java中list或set转map的方法（java list 转 set）

在开发中我们有时需要将list或set转换为map（比如对象属性中的唯一键作为map的key，对象作为map的value），一般的想法就是new一个map，然后把list或set中的值一个个push到map中。

类似下面的代码：

1. 
   
2. List<String> stringList = Lists.newArrayList("t1", "t2", "t3");
3. Map<String, String> map = Maps.newHashMapWithExpectedSize(stringList.size());
4. for (String str : stringList) {
5. map.put(str, str);
6. }

是否还有更优雅的写法呢？答案是有的。

guava提供了集合（实现了Iterables接口或Iterator接口）转map的方法，方法定义如下：

1. 
   
2. /**
3. * Returns an immutable map for which the {@link Map#values} are the given
4. * elements in the given order, and each key is the product of invoking a
5. * supplied function on its corresponding value.
6. *
7. * @param values the values to use when constructing the {@code Map}
8. * @param keyFunction the function used to produce the key for each value
9. * @return a map mapping the result of evaluating the function {@code
10. * keyFunction} on each value in the input collection to that value
11. * @throws IllegalArgumentException if {@code keyFunction} produces the same
12. * key for more than one value in the input collection
13. * @throws NullPointerException if any elements of {@code values} is null, or
14. * if {@code keyFunction} produces {@code null} for any value
15. */
16. public static <K, V> ImmutableMap<K, V> uniqueIndex(
17. Iterable<V> values, Function<? super V, K> keyFunction) {
18. return uniqueIndex(values.iterator(), keyFunction);
19. }
21. /**
22. * Returns an immutable map for which the {@link Map#values} are the given
23. * elements in the given order, and each key is the product of invoking a
24. * supplied function on its corresponding value.
25. *
26. * @param values the values to use when constructing the {@code Map}
27. * @param keyFunction the function used to produce the key for each value
28. * @return a map mapping the result of evaluating the function {@code
29. * keyFunction} on each value in the input collection to that value
30. * @throws IllegalArgumentException if {@code keyFunction} produces the same
31. * key for more than one value in the input collection
32. * @throws NullPointerException if any elements of {@code values} is null, or
33. * if {@code keyFunction} produces {@code null} for any value
34. * @since 10.0
35. */
36. public static <K, V> ImmutableMap<K, V> uniqueIndex(
37. Iterator<V> values, Function<? super V, K> keyFunction) {
38. checkNotNull(keyFunction);
39. ImmutableMap.Builder<K, V> builder = ImmutableMap.builder();
40. while (values.hasNext()) {
41. V value = values.next();
42. builder.put(keyFunction.apply(value), value);
43. }
44. return builder.build();
45. }
46. 这样我们就可以很方便的进行转换了，如下：
47. List<String> stringList = Lists.newArrayList("t1", "t2", "t3");
48. Map<String, String> map = Maps.uniqueIndex(stringList, new Function<String, String>() {
49. @Override
50. public String apply(String input) {
51. return input;
52. }
53. });

需要注意的是，如接口注释所说，如果Function返回的结果产生了重复的key，将会抛出异常。

java8也提供了转换的方法，这里直接照搬别人博客的代码：

1. 
   
2. @Test
3. public void convert_list_to_map_with_java8_lambda () {
5. List<Movie> movies = new ArrayList<Movie>();
6. movies.add(new Movie(1, "The Shawshank Redemption"));
7. movies.add(new Movie(2, "The Godfather"));
9. Map<Integer, Movie> mappedMovies = movies.stream().collect(
10. Collectors.toMap(Movie::getRank, (p) -> p));
12. logger.info(mappedMovies);
14. assertTrue(mappedMovies.size() == 2);
15. assertEquals("The Shawshank Redemption", mappedMovies.get(1).getDescription());