Java：如何将列表转换为地图

Java: How to convert List to Map

提问by Rachel

Recently I have conversation with a colleague about what would be the optimal way to convert Listto Mapin Java and if there any specific benefits of doing so.

最近，我与一位同事讨论了在 Java 中转换List为的最佳方式，Map以及这样做是否有任何具体好处。

I want to know optimal conversion approach and would really appreciate if any one can guide me.

我想知道最佳转换方法，如果有人可以指导我，我将不胜感激。

Is this good approach:

这是好方法吗：

List<Object[]> results;
Map<Integer, String> resultsMap = new HashMap<Integer, String>();
for (Object[] o : results) {
    resultsMap.put((Integer) o[0], (String) o[1]);
}

采纳答案by Jim Garrison

List<Item> list;
Map<Key,Item> map = new HashMap<Key,Item>();
for (Item i : list) map.put(i.getKey(),i);

Assuming of course that each Item has a getKey()method that returns a key of the proper type.

当然，假设每个 Item 都有一个getKey()返回正确类型键的方法。

回答by Daniel

Many solutions come to mind, depending on what you want to achive:

根据您想要实现的目标，您会想到许多解决方案：

Every List item is key and value

每个列表项都是键和值

for( Object o : list ) {
    map.put(o,o);
}

List elements have something to look them up, maybe a name:

列表元素有一些东西可以查找它们，也许是一个名称：

for( MyObject o : list ) {
    map.put(o.name,o);
}

List elements have something to look them up, and there is no guarantee that they are unique: Use Googles MultiMaps

列表元素有一些东西可以查找，并且不能保证它们是唯一的：使用 Googles MultiMaps

for( MyObject o : list ) {
    multimap.put(o.name,o);
}

Giving all the elements the position as a key:

将所有元素的位置作为键：

for( int i=0; i<list.size; i++ ) {
    map.put(i,list.get(i));
}

...

...

It really depends on what you want to achive.

这实际上取决于您想要实现的目标。

As you can see from the examples, a Map is a mapping from a key to a value, while a list is just a series of elements having a position each. So they are simply not automatically convertible.

从示例中可以看出，Map 是从键到值的映射，而列表只是一系列元素，每个元素都有一个位置。所以它们根本不能自动转换。

回答by Steve Kuo

A Listand Mapare conceptually different. A Listis an ordered collection of items. The items can contain duplicates, and an item might not have any concept of a unique identifier (key). A Maphas values mapped to keys. Each key can only point to one value.

AList和Map在概念上是不同的。AList是项目的有序集合。项目可以包含重复项，并且项目可能没有任何唯一标识符（键）的概念。AMap具有映射到键的值。每个键只能指向一个值。

Therefore, depending on your List's items, it may or may not be possible to convert it to a Map. Does your List's items have no duplicates? Does each item have a unique key? If so then it's possible to put them in a Map.

因此，根据您List的项目，可能会也可能不会将其转换为Map. 你List的物品没有重复吗？每个项目都有唯一的键吗？如果是这样，则可以将它们放入Map.

回答by Kango_V

Here's a little method I wrote for exactly this purpose. It uses Validate from Apache Commons.

这是我为此目的编写的一个小方法。它使用来自 Apache Commons 的验证。

Feel free to use it.

随意使用它。

/**
 * Converts a <code>List</code> to a map. One of the methods of the list is called to retrive
 * the value of the key to be used and the object itself from the list entry is used as the
 * objct. An empty <code>Map</code> is returned upon null input.
 * Reflection is used to retrieve the key from the object instance and method name passed in.
 *
 * @param <K> The type of the key to be used in the map
 * @param <V> The type of value to be used in the map and the type of the elements in the
 *            collection
 * @param coll The collection to be converted.
 * @param keyType The class of key
 * @param valueType The class of the value
 * @param keyMethodName The method name to call on each instance in the collection to retrieve
 *            the key
 * @return A map of key to value instances
 * @throws IllegalArgumentException if any of the other paremeters are invalid.
 */
public static <K, V> Map<K, V> asMap(final java.util.Collection<V> coll,
        final Class<K> keyType,
        final Class<V> valueType,
        final String keyMethodName) {

    final HashMap<K, V> map = new HashMap<K, V>();
    Method method = null;

    if (isEmpty(coll)) return map;
    notNull(keyType, Messages.getString(KEY_TYPE_NOT_NULL));
    notNull(valueType, Messages.getString(VALUE_TYPE_NOT_NULL));
    notEmpty(keyMethodName, Messages.getString(KEY_METHOD_NAME_NOT_NULL));

    try {
        // return the Method to invoke to get the key for the map
        method = valueType.getMethod(keyMethodName);
    }
    catch (final NoSuchMethodException e) {
        final String message =
            String.format(
                    Messages.getString(METHOD_NOT_FOUND),
                    keyMethodName,
                    valueType);
        e.fillInStackTrace();
        logger.error(message, e);
        throw new IllegalArgumentException(message, e);
    }
    try {
        for (final V value : coll) {

            Object object;
            object = method.invoke(value);
            @SuppressWarnings("unchecked")
            final K key = (K) object;
            map.put(key, value);
        }
    }
    catch (final Exception e) {
        final String message =
            String.format(
                    Messages.getString(METHOD_CALL_FAILED),
                    method,
                    valueType);
        e.fillInStackTrace();
        logger.error(message, e);
        throw new IllegalArgumentException(message, e);
    }
    return map;
}

回答by cs94njw

I like Kango_V's answer, but I think it's too complex. I think this is simpler - maybe too simple. If inclined, you could replace String with a Generic marker, and make it work for any Key type.

