2012-02-18 61 views
1

我一直想知道现在几天,对我的问题可能是最麻烦的方法。我有一组10个枚举类型,例如{零个,一个,两个,三个,四个,五个,六个,七个,八个,九个}和I使用这些枚举作为键在地图通过添加/减去相应值来合并地图

Map<MyEnumType, Integer> myMap; 

然后我生成100个左右的这些地图,同相同的键但不同的值。将所有这些地图添加到一个地图的最佳做法是什么?我的意思是将我生成的100张地图的所有值加起来。

回答

4

遍历枚举值,并为每个枚举值,遍历地图(反之亦然):

Map<MyEnumType, Integer> sumMap = new EnumMap<MyEnumType, Integer>(MyEnumType.class); 
for (MyEnumType e : MyEnumType.values()) { 
    int sum = 0; 
    for (Map<MyEnumType, Integer> map : maps) { 
     sum += map.get(e); // you might want a null check here 
    } 
    sumMap.put(e, sum); 
} 
1

其他选项是创建特定类总结整数,而不是以前的压倒一切的价值观。这里有一个例子它是如何与匿名类来完成:

public class MapSummer { 

    private final Map<MyEnumType, Integer> sumMap = new HashMap<MyEnumType, Integer>() { 
     @Override 
     public Integer put(MyEnumType key, Integer value) { 
      return super.put(key, get(key) + value); 
     } 

     @Override 
     public Integer get(Object key) { 
      return super.get(key) != null ? super.get(key) : 0; 
     } 
    }; 

    public Map<MyEnumType, Integer> sum(List<Map<MyEnumType, Integer>> mapList) { 
     for (Map<MyEnumType, Integer> map : mapList) { 
      sumMap.putAll(map); 
     } 
     return sumMap; 
    } 

} 

enum MyEnumType { 
    ONE, TWO, THREE, FOUR; 
} 

和单元测试:

public class MapSummerTest { 

    private final MapSummer summer = new MapSummer(); 

    @Test 
    public void shouldSumValuesInMap() { 
     final Map<MyEnumType, Integer> map1 = new HashMap<MyEnumType, Integer>() {{ 
      put(ONE, 1); 
      put(TWO, 2); 
     }}; 

     final Map<MyEnumType, Integer> map2 = new HashMap<MyEnumType, Integer>() {{ 
      put(TWO, 2); 
      put(THREE, 3); 
     }}; 

     final Map<MyEnumType, Integer> sumMap = summer.sum(Arrays.asList(map1, map2)); 

     assertThat(sumMap.get(ONE), equalTo(1)); 
     assertThat(sumMap.get(TWO), equalTo(4)); 
     assertThat(sumMap.get(THREE), equalTo(3)); 
     assertThat(sumMap.get(FOUR), equalTo(0)); 
    } 
}