对于Map，我们有一个元素列表和一个(函数，动作)f，这样:

[a,b,c] f(x) => [f(a),f(b),f(c)]

对于平面映射，我们有一个元素列表，我们有一个(function,action) f，我们希望结果是扁平的:

[[a,b],[c,d,e]] f(x) =>[f(a),f(b),f(c),f(d),f(e)]

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

.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]。

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>

我不太确定我是否应该回答这个问题，但每当我面对不理解这一点的人时，我就用同样的例子。

假设你有一个苹果。例如，地图是将苹果转换为苹果汁或一对一映射。

同样的苹果，只得到种子，这就是flatMap所做的，或者一对多，一个苹果作为输入，许多种子作为输出。

传递给流的函数。Map必须返回一个对象。这意味着输入流中的每个对象都会导致输出流中的一个对象。

传递给流的函数。flatMap为每个对象返回一个流。这意味着该函数可以为每个输入对象返回任意数量的对象(包括none)。然后将结果流连接到一个输出流。