我喜欢 Kango_V 的回答，但我认为它太复杂了。我认为这更简单——也许太简单了。如果愿意，您可以将 String 替换为 Generic 标记，并使其适用于任何 Key 类型。

public static <E> Map<String, E> convertListToMap(Collection<E> sourceList, ListToMapConverterInterface<E> converterInterface) {
    Map<String, E> newMap = new HashMap<String, E>();
    for( E item : sourceList ) {
        newMap.put( converterInterface.getKeyForItem( item ), item );
    }
    return newMap;
}

public interface ListToMapConverterInterface<E> {
    public String getKeyForItem(E item);
}

Used like this:

像这样使用：

        Map<String, PricingPlanAttribute> pricingPlanAttributeMap = convertListToMap( pricingPlanAttributeList,
                new ListToMapConverterInterface<PricingPlanAttribute>() {

                    @Override
                    public String getKeyForItem(PricingPlanAttribute item) {
                        return item.getFullName();
                    }
                } );

回答by xxf

Universal method

通用方法

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);
}

回答by ripper234

Just in case this question isn't closed as a duplicate, the right answer is to use Google Collections:

以防万一这个问题没有作为重复关闭，正确的答案是使用 Google Collections：

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

回答by Andrejs

There is also a simple way of doing this using Maps.uniqueIndex(...)from Google guavalibraries

还有一种使用Google番石榴库中的Maps.uniqueIndex(...)的简单方法

回答by Alexis C.

With java-8, you'll be able to do this in one line using streams, and the Collectorsclass.

使用java-8，您将能够使用流和Collectors类在一行中完成此操作。

Map<String, Item> map = 
    list.stream().collect(Collectors.toMap(Item::getKey, item -> item));

Short demo:

简短演示：

import java.util.Arrays;
import java.util.List;
import java.util.Map;
import java.util.stream.Collectors;

public class Test{
    public static void main (String [] args){
        List<Item> list = IntStream.rangeClosed(1, 4)
                                   .mapToObj(Item::new)
                                   .collect(Collectors.toList()); //[Item [i=1], Item [i=2], Item [i=3], Item [i=4]]

        Map<String, Item> map = 
            list.stream().collect(Collectors.toMap(Item::getKey, item -> item));

        map.forEach((k, v) -> System.out.println(k + " => " + v));
    }
}
class Item {

    private final int i;

    public Item(int i){
        this.i = i;
    }

    public String getKey(){
        return "Key-"+i;
    }

    @Override
    public String toString() {
        return "Item [i=" + i + "]";
    }
}

Output:

输出：

Key-1 => Item [i=1]
Key-2 => Item [i=2]
Key-3 => Item [i=3]
Key-4 => Item [i=4]

As noted in comments, you can use Function.identity()instead of item -> item, although I find i -> irather explicit.

正如评论中所指出的，您可以使用Function.identity()而不是item -> item，尽管我发现i -> i相当明确。

And to be complete note that you can use a binary operator if your function is not bijective. For example let's consider this Listand the mapping function that for an int value, compute the result of it modulo 3:

并且要完整注意，如果您的函数不是双射的，您可以使用二元运算符。例如，让我们考虑这个List和映射函数，对于一个 int 值，计算它的模 3 的结果：

List<Integer> intList = Arrays.asList(1, 2, 3, 4, 5, 6);
Map<String, Integer> map = 
    intList.stream().collect(toMap(i -> String.valueOf(i % 3), i -> i));

When running this code, you'll get an error saying java.lang.IllegalStateException: Duplicate key 1. This is because 1 % 3 is the same as 4 % 3 and hence have the same key value given the key mapping function. In this case you can provide a merge operator.

运行此代码时，您会收到一条错误消息，指出java.lang.IllegalStateException: Duplicate key 1. 这是因为 1 % 3 与 4 % 3 相同，因此在给定键映射函数的情况下具有相同的键值。在这种情况下，您可以提供合并运算符。

Here's one that sum the values; (i1, i2) -> i1 + i2;that can be replaced with the method reference Integer::sum.

这是对值求和的一个；(i1, i2) -> i1 + i2;可以替换为方法引用Integer::sum。

Map<String, Integer> map = 
    intList.stream().collect(toMap(i -> String.valueOf(i % 3), 
                                   i -> i, 
                                   Integer::sum));

which now outputs:

现在输出：

0 => 9 (i.e 3 + 6)
1 => 5 (i.e 1 + 4)
2 => 7 (i.e 2 + 5)

Hope it helps! :)

希望能帮助到你！:)

回答by glts

Since Java 8, the answer by @ZouZouusing the Collectors.toMapcollector is certainly the idiomatic way to solve this problem.

从 Java 8 开始，@ZouZou使用Collectors.toMap收集器的答案当然是解决这个问题的惯用方法。

And as this is such a common task, we can make it into a static utility.

由于这是一项如此常见的任务，我们可以将其变成一个静态实用程序。

That way the solution truly becomes a one-liner.

这样，解决方案就真正变成了单线解决方案。

/**
 * 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()));
}

And here's how you would use it on a List<Student>:

这是您将如何在 a 上使用它的方法List<Student>：

Map<Long, Student> studentsById = toMapBy(students, Student::getId);