blocks|key|1096102|text|List<Item>+list;
Map<Key,Item>+map+=+new+HashMap<Key,Item>();
for+(Item+i+:+list)+map.put(i.getKey(),i);|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1096103|当然，假设每一项都有一个返回正确类型的键的getKey()方法。|unstyled|offset|length|style|CODE|1096104|entityMap^0|0|L|8|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|N|8|@$G|O|H|P|I|J]]|9|@]|A|$]]|$1|K|3|-4|5|F|7|Q|8|@]|9|@]|A|$]]]|L|$]]

<pre><code>List&lt;Item&gt; list;
Map&lt;Key,Item&gt; map = new HashMap&lt;Key,Item&gt;();
for (Item i : list) map.put(i.getKey(),i);
</code></pre>

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

blocks|key|881427|text|脑海中浮现出许多解决方案，具体取决于您想要实现的目标：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|881428|每个列表项都是键和值|881429|for(+Object+o+:+list+)+{
++++map.put(o,o);
}|code-block|syntax|javascript|881430|列表元素有一些东西来查找它们，可能是一个名字：|881431|for(+MyObject+o+:+list+)+{
++++map.put(o.name,o);
}|881432|列表元素有一些东西可以查找它们，并且不能保证它们是唯一的:使用Googles+MultiMaps|offset|length|881433|for(+MyObject+o+:+list+)+{
++++multimap.put(o.name,o);
}|881434|将所有元素的位置指定为键：|881435|for(+int+i=0;+i<list.size;+i%2B%2B+)+{
++++map.put(i,list.get(i));
}|881436|..。|881437|这真的取决于你想要实现什么。|881438|从示例中可以看到，Map是从键到值的映射，而list只是一系列元素，每个元素都有一个位置。所以它们是不能自动转换的。|881439|entityMap|0|LINK|mutability|MUTABLE|url|http://google-collections.googlecode.com/svn/trunk/javadoc/com/google/common/collect/Multimap.html^0|0|0|0|0|0|13|9|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|1A|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|1B|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|1C|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|1D|8|@]|9|@]|A|$]]|$1|K|3|L|5|F|7|1E|8|@]|9|@]|A|$G|H]]|$1|M|3|N|5|6|7|1F|8|@]|9|@$O|1G|P|1H|1|1I]]|A|$]]|$1|Q|3|R|5|F|7|1J|8|@]|9|@]|A|$G|H]]|$1|S|3|T|5|6|7|1K|8|@]|9|@]|A|$]]|$1|U|3|V|5|F|7|1L|8|@]|9|@]|A|$G|H]]|$1|W|3|X|5|6|7|1M|8|@]|9|@]|A|$]]|$1|Y|3|Z|5|6|7|1N|8|@]|9|@]|A|$]]|$1|10|3|11|5|6|7|1O|8|@]|9|@]|A|$]]|$1|12|3|-4|5|6|7|1P|8|@]|9|@]|A|$]]]|13|$14|$5|15|16|17|A|$18|19]]]]

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

Every List item is key and value

<pre><code>for( Object o : list ) {
 map.put(o,o);
}
</code></pre>

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

<pre><code>for( MyObject o : list ) {
 map.put(o.name,o);
}
</code></pre>

List elements have something to look them up, and there is no guarantee that they are unique: Use Googles <a href="http://google-collections.googlecode.com/svn/trunk/javadoc/com/google/common/collect/Multimap.html" rel="nofollow">MultiMaps</a>

<pre><code>for( MyObject o : list ) {
 multimap.put(o.name,o);
}
</code></pre>

Giving all the elements the position as a key:

<pre><code>for( int i=0; i&lt;list.size; i++ ) {
 map.put(i,list.get(i));
}
</code></pre>

...

It really depends on what you want to achive.

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.

blocks|key|881554|text|List和Map在概念上是不同的。List是项的有序集合。项可以包含重复项，并且项可能没有任何唯一标识符(键)的概念。Map具有映射到键的值。每个键只能指向一个值。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|881555|因此，根据你的List的项目，可能有可能也可能不可能将其转换为Map。你的List的项目是否没有重复项?是否每个项目都有唯一的键?如果是，那么可以将它们放在Map中。|881556|entityMap^0|0|4|5|3|H|4|1N|3|0|7|4|V|3|11|4|26|3|0^^$0|@$1|2|3|4|5|6|7|J|8|@$9|K|A|L|B|C]|$9|M|A|N|B|C]|$9|O|A|P|B|C]|$9|Q|A|R|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|S|8|@$9|T|A|U|B|C]|$9|V|A|W|B|C]|$9|X|A|Y|B|C]|$9|Z|A|10|B|C]]|D|@]|E|$]]|$1|H|3|-4|5|6|7|11|8|@]|D|@]|E|$]]]|I|$]]

A <code>List</code> and <code>Map</code> are conceptually different. A <code>List</code> is an ordered collection of items. The items can contain duplicates, and an item might not have any concept of a unique identifier (key). A <code>Map</code> has values mapped to keys. Each key can only point to one value.

