);
            }
        });
    }
}

实例

下面是它的使用示例。请注意,Map的类型由该方法推断。

import java.util.*;

public class MapPrinter {
    public static void main(String[] args) {
        List<Map<?, ?>> maps = new ArrayList<Map<?, ?>>() {
            private static final long serialVersionUID = 1L;
            {
                add(new LinkedHashMap<String, Integer>() {
                    private static final long serialVersionUID = 1L;
                    {
                        put("One", 0);
                        put("Two", 1);
                        put("Three", 3);
                    }
                });

                add(new LinkedHashMap<String, Object>() {
                    private static final long serialVersionUID = 1L;
                    {
                        put("Object", new Object());
                        put("Integer", new Integer(0));
                        put("Double", new Double(0.0));
                    }
                });
            }
        };

        for (Map<?, ?> map : maps) {
            MapUtils.printMap(map);
            System.out.println();
        }
    }
}

输出

One = 0
Two = 1
Three = 3

Object = java.lang.Object@15db9742
Integer = 0
Double = 0.0

我用以下代码将地图数据复制到另一个地图:

HashMap product =(HashMap)shopping_truck.get(i);
HashMap tmp = new HashMap();
for (Iterator it = product.entrySet().iterator(); it.hasNext();) {
    Map.Entry thisEntry = (Map.Entry) it.next();
    tmp.put(thisEntry.getKey(), thisEntry.getValue());
}

是的,顺序取决于特定的Map实现。

@ScArcher2具有更优雅的Java 1.5语法。在1.4中,我会这样做:

Iterator entries = myMap.entrySet().iterator();
while (entries.hasNext()) {
  Entry thisEntry = (Entry) entries.next();
  Object key = thisEntry.getKey();
  Object value = thisEntry.getValue();
  // ...
}

在Java 1.4中尝试:

for( Iterator entries = myMap.entrySet().iterator(); entries.hasNext();){

  Entry entry = (Entry) entries.next();

  System.out.println(entry.getKey() + "/" + entry.getValue());

  //...
}

这些都是迭代HashMap的所有可能方法。

HashMap<Integer,String> map=new HashMap<Integer,String>();
    map.put(1,"David");  //Adding elements in Map
    map.put(2,"John");
    map.put(4,"Samyuktha");
    map.put(3,"jasmin");
    System.out.println("Iterating Hashmap...");

    //way 1 (java 8 Method)
    map.forEach((key, value) -> {
        System.out.println(key+" : "+ value);
    });

    //way 2 (java 7 Method)
    for(Map.Entry me : map.entrySet()){
        System.out.println(me.getKey()+" "+me.getValue());
    }

    //way 3 (Legacy way to iterate HashMap)
    Iterator iterator = map.entrySet().iterator();//map.keySet().iterator()
    while (iterator.hasNext())
    {
        Map.Entry me =(Map.Entry)iterator.next();
        System.out.println(me.getKey()+" : "+ me.getValue());
    }
    
}