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>

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

流操作flatMap和map接受函数作为输入。

flatMap期望该函数为流的每个元素返回一个新的流，并返回一个流，该流结合了该函数为每个元素返回的流的所有元素。换句话说，使用flatMap，对于来自源的每个元素，函数将创建多个元素。http://www.zoftino.com/java-stream-examples#flatmap-operation

Map期望函数返回一个转换后的值，并返回一个包含转换后元素的新流。换句话说，使用map，对于来自源的每个元素，函数将创建一个转换后的元素。 http://www.zoftino.com/java-stream-examples#map-operation

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/