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


当前回答

我有一种感觉,这里的大多数答案都把简单的问题复杂化了。如果你已经理解了地图是如何工作的,那就很容易掌握了。

在使用map()时,有些情况下我们可能会得到不需要的嵌套结构,flatMap()方法的设计是通过避免换行来克服这一问题。


例子:

1

List<List<Integer>> result = Stream.of(Arrays.asList(1), Arrays.asList(2, 3))
  .collect(Collectors.toList());

我们可以使用flatMap来避免使用嵌套列表:

List<Integer> result = Stream.of(Arrays.asList(1), Arrays.asList(2, 3))
  .flatMap(i -> i.stream())
  .collect(Collectors.toList());

2

Optional<Optional<String>> result = Optional.of(42)
      .map(id -> findById(id));

Optional<String> result = Optional.of(42)
      .flatMap(id -> findById(id));

地点:

private Optional<String> findById(Integer id)

其他回答

map() takes a Stream and transform it to another Stream. It applies a function on each element of Stream and store return value into new Stream. It does not flatten the stream. But flatMap() is the combination of a map and a flat operation i.e, it applies a function to elements as well as flatten them. 2) map() is used for transformation only, but flatMap() is used for both transformation and flattening. please read more here. https://javaint4bytes.blogspot.com/2022/11/stream-flatmap-in-java-with-examples.html

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 !

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/

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>

map()和flatMap()

map ()

只接受一个函数<T, R>一个lambda参数,其中T是元素,R是使用T构建的返回元素。最后,我们将有一个带有类型为R的对象的流。一个简单的例子可以是:

Stream
  .of(1,2,3,4,5)
  .map(myInt -> "preFix_"+myInt)
  .forEach(System.out::println);

它只是取Type Integer的元素1到5,使用每个元素从String类型中构建一个值为“prefix_”+integer_value的新元素,并打印出来。

flatMap ()

知道flatMap()接受一个函数F<T, R> where是很有用的

T is a type from which a Stream can be built from/with. It can be a List (T.stream()), an array (Arrays.stream(someArray)), etc.. anything that from which a Stream can be with/or form. in the example below each dev has many languages, so dev. Languages is a List and will use a lambda parameter. R is the resulting Stream that will be built using T. Knowing that we have many instances of T, we will naturally have many Streams from R. All these Streams from Type R will now be combined into one single 'flat' Stream from Type R.

例子

Bachiri Taoufiq的例子(见这里的答案)1简单易懂。为了清晰起见,假设我们有一个开发团队:

dev_team = {dev_1,dev_2,dev_3}

,每个开发人员都懂多种语言:

dev_1 = {lang_a,lang_b,lang_c},
dev_2 = {lang_d},
dev_3 = {lang_e,lang_f}

在dev_team上应用Stream.map()来获取每个开发人员的语言:

dev_team.map(dev -> dev.getLanguages())

会给你这样的结构:

{ 
  {lang_a,lang_b,lang_c},
  {lang_d},
  {lang_e,lang_f}
}

它基本上是一个List<List<Languages>> /Object[Languages[]]。不是很漂亮,也不像java8 !!

使用Stream.flatMap(),你可以“扁平化”的东西,因为它采取上述结构 并将其转换为{lang_a, lang_b, lang_c, lang_d, lang_e, lang_f},基本上可以作为List<Languages>/Language[]/etc…

所以最后,你的代码会像这样更有意义:

dev_team
   .stream()    /* {dev_1,dev_2,dev_3} */
   .map(dev -> dev.getLanguages()) /* {{lang_a,...,lang_c},{lang_d}{lang_e,lang_f}}} */
   .flatMap(languages ->  languages.stream()) /* {lang_a,...,lang_d, lang_e, lang_f} */
   .doWhateverWithYourNewStreamHere();

或者仅仅是:

dev_team
       .stream()    /* {dev_1,dev_2,dev_3} */
       .flatMap(dev -> dev.getLanguages().stream()) /* {lang_a,...,lang_d, lang_e, lang_f} */
       .doWhateverWithYourNewStreamHere();

何时使用map()和flatMap():

Use map() when each element of type T from your stream is supposed to be mapped/transformed to a single element of type R. The result is a mapping of type (1 start element -> 1 end element) and new stream of elements of type R is returned. Use flatMap() when each element of type T from your stream is supposed to mapped/transformed to a Collections of elements of type R. The result is a mapping of type (1 start element -> n end elements). These Collections are then merged (or flattened) to a new stream of elements of type R. This is useful for example to represent nested loops.

Java 8 之前:

List<Foo> myFoos = new ArrayList<Foo>();
    for(Foo foo: myFoos){
        for(Bar bar:  foo.getMyBars()){
            System.out.println(bar.getMyName());
        }
    }

后Java 8

myFoos
    .stream()
    .flatMap(foo -> foo.getMyBars().stream())
    .forEach(bar -> System.out.println(bar.getMyName()));