java - 根据条件过滤并收集 Map<String, List<Object>> 为 Map<String, List<Object>>

Question

让

class Person {
    String name;
    int age;
}

我的变量是

Map<Sring, List<Person>> myVariable;

我想要得到的结果是

Map<String, List<Person>> result;

条件是收集所有满足此条件的人

Predicate<Person> predicate = p -> p.age > 20;

如果 map 中的条目包含 3 个人，而其中只有 2 人满足此条件，则应删除第 3 个人

示例:myVariable =

"1": [{name: "Foo", age: 15}, {name: "Foo", age: 13}, {name: "Foo", age: 14}]
"2": [{name: "Foo", age: 15}, {name: "Foo", age: 13}, {name: "Foo", age: 14}]
"3": [{name: "Foo", age: 15}, {name: "Foo", age: 13}, {name: "Foo", age: 14}]
"4": [{name: "Foo", age: 15}, {name: "Foo", age: 21}, {name: "Foo", age: 27}]

结果=

"4": {name: "Foo", age: 21}, {name: "Foo", age: 27}]

我试过了

Map<String, List<Persin>> result = myVariable.keySet().stream()
            .filter(p -> myVariable.keySet(p).stream().anyMatch(predicate))
            .collect(Collectors.toMap(p -> p, myVariable.get(Function.identity()).stream().filter(predicate).collect(Collectors.toList())));

但是这不起作用。

Answer 1

您可以使用两个filter调用:

Map<String, List<Person>> result = myVariable.entrySet().stream()
        .map(entry -> Map.entry(entry.getKey(), entry.getValue()
                .stream().filter(p -> p.getAge() > 20)
                .collect(Collectors.toList())))
        .filter(entry -> !entry.getValue().isEmpty())
        .collect(Collectors.toMap(Entry::getKey, Entry::getValue));

如果您使用的是 Java 8，请将 Map.entry() 替换为 new AbstractMap.SimpleEntry()。