从一个IntStream生成一个Map [英] Produce a Map from an IntStream of keys

问题描述

我知道我可以初始化 地图 ，然后使用流对其进行填充。例如，在这里我使用 IntStream 随机数来填充预定义地图的键。

I know I can initialize a Map, and then fill it using a stream. For example, here I use an IntStream of random numbers to populate the key of the predefined map.

int initialCapacity = 3; 
Map < Integer, UUID > map = new HashMap <>( initialCapacity );
IntStream numbers = ThreadLocalRandom.current().ints( initialCapacity );
numbers.forEach( number -> map.put( number , UUID.randomUUID() ) );

转储到控制台。

System.out.println( "map = " + map );

map = {-14054365 = ee739423-1200-45e6-80da- d167ce2e2b98，-1079671217 = ba0096fe-b32f-4ebf-a163-114fcb679992，-404094411 = f900052b-8a8d-4e66-b808-618fbc7e115f}

map = {-14054365=ee739423-1200-45e6-80da-d167ce2e2b98, -1079671217=ba0096fe-b32f-4ebf-a163-114fcb679992, -404094411=f900052b-8a8d-4e66-b808-618fbc7e115f}

➥我想知道是否有一种方法可以让流产生 Map 对象，而不是填充先前存在的地图。

➥ I wonder if there is a way to have the stream produce the Map object, rather than filling a pre-existing map.

我不在乎这是否明智。我只是对语法以及如何使用 收藏家 。

I do not care about if this is a wise approach or not. I am just curious about the syntax and how to use a Collector.

我了解 Collectors.toMap 方法。但是我似乎无法正确完成语法或语义。我猜想 Function.identity（）是使用 IntStream 产生的每个数字作为键的正确方法。然后我尝试了 UUID.randomUUID（）或 p-> UUID.randomUUID（）生成每个地图条目的值。

I know about the Collectors.toMap methods. But I cannot seem to get the syntax or semantics done properly. I am guessing that Function.identity() is the right way to use each number produced by our IntStream as the key. Then I tried UUID.randomUUID() or p -> UUID.randomUUID() to generate the value for each map entry.

int initialCapacity = 3;
Map < Integer, UUID > map =
        ThreadLocalRandom
                .current()                              // Returns a random number generator.
                .ints( initialCapacity )                // Returns an `IntStream`. 
                .collect(
                        Collectors.toMap(
                                Function.identity() ,   // Generate map key (?)
                                x -> UUID.randomUUID()  // Generate map value (?)
                        )
                );

但是我在IDE中遇到一个错误，说：

But I get an error in the IDE saying:

期望3个参数，但发现1

Expected 3 arguments but found 1

我找到了的示例具有两个参数的Collectors.toMap ，一个用于键，一个用于值，以产生每个地图项。因此，我不明白为什么要坚持方法调用的第三个参数。

I found examples of Collectors.toMap with two arguments, one for the key, and one for the value, to produce each map entry. So I do not see why the insistence on a third argument to the method call.

推荐答案

IntStream 仅具有一个 collect 方法，该方法需要3个参数。我怀疑您正在寻找 Stream 中的1参数重载。

IntStream only has one collect method, which takes 3 parameters. I suspect you are looking for the 1-parameter overload in Stream.

了解该 IntStream 生成原始 int 数字流，而不是 Integer 对象流。

Understand that IntStream produces a stream of primitive int numbers, not Integer objects.

调用 .boxed（）调用自动装箱（请参见教程）将每个基本 int 包装为 Integer 目的。这将使 Stream< Integer> 而不是 IntStream 。

Calling .boxed() invokes auto-boxing (see tutorial) to wrap each primitive int as an Integer object. This makes a Stream<Integer> rather than IntStream.

Map < Integer, UUID > map =
    ThreadLocalRandom
            .current()
            .ints( initialCapacity )
            .boxed()                                // Add this call
            .collect(
                    Collectors.toMap(
                            Function.identity() ,
                            x -> UUID.randomUUID()
                    )
            );

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