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()));

如果您认为map()是一个迭代(一级for循环)，那么flatmap()是一个两级迭代(类似于嵌套的for循环)。(输入每个迭代元素foo，然后执行foo. getbarlist()并再次迭代该barList)

---

Map():获取一个流，对每个元素做一些事情，收集每个进程的单个结果，输出另一个流。“做一些函数”的定义是隐含的。如果任何元素的处理结果为null，则使用null来组成最终流。因此，结果流中的元素数量将等于输入流的数量。

Flatmap():获取元素/流流和函数(显式定义)的流，将函数应用到每个流的每个元素，并将所有中间结果流收集为更大的流(“扁平化”)。如果任何元素的处理结果为null，则为“扁平化”的最后一步提供空流。如果输入是几个流，则结果流中的元素数量是所有输入中所有参与元素的总和。

流。flatMap，顾名思义，是映射和平面操作的组合。这意味着您首先将一个函数应用到元素上，然后将其压平。流。Map仅将函数应用于流，而不会将流平展。

为了理解流的扁平化是由什么组成的，考虑一个像[[1,2,3]，[4,5,6]，[7,8,9]]这样的结构，它有“两个层次”。扁平化意味着将其转换为“一级”结构:[1,2,3,4,5,6,7,8,9]。

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

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/

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。