在Java 8中,Stream.map()和Stream.flatMap()方法之间有什么区别?


当前回答

Oracle关于Optional的文章强调了map和flatmap的区别:

String version = computer.map(Computer::getSoundcard)
                  .map(Soundcard::getUSB)
                  .map(USB::getVersion)
                  .orElse("UNKNOWN");

Unfortunately, this code doesn't compile. Why? The variable computer is of type Optional<Computer>, so it is perfectly correct to call the map method. However, getSoundcard() returns an object of type Optional. This means the result of the map operation is an object of type Optional<Optional<Soundcard>>. As a result, the call to getUSB() is invalid because the outermost Optional contains as its value another Optional, which of course doesn't support the getUSB() method. With streams, the flatMap method takes a function as an argument, which returns another stream. This function is applied to each element of a stream, which would result in a stream of streams. However, flatMap has the effect of replacing each generated stream by the contents of that stream. In other words, all the separate streams that are generated by the function get amalgamated or "flattened" into one single stream. What we want here is something similar, but we want to "flatten" a two-level Optional into one. Optional also supports a flatMap method. Its purpose is to apply the transformation function on the value of an Optional (just like the map operation does) and then flatten the resulting two-level Optional into a single one. So, to make our code correct, we need to rewrite it as follows using flatMap:

String version = computer.flatMap(Computer::getSoundcard)
                   .flatMap(Soundcard::getUSB)
                   .map(USB::getVersion)
                   .orElse("UNKNOWN");

第一个flatMap确保返回Optional<Soundcard> 而不是一个Optional<Optional<Soundcard>>,和第二个flatMap 实现相同的目的,返回Optional<USB>。注意 第三个调用只需要一个map(),因为getVersion()返回一个 字符串而不是可选对象。

http://www.oracle.com/technetwork/articles/java/java8-optional-2175753.html

其他回答

我想举两个例子来说明更实际的观点: 第一个使用地图的例子:

@Test
public void convertStringToUpperCaseStreams() {
    List<String> collected = Stream.of("a", "b", "hello") // Stream of String 
            .map(String::toUpperCase) // Returns a stream consisting of the results of applying the given function to the elements of this stream.
            .collect(Collectors.toList());
    assertEquals(asList("A", "B", "HELLO"), collected);
}

在第一个例子中没有什么特别的,一个函数被应用来返回大写的String。

第二个使用flatMap的例子:

@Test
public void testflatMap() throws Exception {
    List<Integer> together = Stream.of(asList(1, 2), asList(3, 4)) // Stream of List<Integer>
            .flatMap(List::stream)
            .map(integer -> integer + 1)
            .collect(Collectors.toList());
    assertEquals(asList(2, 3, 4, 5), together);
}

在第二个例子中,传递了一个List流。它不是一个整数流! 如果必须使用转换函数(通过map),则首先必须将流平展为其他类型的流(整数流)。 如果flatMap被移除,则返回以下错误:对于参数类型List, int,操作符+未定义。 不可能在整数列表上应用+ 1 !

这对初学者来说是很困惑的。基本的区别是map为列表中的每个条目发出一个项,而flatMap基本上是一个map + flatten操作。更清楚地说,当你需要多个值时使用flatMap,例如当你期望一个循环返回数组时,flatMap在这种情况下非常有用。

我写了一篇关于这方面的博客,你可以在这里查看。

.map用于A -> B映射

Stream.of("dog", "cat")              // stream of 2 Strings
    .map(s -> s.length())            // stream of 2 Integers: [3, 3]

它将任意项A转换为任意项b


.flatMap用于A ->流< B>连接

Stream.of("dog", "cat")             // stream of 2 Strings
    .flatMapToInt(s -> s.chars())   // stream of 6 ints:      [d, o, g, c, a, t]

it——1将任何项A转换为Stream< B>,然后——2将所有流连接到一个(平面)流。Javadoc


注1:虽然后面的例子是一个原语流(IntStream),而不是一个对象流(stream),但它仍然说明了. flatmap的思想。

注意2:尽管有这个名字,String.chars()方法返回的是整数。所以实际的集合将是:[100,111,103,99,97,116] ,其中100是“d”的代码,111是“o”的代码,等等。同样,为了说明目的,它被表示为[d, o, g, c, a, t]。

flatMap()还利用了流的部分延迟求值。它将读取第一个流,只有在需要时才会进入下一个流。这里详细解释了这种行为:flatMap保证是懒惰的吗?

通过阅读所有的信息,简单的理解方法是:

如果你有一个元素的平面列表,请使用map: [0,1,2,3,4,5] 如果你有一个元素的列表,请使用flatMap:[[1,3,5],[2,4,6]]。这意味着,在映射操作应用于每个元素之前,您的列表需要被平铺