Java 8 Streams:根据不同的属性多次映射同一个对象 [英] Java 8 Streams: Map the same object multiple times based on different properties

问题描述

我的一位同事向我提出了一个有趣的问题，我无法找到一个简洁漂亮的 Java 8 解决方案.问题是流过一个 POJO 列表，然后根据多个属性将它们收集到一个映射中——映射导致 POJO 出现多次

I was presented with an interesting problem by a colleague of mine and I was unable to find a neat and pretty Java 8 solution. The problem is to stream through a list of POJOs and then collect them in a map based on multiple properties - the mapping causes the POJO to occur multiple times

想象以下 POJO:

private static class Customer {
    public String first;
    public String last;

    public Customer(String first, String last) {
        this.first = first;
        this.last = last;
    }

    public String toString() {
        return "Customer(" + first + " " + last + ")";
    }
}

将其设置为List:

// The list of customers
List<Customer> customers = Arrays.asList(
        new Customer("Johnny", "Puma"),
        new Customer("Super", "Mac"));

备选方案 1:在流"之外(或者更确切地说是在 forEach 之外)使用 Map.

Alternative 1: Use a Map outside of the "stream" (or rather outside forEach).

// Alt 1: not pretty since the resulting map is "outside" of
// the stream. If parallel streams are used it must be
// ConcurrentHashMap
Map<String, Customer> res1 = new HashMap<>();
customers.stream().forEach(c -> {
    res1.put(c.first, c);
    res1.put(c.last, c);
});

备选方案 2:创建地图条目并流式传输它们，然后 flatMap 它们.IMO 有点过于冗长，不太容易阅读.

Alternative 2: Create map entries and stream them, then flatMap them. IMO it is a bit too verbose and not so easy to read.

// Alt 2: A bit verbose and "new AbstractMap.SimpleEntry" feels as
// a "hard" dependency to AbstractMap
Map<String, Customer> res2 =
        customers.stream()
                .map(p -> {
                    Map.Entry<String, Customer> firstEntry = new AbstractMap.SimpleEntry<>(p.first, p);
                    Map.Entry<String, Customer> lastEntry = new AbstractMap.SimpleEntry<>(p.last, p);
                    return Stream.of(firstEntry, lastEntry);
                })
                .flatMap(Function.identity())
                .collect(Collectors.toMap(
                        Map.Entry::getKey, Map.Entry::getValue));

替代方案 3:这是我迄今为止提出的最漂亮"代码的另一种方案，但它使用了 reduce 的三参数版本和第三个参数在这个问题中发现有点狡猾:Java 8 函数式编程中reduce"函数的第三个参数的目的.此外，reduce 似乎不太适合这个问题，因为它正在发生变化，并且并行流可能不适用于下面的方法.

Alternative 3: This is another one that I came up with the "prettiest" code so far but it uses the three-arg version of reduce and the third parameter is a bit dodgy as found in this question: Purpose of third argument to 'reduce' function in Java 8 functional programming. Furthermore, reduce does not seem like a good fit for this problem since it is mutating and parallel streams may not work with the approach below.

// Alt 3: using reduce. Not so pretty
Map<String, Customer> res3 = customers.stream().reduce(
        new HashMap<>(),
        (m, p) -> {
            m.put(p.first, p);
            m.put(p.last, p);
            return m;
        }, (m1, m2) -> m2 /* <- NOT USED UNLESS PARALLEL */);

如果上面的代码是这样打印的:

If the above code is printed like this:

System.out.println(res1);
System.out.println(res2);
System.out.println(res3);

结果是:

{Super=Customer(Super Mac), Johnny=Customer(Johnny Puma), Mac=Customer(Super Mac), Puma=Customer(Johnny Puma)}
{Super=Customer(Super Mac), Johnny=Customer(Johnny Puma), Mac=Customer(Super Mac), Puma=Customer(Johnny Puma)}
{Super=Customer(Super Mac), Johnny=Customer(Johnny Puma), Mac=Customer(Super Mac), Puma=Customer(Johnny Puma)}

