使用Java 8即时创建带有复合键的地图 [英] Create a map with composite key with Java 8 on the fly

问题描述

我想创建一个如下图所示的地图->

I want to create a map like below->

Map<Pair<MyClass.a, MyClass.b>, MyClass>>.

我有一个对象列表->

I have a list of object ->

List<MyClass>

Pair是我的项目中已经存在的一类，所以我想使用它.

Here Pair is a class, already in my project, so I wanted to use it.

我需要帮助才能从Java 8流中创建它.

I need help to create it from a Java 8 stream.

我确实尝试了::

ls.stream().collect(Collectors.toMap(new Pair(MyClass.a, MyClass.b), MyClass));

但是我遇到一个错误.我是Java 8的新手，正在尝试学习它.

But I am getting an error. I am new to Java 8 and trying to learn it.

添加示例:

class Person {
   String name ;
   int age ;
   // Some other variables
}

我有一个List<Person>列表.

根据我的要求，我需要使用配对类的键= {name，age}.

In my requirement I need a key = {name, age}, using the pair class.

class Pair<T,U> {     
    Pair(T t, U u) {
        this.t = t
        this.u = u
    }

    // Overridden hashCode && equals methods
}

现在我想创建一个像Map<Pair<String, Int>, Person>

我遇到了一个编译器错误，提示不是功能接口".

I was getting a compiler error that said "Not a functional interface".

我确定必须有一种通过java 8流和收集的方法.

I am sure there must be a way via java 8 stream and collect.

推荐答案

要创建函数，必须使用 lambda表达式.编写new Pair(MyClass.a, MyClass.b)之类的表达式还不够，相反，您需要指定一个带有参数的函数，该函数将是Person实例，即p -> new Pair<>(p.name, p.age).或者，您可以使参数明确:(Person p) -> new Pair<>(p.name, p.age).

In order to create a function, you must use a lambda expression. It’s not sufficient an write an expression like new Pair(MyClass.a, MyClass.b), instead, you specify a function having a parameter, that will be a Person instance, i.e. p -> new Pair<>(p.name, p.age). Alternatively you may make the parameter explicit: (Person p) -> new Pair<>(p.name, p.age).

对于Map创建操作，您必须确定所需的内容.例如，

For your Map creation operation, you have to decide, what you want. E.g.,

Map<Pair<String, Integer>, List<Person>> map
    = list.stream().collect(Collectors.groupingBy(p -> new Pair<>(p.name, p.age)));

会将每个键映射到具有该名称/年龄组合的所有Person实例的列表.

will map each key to a list of all Person instances having that name/age combination.

相反

Map<Pair<String, Integer>, Person> map
    = list.stream().collect(Collectors.toMap(p -> new Pair<>(p.name, p.age), p -> p));

会将名称/年龄对映射到单个Person实例，但如果有多个具有相同密钥的实例，则会引发异常.您可以指定一个函数来解决此类冲突，例如

will map the name/age pairs to a single Person instance, but throw an exception, if there is more than one with the same key. You can specify a function to resolve such conflicts, e.g.

Map<Pair<String, Integer>, Person> map = list.stream().collect(
    Collectors.toMap(p -> new Pair<>(p.name, p.age), p -> p, (first, next) -> first));

将保留第一个，而

Map<Pair<String, Integer>, Person> map = list.stream().collect(
    Collectors.toMap(p -> new Pair<>(p.name, p.age), p -> p, (prev, last) -> last));

将覆盖先前的出现，以每个名称/年龄组合的最后一个Person实例结束.

will overwrite the previous occurrence, ending up with the last Person instance for each name/age combination.

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