从 List<Foo> 到 Map<String, List<Foo>>:寻找更好的实现

From List<Foo> to Map<String, List<Foo>>: looking for a better implementation

让我给你看我的代码:

Class 富

public class Foo {

    String code;
    String value;

    public Foo(String code, String value) {
        super();
        this.code = code;
        this.value = value;
    }
    // getters/setters
}

主要方法(关注getFooMultiMapCode()方法):

public class FooMain {

public static void main(String[] args) {
    Foo foo1 = new Foo("100","foo1");
    Foo foo2 = new Foo("200","foo2");
    Foo foo3 = new Foo("300","foo3");
    Foo foo4 = new Foo("100","foo4");
    Foo foo5 = new Foo("100","foo5");
    Foo foo6 = new Foo("200","foo6");
    List<Foo> foos = Arrays.asList(foo1,foo2,foo3,foo4,foo5,foo6);  
    Map<String,List<Foo>> fooCodeMap = getFooMultiMapCode(foos);
    System.out.println(fooCodeMap);
}

private static Map<String, List<Foo>> getFooMultiMapCode(List<Foo> foos) {
    Map<String, List<Foo>> fooMultiMapCode = new HashMap<String, List<Foo>>();
    for(Foo foo:foos){
        List<Foo> list = fooMultiMapCode.get(foo.getCode());
        if(list==null){
            list = new ArrayList<Foo>();
            list.add(foo);
            fooMultiMapCode.put(foo.getCode(), list);
        }
        else {
            list.add(foo);
        }
    }
    return fooMultiMapCode;
}
}

Main 正确打印此字符串:

{100=[foo1, foo4, foo5], 200=[foo2, foo6], 300=[foo3]}

我想以更简洁的方式重写 getFooMultiMapCode 方法,例如使用 java8 或 lambdaj、guava 等库,但我 不想更改方法签名.

Java 8 和 groupingBy(classifier, downstream) 的解决方案:

return foos.stream().collect(Collectors.groupingBy(Foo::getCode, Collectors.toList()));

或者只是注意到groupingBy(classifier) as @Boris the Spider

return foos.stream().collect(Collectors.groupingBy(Foo::getCode));

扩展一下@Holgers 的评论:非流版本如果你喜欢循环 - 仍然 java8,但是:

private static Map<String, List<Foo>> getFooMultiMapCode(List<Foo> foos) {
    Map<String, List<Foo>> fooMultiMapCode = new HashMap<String, List<Foo>>();
    for(Foo foo : foos){
        fooMultiMapCode.computeIfAbsent(foo.getCode(), x -> new ArrayList<>()).add(foo);
    }
    return fooMultiMapCode;
}

甚至

private static Map<String, List<Foo>> getFooMultiMapCode(List<Foo> foos) {
    Map<String, List<Foo>> fooMultiMapCode = new HashMap<String, List<Foo>>();
    foos.forEach(foo -> fooMultiMapCode.computeIfAbsent(foo.getCode(), x -> new ArrayList<>()).add(foo));
    return fooMultiMapCode;
}

不过我还是更喜欢流媒体版本。