{Super=Customer(Super Mac), Johnny=Customer(Johnny Puma), Mac=Customer(Super Mac), Puma=Customer(Johnny Puma)}
{Super=Customer(Super Mac), Johnny=Customer(Johnny Puma), Mac=Customer(Super Mac), Puma=Customer(Johnny Puma)}
{Super=Customer(Super Mac), Johnny=Customer(Johnny Puma), Mac=Customer(Super Mac), Puma=Customer(Johnny Puma)}

那么，现在我的问题是:我应该如何以 Java 8 有序的方式流过 List，然后以某种方式将其收集为 Map; 将整个事物拆分为两个键(first AND last)，即 Customer 被映射两次.我不想使用任何 3rd 方库，我不想像 alt 1 那样在流之外使用地图.还有其他不错的选择吗?

So, now to my question: How should I, in a Java 8 orderly fashion, stream through the List<Customer> and then somehow collect it as a Map<String, Customer> where you split the whole thing as two keys (first AND last) i.e. the Customer is mapped twice. I do not want to use any 3rd party libraries, I do not want to use a map outside of the stream as in alt 1. Are there any other nice alternatives?

完整代码可以在在hastebin 上找到 以进行简单的复制粘贴以让整个程序运行.

The full code can be found on hastebin for simple copy-paste to get the whole thing running.

推荐答案

我认为您的备选方案 2 和 3 可以重新编写以更加清晰:

I think your alternatives 2 and 3 can be re-written to be more clear:

备选方案 2:

Map<String, Customer> res2 = customers.stream()
    .flatMap(
        c -> Stream.of(c.first, c.last)
        .map(k -> new AbstractMap.SimpleImmutableEntry<>(k, c))
    ).collect(toMap(Map.Entry::getKey, Map.Entry::getValue));

备选方案 3:您的代码通过改变 HashMap 来滥用 reduce.要进行可变归约，请使用 collect:

Alternative 3: Your code abuses reduce by mutating the HashMap. To do mutable reduction, use collect:

Map<String, Customer> res3 = customers.stream()
    .collect(
        HashMap::new, 
        (m,c) -> {m.put(c.first, c); m.put(c.last, c);}, 
        HashMap::putAll
    );

请注意，这些并不相同.如果存在重复的键，则替代方案 2 将引发异常，而替代方案 3 将静默覆盖条目.

Note that these are not identical. Alternative 2 will throw an exception if there are duplicate keys while Alternative 3 will silently overwrite the entries.

如果在重复键的情况下覆盖条目是你想要的，我个人更喜欢备选方案 3.我很清楚它的作用.它最类似于迭代解决方案.我希望它的性能更高，因为替代方案 2 必须使用所有平面映射为每个客户进行大量分配.

If overwriting entries in case of duplicate keys is what you want, I would personally prefer Alternative 3. It is immediately clear to me what it does. It most closely resembles the iterative solution. I would expect it to be more performant as Alternative 2 has to do a bunch of allocations per customer with all that flatmapping.

然而，替代方案 2 比替代方案 3 具有巨大的优势，因为它将条目的产生与其聚合分开.这为您提供了很大的灵活性.例如，如果您想更改备选方案 2 以覆盖重复键上的条目而不是抛出异常，您只需添加 (a,b) ->b 到 toMap(...).如果您决定要将匹配的条目收集到列表中，您所要做的就是将 toMap(...) 替换为 groupingBy(...) 等.

However, Alternative 2 has a huge advantage over Alternative 3 by separating the production of entries from their aggregation. This gives you a great deal of flexibility. For example, if you want to change Alternative 2 to overwrite entries on duplicate keys instead of throwing an exception, you would simply add (a,b) -> b to toMap(...). If you decide you want to collect matching entries into a list, all you would have to do is replace toMap(...) with groupingBy(...), etc.

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