Java8 List转Map（List归约到Map）

喜欢ヅ旅行 2024-03-23 11:21 143阅读 0赞

#### 目录标题 ####

*   *  \`List<Dog> → Map<name, List<Dog>>\`
     *  \`List<Dog> → Map<name, List<Age>>\`
     *  \`List<Dog> → Map<name, Dog>\`
     *  \`List<Dog>\` → \`Map<name, age>\`

### `List<Dog> → Map<name, List<Dog>>` ###

Map<String, List<Dog>> collect1 = list.stream().collect(Collectors.groupingBy(Dog::getName));
    // list为null     →  NPE
    // list为empty    →  {}
    // model存在null  →  NPE
    // key存在null    →  NPE
    // {Tom=[Dog(name=Tom, age=18), Dog(name=Tom, age=19)], John=[Dog(name=John, age=18)]}
    
    综上,最安全方式
    Map<String, List<Dog>> collect1 = list.stream().filter(a -> a != null && a.getName() != null).collect(Collectors.groupingBy(Dog::getName));
    //{Tom=[Dog(name=Tom, age=18), Dog(name=Tom, age=19)], John=[Dog(name=John, age=18)]}

Map<String, List<Dog>> collect2 = list.stream().collect(Collectors.groupingBy(Dog::getName, HashMap::new, Collectors.toList()));
    // list为null     →  NPE
    // list为empty    →  {}
    // model存在null  →  NPE
    // key存在null    →  NPE
    // {Tom=[Dog(name=Tom, age=18), Dog(name=Tom, age=19)], John=[Dog(name=John, age=18)]}
    
    综上,最安全方式
    Map<String, List<Dog>> collect2 = list.stream().filter(a -> a != null && a.getName() != null).collect(Collectors.groupingBy(Dog::getName, HashMap::new, Collectors.toList()));
    // {Tom=[Dog(name=Tom, age=18), Dog(name=Tom, age=19)], John=[Dog(name=John, age=18)]}

Map<String, List<Dog>> collect3 = list.stream().collect(Collectors.groupingBy(Dog::getName, Collectors.mapping(Function.identity(), Collectors.toList())));
    // list为null     →  NPE
    // list为empty    →  {}
    // model存在null  →  NPE
    // key存在null    →  NPE
    // {Tom=[Dog(name=Tom, age=18), Dog(name=Tom, age=19)], John=[Dog(name=John, age=18)]}
    
    综上,最安全方式
    Map<String, List<Dog>> collect3 = list.stream().filter(a -> a != null && a.getName() != null).collect(Collectors.groupingBy(Dog::getName, Collectors.mapping(Function.identity(), Collectors.toList())));
    // {Tom=[Dog(name=Tom, age=18), Dog(name=Tom, age=19)], John=[Dog(name=John, age=18)]}

### `List<Dog> → Map<name, List<Age>>` ###

Map<String, List<Integer>> res = list.stream().collect(Collectors.groupingBy(Dog::getName, Collectors.mapping(Dog::getAge, Collectors.toList())));
    
    // list为null           →  NPE
    // list为empty          →  {}
    // model存在null        →  NPE
    // key存在null          →  NPE
    // value为null依旧收集   →  {Tom=[18, 18], John=[18, null]}
    // value重复收集不去重   →  {Tom=[18, 18], John=[18, null]}
    
    综上,最安全方式
    Map<String, List<Integer>> collect4 = list.stream().filter(a -> a != null && a.getName() != null).collect(Collectors.groupingBy(Dog::getName, Collectors.mapping(Dog::getAge, Collectors.toList())));
    // {Tom=[18, 18], John=[18, null]}

### `List<Dog> → Map<name, Dog>` ###

Map<String, Dog> res = list.stream().collect(Collectors.toMap(Dog::getName, Function.identity()));
    
    // list为null             →  NPE
    // list为empty            →  {}
    // model存在null          →  NPE
    // key为null则null作为key  →  {null=Dog(name=null, age=20), Tom=Dog(name=Tom, age=18)}
    // value为null依旧收集     →  {Tom=Dog(name=Tom, age=18), John=Dog(name=John, age=null)}
    // Key重复报错(包括null)   →  IllegalStateException: Duplicate key
    // value重复收集不去重     →  {Tom=Dog(name=Tom, age=18), John=Dog(name=John, age=18)}
    
    Map<String, Dog> collect8 = list.stream().collect(Collectors.toMap(Dog::getName, Function.identity(), (existing, replacement) -> replacement));
    // Key重复时覆盖(e1, e2) -> e2      →  {null=Dog(name=null, age=21), Tom=Dog(name=Tom, age=22), John=Dog(name=John, age=18)}
    
    Map<String, Dog> collect9 = list.stream().collect(Collectors.toMap(Dog::getName, Function.identity(), (existing, replacement) -> existing));
    // Key重复时丢弃(e1, e2) -> e1      →  {null=Dog(name=null, age=20), Tom=Dog(name=Tom, age=18), John=Dog(name=John, age=18)}

### `List<Dog>` → `Map<name, age>` ###

Map<String, Integer> res = list.stream().collect(Collectors.toMap(Dog::getName, Dog::getAge));
    
    // list为null             →  NPE
    // list为empty            →  {}
    // model存在null          →  NPE
    // key为null则null作为key  →  {null=20, John=18}
    // value为null            →  NPE
    // Key重复报错             →  IllegalStateException: Duplicate key
    // value重复收集不去重      →  {Tom=18, John=18}
    
    Map<String, Integer> collect9 = list.stream().collect(Collectors.toMap(Dog::getName, Dog::getAge, (existing, replacement) -> replacement));
    // Key重复时覆盖(e1, e2) -> e2      →  {Tom=23, John=22}
    
    Map<String, Integer> collect9 = list.stream().collect(Collectors.toMap(Dog::getName, Dog::getAge, (existing, replacement) -> existing));
    // Key重复时丢弃(e1, e2) -> e1      →  {Tom=22, John=22}
    
    综上,最安全方式
    Map<String, Integer> res = list.stream().filter(a -> a != null && a.getAge() != null).collect(Collectors.toMap(Dog::getName, Dog::getAge, (existing, replacement) -> replacement));
    // {null=22, Tom=23, John=22}

Map<String, Integer> res = list.stream().collect(HashMap::new, (map1, per) -> map1.put(per.getName(), per.getAge()), HashMap::putAll);per.getAge()), HashMap::putAll);
    
    // list为null                      →  NPE
    // list为empty                     →  {}
    // model存在null                   →  NPE
    // key为null则null作为key           →  {null=20, John=18}
    // value存在null则null作为value     →  {Tom=18, John=null}
    // Key重复时value被后面的值替换      →  {Tom=22}
    // value重复收集不去重              →  {Tom=18, John=18}