展平地图<整数，列表<字符串>>映射<字符串，整数>带流和 lambda [英] Flatten a Map<Integer, List<String>> to Map<String, Integer> with stream and lambda

问题描述

我想展平一个 Map，它将 Integer 键关联到 String 列表，而不会丢失键映射.我很好奇，好像使用 stream 和 lambda 这样做是可能和有用的.

I would like to flatten a Map which associates an Integer key to a list of String, without losing the key mapping. I am curious as though it is possible and useful to do so with stream and lambda.

我们从这样的开始:

Map<Integer, List<String>> mapFrom = new HashMap<>();

让我们假设 mapFrom 填充在某个地方，看起来像:

Let's assume that mapFrom is populated somewhere, and looks like:

1: a,b,c
2: d,e,f
etc.

我们还假设列表中的值是唯一的.

Let's also assume that the values in the lists are unique.

现在，我想展开"它以获得第二张地图，例如:

Now, I want to "unfold" it to get a second map like:

a: 1
b: 1
c: 1
d: 2
e: 2
f: 2
etc.

我可以这样做(或者非常类似，使用 foreach):

I could do it like this (or very similarly, using foreach):

Map<String, Integer> mapTo = new HashMap<>();
for (Map.Entry<Integer, List<String>> entry: mapFrom.entrySet()) {
    for (String s: entry.getValue()) {
        mapTo.put(s, entry.getKey());
    }
}

现在让我们假设我想使用 lambda 而不是嵌套的 for 循环.我可能会这样做:

Now let's assume that I want to use lambda instead of nested for loops. I would probably do something like this:

Map<String, Integer> mapTo = mapFrom.entrySet().stream().map(e -> {
    e.getValue().stream().?
    // Here I can iterate on each List, 
    // but my best try would only give me a flat map for each key, 
    // that I wouldn't know how to flatten.
}).collect(Collectors.toMap(/*A String value*/,/*An Integer key*/))

我也尝试了flatMap，但我认为这不是正确的方法，因为虽然它可以帮助我摆脱维度问题，但我失去了关键过程.

I also gave a try to flatMap, but I don't think that it is the right way to go, because although it helps me get rid of the dimensionality issue, I lose the key in the process.

简而言之，我的两个问题是:

In a nutshell, my two questions are :

• 是否可以使用 streams 和 lambda 来实现?
• 这样做是否有用(性能、可读性)?

推荐答案

您需要使用 flatMap 将值扁平化到一个新的流中，但是由于您仍然需要原始键来收集到一个Map，你必须映射到一个持有key和value的临时对象，例如

You need to use flatMap to flatten the values into a new stream, but since you still need the original keys for collecting into a Map, you have to map to a temporary object holding key and value, e.g.

Map<String, Integer> mapTo = mapFrom.entrySet().stream()
       .flatMap(e->e.getValue().stream()
                    .map(v->new AbstractMap.SimpleImmutableEntry<>(e.getKey(), v)))
       .collect(Collectors.toMap(Map.Entry::getValue, Map.Entry::getKey));

Map.Entry 是不存在的元组类型的替代，任何其他能够容纳两个不同类型对象的类型都足够了.

The Map.Entry is a stand-in for the nonexistent tuple type, any other type capable of holding two objects of different type is sufficient.

另一种不需要这些临时对象的方法是自定义收集器:

An alternative not requiring these temporary objects, is a custom collector:

Map<String, Integer> mapTo = mapFrom.entrySet().stream().collect(
    HashMap::new, (m,e)->e.getValue().forEach(v->m.put(v, e.getKey())), Map::putAll);

这与 toMap 的不同之处在于静默覆盖重复键，而没有合并函数的 toMap 如果存在重复键，则会抛出异常.基本上，这个自定义收集器是一个具有并行能力的变体

This differs from toMap in overwriting duplicate keys silently, whereas toMap without a merger function will throw an exception, if there is a duplicate key. Basically, this custom collector is a parallel capable variant of

Map<String, Integer> mapTo = new HashMap<>();
mapFrom.forEach((k, l) -> l.forEach(v -> mapTo.put(v, k)));

但请注意，即使输入图非常大，此任务也不会从并行处理中受益.只有当流管道中有额外的计算密集型任务可以从 SMP 中受益时，才有机会从并行流中受益.因此，简洁、顺序的 Collection API 解决方案或许更可取.

But note that this task wouldn’t benefit from parallel processing, even with a very large input map. Only if there were additional computational intense task within the stream pipeline that could benefit from SMP, there was a chance of getting a benefit from parallel streams. So perhaps, the concise, sequential Collection API solution is preferable.

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