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

blocks|key|1096554|text|下面是我写的一个小方法，就是为了这个目的。它使用Apache+Commons中的验证。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1096555|请随意使用它。|1096556|/**
+*+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;
}|code-block|syntax|javascript|1096557|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|L|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|M|8|@]|9|@]|A|$G|H]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

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

Feel free to use it.

<pre><code>/**
 * Converts a &lt;code&gt;List&lt;/code&gt; 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 &lt;code&gt;Map&lt;/code&gt; is returned upon null input.
 * Reflection is used to retrieve the key from the object instance and method name passed in.
 *
 * @param &lt;K&gt; The type of the key to be used in the map
 * @param &lt;V&gt; 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 &lt;K, V&gt; Map&lt;K, V&gt; asMap(final java.util.Collection&lt;V&gt; coll,
 final Class&lt;K&gt; keyType,
 final Class&lt;V&gt; valueType,
 final String keyMethodName) {

 final HashMap&lt;K, V&gt; map = new HashMap&lt;K, V&gt;();
 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;
}
</code></pre>

blocks|key|881752|text|我喜欢Kango_V的答案，但我认为它太复杂了。我认为这更简单--也许太简单了。如果需要，您可以用通用标记替换字符串，并使其适用于任何键类型。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|881753|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);
}|code-block|syntax|javascript|881754|如下所示：|881755|++++++++Map<String,+PricingPlanAttribute>+pricingPlanAttributeMap+=+convertListToMap(+pricingPlanAttributeList,
++++++++++++++++new+ListToMapConverterInterface<PricingPlanAttribute>()+{

++++++++++++++++++++@Override
++++++++++++++++++++public+String+getKeyForItem(PricingPlanAttribute+item)+{
++++++++++++++++++++++++return+item.getFullName();
++++++++++++++++++++}
++++++++++++++++}+);|881756|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

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.

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

public interface ListToMapConverterInterface&lt;E&gt; {
 public String getKeyForItem(E item);
}
</code></pre>

Used like this:

<pre><code> Map&lt;String, PricingPlanAttribute&gt; pricingPlanAttributeMap = convertListToMap( pricingPlanAttributeList,
 new ListToMapConverterInterface&lt;PricingPlanAttribute&gt;() {

 @Override
 public String getKeyForItem(PricingPlanAttribute item) {
 return item.getFullName();
 }
 } );
</code></pre>

blocks|key|881772|text|通用方法|type|unstyled|depth|inlineStyleRanges|entityRanges|data|881773|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);
}|code-block|syntax|javascript|881774|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Universal method

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

public static interface ListToMapConverter&lt;K, V&gt; {
 public K getKey(V item);
}
</code></pre>

blocks|key|881830|text|以防此问题不会作为重复问题结束，the+right+answer+is+to+use+Google+Collections|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|881831|Map<String,Role>+mappedRoles+=+Maps.uniqueIndex(yourList,+new+Function<Role,String>()+{
++public+String+apply(Role+from)+{
++++return+from.getName();+//+or+something+else
++}});|code-block|syntax|javascript|881832|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/a/3199899/11236^0|G|19|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@$A|R|B|S|1|T]]|C|$]]|$1|D|3|E|5|F|7|U|8|@]|9|@]|C|$G|H]]|$1|I|3|-4|5|6|7|V|8|@]|9|@]|C|$]]]|J|$K|$5|L|M|N|C|$O|P]]]]

Just in case this question isn't closed as a duplicate, <a href="https://stackoverflow.com/a/3199899/11236">the right answer is to use Google Collections</a>:

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

blocks|key|1096744|text|还有一种简单的方法可以使用谷歌guava库中的Maps.uniqueIndex(...)来实现这一点|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1096745|entityMap|0|LINK|mutability|MUTABLE|url|/questions/tagged/guava|1|https://guava.dev/releases/snapshot-jre/api/docs/com/google/common/collect/Maps.html#uniqueIndex-java.lang.Iterable-com.google.common.base.Function-^0|F|5|0|N|L|1|0^^$0|@$1|2|3|4|5|6|7|N|8|@]|9|@$A|O|B|P|1|Q]|$A|R|B|S|1|T]]|C|$]]|$1|D|3|-4|5|6|7|U|8|@]|9|@]|C|$]]]|E|$F|$5|G|H|I|C|$J|K]]|L|$5|G|H|I|C|$J|M]]]]

