JDK1.8 Stream之toMap和groupingBy全参数的使用

现在有一个商品地址对象的集合List<ProductAddress> productAddresses，该集合中productId和addressId是一对多的关系

@Data
@AllArgsConstructor
class ProductAddress {
    private int productId;

    private int addressId;

    private static List<ProductAddress> productAddresses;

    static {
        productAddresses = new ArrayList<>();
        productAddresses.add(new ProductAddress(1001, 1));
        productAddresses.add(new ProductAddress(1001, 2));
        productAddresses.add(new ProductAddress(1001, 3));
        productAddresses.add(new ProductAddress(1002, 3));
        productAddresses.add(new ProductAddress(1002, 4));
        productAddresses.add(new ProductAddress(1003, 5));
    }
}

需求

以productId分组变成一个Map<Integer,List<ProductAddress>>

传统方法

public static void main(String[] args) {
    Map<Integer, List<ProductAddress>> map = new HashMap<>();
    for (ProductAddress productAddress : productAddresses) {
        if (map.get(productAddress.getProductId()) == null) {
            map.put(productAddress.getProductId(), new ArrayList<>(Arrays.asList(productAddress)));
        } else {
            map.get(productAddress.getProductId()).add(productAddress);
        }
    }
    System.out.println(JSONObject.toJSON(map));
}

lambda表达式实现

1 public static void main(String[] args) {
    Map<Integer, List<ProductAddress>> map = productAddresses.stream().collect(Collectors.groupingBy(ProductAddress::getProductId));
    System.out.println(JSONObject.toJSON(map));
}

新的需求

以productId分组变成一个Map<Integer,List<Integer>>，value是其对应的addressId的集合

传统方法

public static void main(String[] args) {
    Map<Integer, List<Integer>> map = new HashMap<>();
    for (ProductAddress productAddress : productAddresses) {
        if (map.get(productAddress.getProductId()) == null) {
            map.put(productAddress.getProductId(), new ArrayList<>(Arrays.asList(productAddress.getAddressId())));
        } else {
            map.get(productAddress.getProductId()).add(productAddress.getAddressId());
        }
    }
    System.out.println(JSONObject.toJSON(map));
}

lambda表达式实现

首先想到的是使用toMap把value转成list并使用toMap提供的key冲突解决办法。

public static void main(String[] args) {
    Map<Integer, List<Integer>> map = productAddresses.stream()
            .collect(Collectors.groupingBy(ProductAddress::getProductId, HashMap::new,
                    Collectors.mapping(ProductAddress::getAddressId, Collectors.toList())));
    productAddresses.stream().collect(Collectors.toMap(ProductAddress::getProductId, v -> new ArrayList(Arrays.asList(v.getAddressId())), (existValue, newValue) -> {
        existValue.addAll(newValue);
        return existValue;
    }));
    System.out.println(map);
}

但感觉这个写法看起来怪怪的，琢磨groupingBy发现，使用groupingBy的后两个参数完成对象到具体字段的映射可以优雅实现

public static void main(String[] args) {
    Map<Integer, List<Integer>> map = productAddresses.stream()
            .collect(Collectors.groupingBy(ProductAddress::getProductId, HashMap::new,
                    Collectors.mapping(ProductAddress::getAddressId, Collectors.toList())));
    System.out.println(map);
}