Java - Collectors.toMap() 使用

news2024/12/25 9:28:21

List 转 Map

一、Collectors.toMap() 函数源码定义：

	// 两个参数
    public static <T, K, U>
    Collector<T, ?, Map<K,U>> toMap(Function<? super T, ? extends K> keyMapper,
                                    Function<? super T, ? extends U> valueMapper) {
        return toMap(keyMapper, valueMapper, throwingMerger(), HashMap::new);
    }
	
	// 三个参数
    public static <T, K, U>
    Collector<T, ?, Map<K,U>> toMap(Function<? super T, ? extends K> keyMapper,
                                    Function<? super T, ? extends U> valueMapper,
                                    BinaryOperator<U> mergeFunction) {
        return toMap(keyMapper, valueMapper, mergeFunction, HashMap::new);
    }
    
    // 四个参数
    public static <T, K, U, M extends Map<K, U>>
    Collector<T, ?, M> toMap(Function<? super T, ? extends K> keyMapper,
                                Function<? super T, ? extends U> valueMapper,
                                BinaryOperator<U> mergeFunction,
                                Supplier<M> mapSupplier) {
        BiConsumer<M, T> accumulator
                = (map, element) -> map.merge(keyMapper.apply(element),
                                              valueMapper.apply(element), mergeFunction);
        return new CollectorImpl<>(mapSupplier, accumulator, mapMerger(mergeFunction), CH_ID);
    }

参数含义分别是：

keyMapper：Key 的映射函数。
valueMapper：Value 的映射函数。
mergeFunction：当 Key 冲突时，调用的合并方法。
mapSupplier：Map 构造器，在需要返回特定的 Map 时使用。

二、使用方法

List<People> arrayList = new ArrayList<>();
arrayList.add(new People("张三", "上海", "演员"));
arrayList.add(new People("李四", "北京", "IT"));
arrayList.add(new People("王五", "深圳", "医生"));
arrayList.add(new People("张三", "成都", "律师"));

1.name作为key，整个People对象作为value

Map<String, People> map = arrayList.stream().collect(Collectors.toMap(People::getName, a -> a, (oldV, newV) -> newV));
System.out.println(map);

// 运行结果
{李四=People{name='李四', city='北京', job='IT'}, 张三=People{name='张三', city='成都', job='律师'}, 王五=People{name='王五', city='深圳', job='医生'}}

这里使用传入 mergeFunction 参数，当 Key 冲突时，调用的合并方法。当Key出现重时如果想使用新值使用 (oldV, newV) -> newV，如果想使用老值使用 (oldV, newV) -> oldV。

如果不传 mergeFunction 参数，结果会报错：

Map<String, People> map = arrayList.stream().collect(Collectors.toMap(People::getName, a -> a));

// 报错结果
Exception in thread "main" java.lang.IllegalStateException: Duplicate key People{name='张三', city='上海', job='演员'}
	at java.util.stream.Collectors.lambda$throwingMerger$0(Collectors.java:133)
	at java.util.HashMap.merge(HashMap.java:1255)
	at java.util.stream.Collectors.lambda$toMap$58(Collectors.java:1320)
	at java.util.stream.ReduceOps$3ReducingSink.accept(ReduceOps.java:169)
	at java.util.ArrayList$ArrayListSpliterator.forEachRemaining(ArrayList.java:1384)
	at java.util.stream.AbstractPipeline.copyInto(AbstractPipeline.java:482)
	at java.util.stream.AbstractPipeline.wrapAndCopyInto(AbstractPipeline.java:472)
	at java.util.stream.ReduceOps$ReduceOp.evaluateSequential(ReduceOps.java:708)
	at java.util.stream.AbstractPipeline.evaluate(AbstractPipeline.java:234)
	at java.util.stream.ReferencePipeline.collect(ReferencePipeline.java:499)
	at org.example.Main.collectorsToMap(Main.java:132)
	at org.example.Main.main(Main.java:69)

所以除非你能确保Key不会出现重复（这个很难保证），否则就需要传入 mergeFunction 参数，确定value合并策略。

2.name作为key，整个job作为value

Map<String, String> map3 = arrayList.stream().collect(Collectors.toMap(People::getName, People::getJob, (oldV, newV) -> newV));
System.out.println(map3);

// 运行结果
{李四=IT, 张三=律师, 王五=医生}

如果我们将数据改下，再加一条数据

arrayList.add(new People("张三", "云南", null));

Map<String, String> map3 = arrayList.stream().collect(Collectors.toMap(People::getName, People::getJob, (oldV, newV) -> newV));

// 运行报错
Exception in thread "main" java.lang.NullPointerException
	at java.util.HashMap.merge(HashMap.java:1226)
	at java.util.stream.Collectors.lambda$toMap$58(Collectors.java:1320)
	at java.util.stream.ReduceOps$3ReducingSink.accept(ReduceOps.java:169)
	at java.util.ArrayList$ArrayListSpliterator.forEachRemaining(ArrayList.java:1384)
	at java.util.stream.AbstractPipeline.copyInto(AbstractPipeline.java:482)
	at java.util.stream.AbstractPipeline.wrapAndCopyInto(AbstractPipeline.java:472)
	at java.util.stream.ReduceOps$ReduceOp.evaluateSequential(ReduceOps.java:708)
	at java.util.stream.AbstractPipeline.evaluate(AbstractPipeline.java:234)
	at java.util.stream.ReferencePipeline.collect(ReferencePipeline.java:499)
	at org.example.Main.collectorsToMap(Main.java:135)
	at org.example.Main.main(Main.java:69)

我们要确保value不能null，所以需要加安全判断：

Map<String, String> map3 = arrayList.stream().collect(Collectors.toMap(People::getName, a -> a.getJob() == null ? "" : a.getJob(), (oldV, newV) -> newV));

3.假如存在name重复，两个vaue可以这样映射到同一个name

Map<String, String> map3 = arrayList.stream().collect(Collectors.toMap(People::getName, a -> a.getJob() == null ? "" : a.getJob(), (oldV, newV) -> (newV + "," + oldV)));
System.out.println(map3);
       
// 运行结果
{李四=IT, 张三=律师,演员, 王五=医生}

4.把People集合按照group分组到map中

需要注意的是group分组后value对应的是个数组

Map<String, List<People>> map = arrayList.stream().collect(Collectors.groupingBy(People::getName));
System.out.println(map);

// 运行结果
{李四=[People{name='李四', city='北京', job='IT'}], 张三=[People{name='张三', city='上海', job='演员'}, People{name='张三', city='成都', job='律师'}], 王五=[People{name='王五', city='深圳', job='医生'}]}