There is also a simple way of doing this using <a href="https://guava.dev/releases/snapshot-jre/api/docs/com/google/common/collect/Maps.html#uniqueIndex-java.lang.Iterable-com.google.common.base.Function-" rel="nofollow noreferrer">Maps.uniqueIndex(...)</a> from Google <a href="/questions/tagged/guava" class="post-tag" title="show questions tagged &#39;guava&#39;" rel="tag">guava</a> libraries

blocks|key|1096798|text|使用java-8，您可以使用streams和Collectors类在一行中完成此操作。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1096799|Map<String,+Item>+map+=+
++++list.stream().collect(Collectors.toMap(Item::getKey,+item+->+item));|code-block|syntax|javascript|1096800|简短的演示：|1096801|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+%2B+"+=>+"+%2B+v));
++++}
}
class+Item+{

++++private+final+int+i;

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

++++public+String+getKey(){
++++++++return+"Key-"%2Bi;
++++}

++++@Override
++++public+String+toString()+{
++++++++return+"Item+[i="+%2B+i+%2B+"]";
++++}
}|1096802|输出：|1096803|Key-1+=>+Item+[i=1]
Key-2+=>+Item+[i=2]
Key-3+=>+Item+[i=3]
Key-4+=>+Item+[i=4]|1096804|正如注释中所指出的，您可以使用Function.identity()而不是item+->+item，尽管我发现i+->+i相当明确。|1096805|为了完整，请注意，如果您的函数不是双射的，则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，计算它的结果为模3：|1096806|List<Integer>+intList+=+Arrays.asList(1,+2,+3,+4,+5,+6);
Map<String,+Integer>+map+=+
++++intList.stream().collect(toMap(i+->+String.valueOf(i+%25+3),+i+->+i));|1096807|当运行这段代码时，你会得到一个错误提示java.lang.IllegalStateException:+Duplicate+key+1。这是因为1%253与4%253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。|1096808|下面是一个求和值的方法；可以用方法引用Integer::sum替换的(i1,+i2)+->+i1+%2B+i2;。|1096809|Map<String,+Integer>+map+=+
++++intList.stream().collect(toMap(i+->+String.valueOf(i+%25+3),+
+++++++++++++++++++++++++++++++++++i+->+i,+
+++++++++++++++++++++++++++++++++++Integer::sum));|1096810|它现在输出：|1096811|0+=>+9+(i.e+3+%2B+6)
1+=>+5+(i.e+1+%2B+4)
2+=>+7+(i.e+2+%2B+5)|1096812|希望它能有所帮助！:)|1096813|entityMap|0|LINK|mutability|MUTABLE|url|/questions/tagged/java-8|1|https://docs.oracle.com/javase/8/docs/api/java/util/stream/Stream.html|2|https://docs.oracle.com/javase/8/docs/api/java/util/stream/Collectors.html^0|M|A|2|6|0|E|7|1|M|A|2|0|0|0|0|0|0|F|J|11|C|1J|6|0|16|4|0|0|J|1C|0|J|C|Y|K|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|1M|8|@$9|1N|A|1O|B|C]]|D|@$9|1P|A|1Q|1|1R]|$9|1S|A|1T|1|1U]|$9|1V|A|1W|1|1X]]|E|$]]|$1|F|3|G|5|H|7|1Y|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|1Z|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|20|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|21|8|@]|D|@]|E|$]]|$1|Q|3|R|5|H|7|22|8|@]|D|@]|E|$I|J]]|$1|S|3|T|5|6|7|23|8|@$9|24|A|25|B|C]|$9|26|A|27|B|C]|$9|28|A|29|B|C]]|D|@]|E|$]]|$1|U|3|V|5|6|7|2A|8|@$9|2B|A|2C|B|C]]|D|@]|E|$]]|$1|W|3|X|5|H|7|2D|8|@]|D|@]|E|$I|J]]|$1|Y|3|Z|5|6|7|2E|8|@$9|2F|A|2G|B|C]]|D|@]|E|$]]|$1|10|3|11|5|6|7|2H|8|@$9|2I|A|2J|B|C]|$9|2K|A|2L|B|C]]|D|@]|E|$]]|$1|12|3|13|5|H|7|2M|8|@]|D|@]|E|$I|J]]|$1|14|3|15|5|6|7|2N|8|@]|D|@]|E|$]]|$1|16|3|17|5|H|7|2O|8|@]|D|@]|E|$I|J]]|$1|18|3|19|5|6|7|2P|8|@]|D|@]|E|$]]|$1|1A|3|-4|5|6|7|2Q|8|@]|D|@]|E|$]]]|1B|$1C|$5|1D|1E|1F|E|$1G|1H]]|1I|$5|1D|1E|1F|E|$1G|1J]]|1K|$5|1D|1E|1F|E|$1G|1L]]]]

With <a href="/questions/tagged/java-8" class="post-tag" title="show questions tagged &#39;java-8&#39;" rel="tag">java-8</a>, you'll be able to do this in one line using <a href="https://docs.oracle.com/javase/8/docs/api/java/util/stream/Stream.html">streams</a>, and the <a href="https://docs.oracle.com/javase/8/docs/api/java/util/stream/Collectors.html"><code>Collectors</code></a> class.

