最近,我和一位同事讨论了在Java中将List转换为Map的最佳方法,以及这样做是否有任何具体的好处。
我想知道最佳的转换方法,如果有人能指导我,我将非常感激。
这是一个好方法吗?
List<Object[]> results;
Map<Integer, String> resultsMap = new HashMap<Integer, String>();
for (Object[] o : results) {
resultsMap.put((Integer) o[0], (String) o[1]);
}
一个Java 8转换List<?>的对象到Map<k, v>:
List<Hosting> list = new ArrayList<>();
list.add(new Hosting(1, "liquidweb.com", new Date()));
list.add(new Hosting(2, "linode.com", new Date()));
list.add(new Hosting(3, "digitalocean.com", new Date()));
//example 1
Map<Integer, String> result1 = list.stream().collect(
Collectors.toMap(Hosting::getId, Hosting::getName));
System.out.println("Result 1 : " + result1);
//example 2
Map<Integer, String> result2 = list.stream().collect(
Collectors.toMap(x -> x.getId(), x -> x.getName()));
从下面复制的代码:
https://www.mkyong.com/java8/java-8-convert-list-to-map/
一个Java 8转换List<?>的对象到Map<k, v>:
List<Hosting> list = new ArrayList<>();
list.add(new Hosting(1, "liquidweb.com", new Date()));
list.add(new Hosting(2, "linode.com", new Date()));
list.add(new Hosting(3, "digitalocean.com", new Date()));
//example 1
Map<Integer, String> result1 = list.stream().collect(
Collectors.toMap(Hosting::getId, Hosting::getName));
System.out.println("Result 1 : " + result1);
//example 2
Map<Integer, String> result2 = list.stream().collect(
Collectors.toMap(x -> x.getId(), x -> x.getName()));
从下面复制的代码:
https://www.mkyong.com/java8/java-8-convert-list-to-map/
从Java 8开始,答案由@ZouZou使用收集器。toMap收集器当然是解决这个问题的惯用方法。
由于这是一个非常常见的任务,我们可以将其变成一个静态实用程序。
这样解决方案就变成了一行程序。
/**
* Returns a map where each entry is an item of {@code list} mapped by the
* key produced by applying {@code mapper} to the item.
*
* @param list the list to map
* @param mapper the function to produce the key from a list item
* @return the resulting map
* @throws IllegalStateException on duplicate key
*/
public static <K, T> Map<K, T> toMapBy(List<T> list,
Function<? super T, ? extends K> mapper) {
return list.stream().collect(Collectors.toMap(mapper, Function.identity()));
}
下面是如何在List<Student>中使用它:
Map<Long, Student> studentsById = toMapBy(students, Student::getId);
就像已经说过的,在java-8中,我们有收藏家的简洁解决方案:
list.stream().collect(
groupingBy(Item::getKey)
)
同时,你可以通过另一个groupingBy方法作为第二个参数来嵌套多个组:
list.stream().collect(
groupingBy(Item::getKey, groupingBy(Item::getOtherKey))
)
这样,我们就有了多级映射,就像这样:map <key, map <key, List<Item>>>