关于java8:ListT-转-MapK-T通用方法

咱们开发过程中常常遇到把List<T>转成map对象的场景，同时须要对key值雷同的对象做个合并，lambda曾经做得很好了。
定义两个实体类别离命名为A、B。

@Dataclass A {    private String a1;    private String a2;    private String a3;    public A(String a1, String a2, String a3) {        this.a1 = a1;        this.a2 = a2;        this.a3 = a3;    }}@Dataclass B {    private String b1;    private String b2;    private String b3;    public B(String b1, String b2, String b3) {        this.b1 = b1;        this.b2 = b2;        this.b3 = b3;    }}

lambda转换代码：

@Testpublic void test1() {    List<A> aList = new ArrayList<>();    aList.add(new A("a1", "a21", "a3"));    aList.add(new A("a1", "a22", "a3"));    aList.add(new A("a11", "a23", "a3"));    aList.add(new A("a11", "a24", "a3"));    aList.add(new A("a21", "a25", "a3"));    System.out.println(aList);    Map<String, A> tagMap = CollectionUtils.isEmpty(aList) ? new HashMap<>() :            aList.stream().collect(Collectors.toMap(A::getA1, a -> a, (k1, k2) -> k1));    System.out.println("----------------------");    System.out.println(tagMap);    System.out.println("----------------------");}

能不能把转换的过程提取成公共办法呢？
我做了个尝试，公共的转换方法如下：

public <K, T> Map<K, T> convertList2Map(List<T> list, Function<T, K> function) {    Map<K, T> map = CollectionUtils.isEmpty(list) ? new HashMap<>() :            list.stream().collect(Collectors.toMap(function, a -> a, (k1, k2) -> k1));    return map;}

上面是验证过程, 别离应用空集合与有数据的汇合做比照：

@Testpublic void test1() {    List<A> aList = new ArrayList<>();    System.out.println(aList);    Map<String, A> tagMap = CollectionUtils.isEmpty(aList) ? new HashMap<>() :            aList.stream().collect(Collectors.toMap(A::getA1, a -> a, (k1, k2) -> k1));    System.out.println("----------------------");    System.out.println(tagMap);    System.out.println("----------------------");    Map<String, A> tagMap1 = convertList2Map(aList, A::getA1);;    System.out.println(tagMap1);    List<B> bList = new ArrayList<>();    bList.add(new B("b1", "a21", "a3"));    bList.add(new B("b1", "a22", "a3"));    bList.add(new B("b11", "a23", "a3"));    bList.add(new B("b11", "a24", "a3"));    bList.add(new B("b21", "a25", "a3"));    System.out.println("----------------------");    System.out.println(bList);    Map<String, B> bMap = CollectionUtils.isEmpty(bList) ? new HashMap<>() :            bList.stream().collect(Collectors.toMap(B::getB1, a -> a, (k1, k2) -> k1));    System.out.println("----------bMap------------");    System.out.println(bMap);    Map<String, B> bMap1 = convertList2Map(bList, B::getB1);    System.out.println("----------bMap1------------");    System.out.println(bMap1);}

通过比对两种转换形式的打印后果，论断是通用的转换方法是可行的。效率上没有进步，就是代码会简短一点，码农不必记那么多代码了！