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>

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)

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]。

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。

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