Apache Commons MapUtils.populateMap

如果您不使用Java 8，并且出于某种原因不想使用显式循环，可以尝试MapUtils。populateMap来自Apache Commons。

MapUtils.populateMap

假设您有一个巴黎的列表。

List<ImmutablePair<String, String>> pairs = ImmutableList.of(
    new ImmutablePair<>("A", "aaa"),
    new ImmutablePair<>("B", "bbb")
);

现在需要Pair对象的Pair键的Map。

Map<String, Pair<String, String>> map = new HashMap<>();
MapUtils.populateMap(map, pairs, new Transformer<Pair<String, String>, String>() {

  @Override
  public String transform(Pair<String, String> input) {
    return input.getKey();
  }
});

System.out.println(map);

给输出:

{A=(A,aaa), B=(B,bbb)}

也就是说，for循环可能更容易理解。(下面给出了相同的输出):

Map<String, Pair<String, String>> map = new HashMap<>();
for (Pair<String, String> pair : pairs) {
  map.put(pair.getKey(), pair);
}
System.out.println(map);

为了防止这个问题没有重复，正确的答案是使用谷歌Collections:

Map<String,Role> mappedRoles = Maps.uniqueIndex(yourList, new Function<Role,String>() {
  public String apply(Role from) {
    return from.getName(); // or something else
  }});

如果你使用Kotlin，这里有一个例子:

listOf("one", "two").mapIndexed { i, it -> i to it }.toMap()

普遍的方法

public static <K, V> Map<K, V> listAsMap(Collection<V> sourceList, ListToMapConverter<K, V> converter) {
    Map<K, V> newMap = new HashMap<K, V>();
    for (V item : sourceList) {
        newMap.put( converter.getKey(item), item );
    }
    return newMap;
}

public static interface ListToMapConverter<K, V> {
    public K getKey(V item);
}

public class EmployeeDetailsFetchListToMap {
  public static void main(String[] args) {
    List<EmployeeDetailsFetch> list = new ArrayList<>();
    list.add(new EmployeeDetailsFetch(1L, "vinay", 25000F));
    list.add(new EmployeeDetailsFetch(2L, "kohli", 5000000F));
    list.add(new EmployeeDetailsFetch(3L, "dhoni", 20000000F));

    //adding id as key and map of id and student name
    Map<Long, Map<Long, String>> map1 = list.stream()
        .collect(
            Collectors.groupingBy(
                EmployeeDetailsFetch::getEmpId,
                Collectors.toMap(
                    EmployeeDetailsFetch::getEmpId,
                    EmployeeDetailsFetch::getEmployeeName
                )
            )
        );
    System.out.println(map1);

    //converting list into map of Student
    //Adding id as Key and Value as Student into a map
    Map<Long, EmployeeDetailsFetch> map = list.stream()
        .collect(
            Collectors.toMap(
                EmployeeDetailsFetch::getEmpId, 
                EmployeeDetailsFetch -> EmployeeDetailsFetch
            )
        );

    for(Map.Entry<Long, EmployeeDetailsFetch> m : map.entrySet()) {
      System.out.println("key :" + m.getKey() + "  Value : " + m.getValue());
    }
  }
}

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);