如何在Java 8中组合/合并两个列表映射(映射<;字符串、映射<;枚举、列表<;字符串>;)
鉴于: 包含水果的地图如何在Java 8中组合/合并两个列表映射(映射<;字符串、映射<;枚举、列表<;字符串>;),java,collections,java-8,java-stream,Java,Collections,Java 8,Java Stream,鉴于: 包含水果的地图 enum Food{ FRUITS, VEGGIES; } Map<String, Map<Food, List<String>>> fruitBasket= new HashMap<>(); fruitBasket.put("basket1", Collections.singletonMap(Food.FRUITS, Arrays.asList("apple","banana"))); fruitBasket.put(
enum Food{
FRUITS, VEGGIES;
}
Map<String, Map<Food, List<String>>> fruitBasket= new HashMap<>();
fruitBasket.put("basket1", Collections.singletonMap(Food.FRUITS, Arrays.asList("apple","banana")));
fruitBasket.put("basket2", Collections.singletonMap(Food.FRUITS, Arrays.asList"orange", "kiwi")));
fruitBasket.put("basket3", Collections.singletonMap(Food.FRUITS, Arrays.asList("banana", "orange")));
fruitBasket:
[
basket1, [Food.FRUITS, {"apple", "banana"}],
basket2, [Food.FRUITS, {"orange", "kiwi"}],
basket3, [Food.FRUITS, {"banana", "orange"}]
]
我的解决方案:
Solution 1:
Map<String, Map<Food, List<String>>> groceryBasket= new HashMap<>();
grocery basket = Stream.concat(fruitBasket.entrySet().stream(), veggieBasket.entrySet().stream())
.collect(Collectors.toMap(Map.Entry::getKey, Map.Entry::getValue, (fruitList, veggieList ) ->
{
final List<String> groceryList = new ArrayList<>();
groceryList .addAll(fruitList);
groceryList .addAll(veggieList);
return groceryList;
}));
Solution 2:
Map<String, Map<Food, List<String>>> groceryBasket= new HashMap<>();
grocery basket = Stream.concat(fruitBasket.entrySet().stream(), veggieBasket.entrySet().stream())
.collect(Collectors.toMap(Map.Entry::getKey, Map.Entry::getValue, (fruitList, veggieList ) ->
{
return Stream.of(fruitList, veggieList).flatMap(x -> x.stream()).collect(Collectors.toList());
}));
解决方案1:
Map groceryBasket=newhashmap();
杂货篮=Stream.concat(fruitBasket.entrySet().Stream(),vegiebasket.entrySet().Stream())
.collect(Collectors.toMap(Map.Entry::getKey,Map.Entry::getValue,(水果列表,蔬菜列表)->
{
最终列表groceryList=新的ArrayList();
杂货店名录(水果名录);
杂货店名录(蔬菜名录);
退货杂货店;
}));
解决方案2:
Map groceryBasket=newhashmap();
杂货篮=Stream.concat(fruitBasket.entrySet().Stream(),vegiebasket.entrySet().Stream())
.collect(Collectors.toMap(Map.Entry::getKey,Map.Entry::getValue,(水果列表,蔬菜列表)->
{
返回Stream.of(水果列表,蔬菜列表).flatMap(x->x.Stream()).collect(Collectors.toList());
}));
我尝试了解决方案1和解决方案2,我想知道是否有更好的/优化的方法来处理此问题?为什么不尝试将合并解析为:
Map<String, Map<Food, List<String>>> groceryBasket =
Stream.concat(fruitBasket.entrySet().stream(), veggieBasket.entrySet().stream())
.collect(Collectors.toMap(Map.Entry::getKey, Map.Entry::getValue, (a, b) -> {
Map<Food, List<String>> innerMap = new HashMap<>(a);
innerMap.putAll(b);
return innerMap;
}));
Map groceryBasket=
Stream.concat(fruitBasket.entrySet().Stream(),vegiebasket.entrySet().Stream())
.collect(Collectors.toMap(Map.Entry::getKey,Map.Entry::getValue,(a,b)->{
Map innerMap=新的HashMap(a);
innerMap.putAll(b);
返回内部映射;
}));
或者,如果内部映射是可变的,那么
Map<String, Map<Food, List<String>>> groceryBasket = Stream.concat(fruitBasket.entrySet().stream(),
veggieBasket.entrySet().stream())
.collect(Collectors.toMap(Map.Entry::getKey, Map.Entry::getValue, (a, b) -> {
a.putAll(b);
return a;
}));
Map groceryBasket=Stream.concat(fruitBasket.entrySet().Stream(),
vegiebasket.entrySet().stream())
.collect(Collectors.toMap(Map.Entry::getKey,Map.Entry::getValue,(a,b)->{
a、 普塔尔(b);
返回a;
}));
您必须创建可修改列表以合并重复密钥
Stream.concat(fruitBasket.entrySet().stream(), veggieBasket.entrySet().stream())
.collect(Collectors.toMap(Map.Entry::getKey, e -> listOf(e.getValue()),
(left, right) -> {
left.addAll(right);
return left;
}));
要创建可修改列表
public static List<Map<Food, List<String>>> listOf(Map<Food, List<String>> a) {
final ArrayList<Map<Food, List<String>>> list = new ArrayList<>();
list.add(a);
return list;
}
公共静态列表列表(映射a){
最终ArrayList=新ArrayList();
列表.添加(a);
退货清单;
}
输出:
{basket3=[{水果=[香蕉,橙子]},{蔬菜=[土豆,番茄]}
篮子2=[{水果=[橘子,猕猴桃]},{蔬菜=[洋葱,土豆]}
篮子1=[{水果=[苹果,香蕉]},{蔬菜=[番茄,洋葱]}
您可以这样做:
Map<String, Map<Food, List<String>>> groceryBasket =
Stream.concat(fruitBasket.entrySet().stream(),veggieBasket.entrySet().stream())
.collect(Collectors.toMap(Map.Entry::getKey, v -> v.getValue().entrySet().stream()
.collect(Collectors.toMap(Map.Entry::getKey, Map.Entry::getValue)),
(a, b) -> { a.putAll(b);return a; }
)
);
Map groceryBasket=
Stream.concat(fruitBasket.entrySet().Stream(),vegiebasket.entrySet().Stream())
.collect(Collectors.toMap(Map.Entry::getKey,v->v.getValue().entrySet().stream())
.collect(Collectors.toMap(Map.Entry::getKey,Map.Entry::getValue)),
(a,b)->{a.putAll(b);返回a;}
)
);
您的解决方案的返回类型是什么?因为它是一个singletonMap-不可变的
@HadiJ correct,在答案中编辑。谢谢你指出。
public static List<Map<Food, List<String>>> listOf(Map<Food, List<String>> a) {
final ArrayList<Map<Food, List<String>>> list = new ArrayList<>();
list.add(a);
return list;
}
Map<String, Map<Food, List<String>>> groceryBasket =
Stream.concat(fruitBasket.entrySet().stream(),veggieBasket.entrySet().stream())
.collect(Collectors.toMap(Map.Entry::getKey, v -> v.getValue().entrySet().stream()
.collect(Collectors.toMap(Map.Entry::getKey, Map.Entry::getValue)),
(a, b) -> { a.putAll(b);return a; }
)
);