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


当前回答

地图: 该方法以一个Function作为参数,并返回一个新的流,该流由将传递的函数应用于流的所有元素所生成的结果组成。

让我们想象一下,我有一个整数值列表(1,2,3,4,5)和一个函数接口,其逻辑是传递的整数值的平方。(e -> e * e)。

List<Integer> intList = Arrays.asList(1, 2, 3, 4, 5);

List<Integer> newList = intList.stream().map( e -> e * e ).collect(Collectors.toList());

System.out.println(newList);

输出:

[1, 4, 9, 16, 25]

如您所见,输出是一个新流,其值是输入流值的平方。

[1, 2, 3, 4, 5] -> apply e -> e * e -> [ 1*1, 2*2, 3*3, 4*4, 5*5 ] -> [1, 4, 9, 16, 25 ]

http://codedestine.com/java-8-stream-map-method/

FlatMap: - 该方法以一个函数作为参数,该函数接受一个参数T作为输入参数,并返回一个参数R的流作为返回值。当此函数应用于此流的每个元素时,它将生成一个新值流。然后,每个元素生成的这些新流的所有元素被复制到一个新流,该新流将是该方法的返回值。

让我们想象一下,我有一个学生对象列表,每个学生可以选择多个科目。

List<Student> studentList = new ArrayList<Student>();

  studentList.add(new Student("Robert","5st grade", Arrays.asList(new String[]{"history","math","geography"})));
  studentList.add(new Student("Martin","8st grade", Arrays.asList(new String[]{"economics","biology"})));
  studentList.add(new Student("Robert","9st grade", Arrays.asList(new String[]{"science","math"})));

  Set<Student> courses = studentList.stream().flatMap( e -> e.getCourse().stream()).collect(Collectors.toSet());

  System.out.println(courses);

输出:

[economics, biology, geography, science, history, math]

如您所见,输出是一个新流,其值是输入流的每个元素返回的流的所有元素的集合。

[s1, s2, s3] -> [{“历史”,“数学”,“地理”},{“经济学”、“生物学”},{“科学”,“数学”}]- >采取独特的主题- - - > [经济、生物、地理、科学、历史、数学]

http://codedestine.com/java-8-stream-flatmap-method/

其他回答

.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]。

如果你熟悉c#也可以很好的类比。基本上c# Select类似于java map和c# SelectMany java flatMap。对于集合,同样适用于Kotlin。

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 !

简单的答案。

映射操作可以生成流的流。前流<流<整数> >

flatMap操作只会产生流。前流<整数>

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