在Stream :: flatMap中使用Java 8的Optional

问题描述

新的Java 8流框架和朋友们提供了一些非常简洁的Java代码，但是我遇到了一个看似简单的情况，这个问题很简洁。

The new Java 8 stream framework and friends make for some very concise java code, but I have come across a seemingly-simple situation that is tricky to do concisely.

考虑列表< Thing>东西和方法可选<其他>解决（事情）。我想将 Thing 映射到可选< Other> s并获取第一个其他。显而易见的解决方案是使用 things.stream（）。flatMap（this :: resolve）.findFirst（），但 flatMap 要求您返回一个流，并且可选没有 stream（）方法（或者是它是集合或提供一种方法将其转换为或以集合的形式查看。

Consider a List<Thing> things and method Optional<Other> resolve(Thing thing). I want to map the Things to Optional<Other>s and get the first Other. The obvious solution would be to use things.stream().flatMap(this::resolve).findFirst(), but flatMap requires that you return a stream, and Optional doesn't have a stream() method (or is it a Collection or provide a method to convert it to or view it as a Collection).

我能想到的最好的是：

things.stream()
    .map(this::resolve)
    .filter(Optional::isPresent)
    .map(Optional::get)
    .findFirst();

但这似乎是一个非常常见的案例。任何人都有更好的主意吗？

But that seems awfully long-winded for what seems like a very common case. Anyone have a better idea?

推荐答案

Java 9

已添加到JDK 9.这使您可以执行以下操作，而无需任何帮助方法：

Java 9

Optional.stream has been added to JDK 9. This enables you to do the following, without the need of any helper method:

Optional<Other> result =
    things.stream()
          .map(this::resolve)
          .flatMap(Optional::stream)
          .findFirst();

Java 8

是的，这是API中的一个小洞，因为将Optional变为零或一个长度的流有点不方便。你可以这样做：

Java 8

Yes, this was a small hole in the API, in that it's somewhat inconvenient to turn an Optional into a zero-or-one length Stream. You could do this:

Optional<Other> result =
    things.stream()
          .map(this::resolve)
          .flatMap(o -> o.isPresent() ? Stream.of(o.get()) : Stream.empty())
          .findFirst();

在flatMap中使用三元运算符有点麻烦，所以写起来可能会更好一个小帮手功能：

Having the ternary operator inside the flatMap is a bit cumbersome, though, so it might be better to write a little helper function to do this:

/**
 * Turns an Optional<T> into a Stream<T> of length zero or one depending upon
 * whether a value is present.
 */
static <T> Stream<T> streamopt(Optional<T> opt) {
    if (opt.isPresent())
        return Stream.of(opt.get());
    else
        return Stream.empty();
}

Optional<Other> result =
    things.stream()
          .flatMap(t -> streamopt(resolve(t)))
          .findFirst();

在这里，我已经内联调用resolve（）而不是单独的map（）操作，但这是一个品味问题。

Here, I've inlined the call to resolve() instead of having a separate map() operation, but this is a matter of taste.

这篇关于在Stream :: flatMap中使用Java 8的Optional的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！