<pre><code>Map&lt;String, Item&gt; map = 
 list.stream().collect(Collectors.toMap(Item::getKey, item -&gt; item));
</code></pre>

Short demo:

<pre><code>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&lt;Item&gt; list = IntStream.rangeClosed(1, 4)
 .mapToObj(Item::new)
 .collect(Collectors.toList()); //[Item [i=1], Item [i=2], Item [i=3], Item [i=4]]

 Map&lt;String, Item&gt; map = 
 list.stream().collect(Collectors.toMap(Item::getKey, item -&gt; item));

 map.forEach((k, v) -&gt; System.out.println(k + " =&gt; " + 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 + "]";
 }
}
</code></pre>

Output:

<pre><code>Key-1 =&gt; Item [i=1]
Key-2 =&gt; Item [i=2]
Key-3 =&gt; Item [i=3]
Key-4 =&gt; Item [i=4]
</code></pre>

As noted in comments, you can use <code>Function.identity()</code> instead of <code>item -&gt; item</code>, although I find <code>i -&gt; i</code> rather explicit.

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

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

When running this code, you'll get an error saying <code>java.lang.IllegalStateException: Duplicate key 1</code>. 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.

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

<pre><code>Map&lt;String, Integer&gt; map = 
 intList.stream().collect(toMap(i -&gt; String.valueOf(i % 3), 
 i -&gt; i, 
 Integer::sum));
</code></pre>

which now outputs:

<pre><code>0 =&gt; 9 (i.e 3 + 6)
1 =&gt; 5 (i.e 1 + 4)
2 =&gt; 7 (i.e 2 + 5)
</code></pre>

Hope it helps! :)

blocks|key|1096867|text|从Java8开始，使用Collectors.toMap收集器的answer+by+@ZouZou无疑是解决这个问题的惯用方法。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1096868|由于这是一项非常常见的任务，我们可以将其转换为静态实用程序。|1096869|这样，解决方案就真正变成了一行程序。|1096870|/**
+*+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()));
}|code-block|syntax|javascript|1096871|下面是如何在List<Student>上使用它|1096872|Map<Long,+Student>+studentsById+=+toMapBy(students,+Student::getId);|1096873|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/a/20887747^0|B|G|V|H|0|0|0|0|0|6|D|0|0^^$0|@$1|2|3|4|5|6|7|10|8|@$9|11|A|12|B|C]]|D|@$9|13|A|14|1|15]]|E|$]]|$1|F|3|G|5|6|7|16|8|@]|D|@]|E|$]]|$1|H|3|I|5|6|7|17|8|@]|D|@]|E|$]]|$1|J|3|K|5|L|7|18|8|@]|D|@]|E|$M|N]]|$1|O|3|P|5|6|7|19|8|@$9|1A|A|1B|B|C]]|D|@]|E|$]]|$1|Q|3|R|5|L|7|1C|8|@]|D|@]|E|$M|N]]|$1|S|3|-4|5|6|7|1D|8|@]|D|@]|E|$]]]|T|$U|$5|V|W|X|E|$Y|Z]]]]

Since Java 8, the <a href="https://stackoverflow.com/a/20887747">answer by @ZouZou</a> using the <code>Collectors.toMap</code> collector is certainly the idiomatic way to solve this problem.

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

That way the solution truly becomes a one-liner.

<pre><code>/**
 * 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 &lt;K, T&gt; Map&lt;K, T&gt; toMapBy(List&lt;T&gt; list,
 Function&lt;? super T, ? extends K&gt; mapper) {
 return list.stream().collect(Collectors.toMap(mapper, Function.identity()));
}
</code></pre>

And here's how you would use it on a <code>List&lt;Student&gt;</code>:

<pre><code>Map&lt;Long, Student&gt; studentsById = toMapBy(students, Student::getId);
</code></pre>

blocks|key|1096918|text|没有java-8，您将能够在一行Commons集合和Closure类中完成此操作|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1096919|List<Item>+list;
@SuppressWarnings("unchecked")
Map<Key,+Item>+map++=+new+HashMap<Key,+Item>>(){{
++++CollectionUtils.forAllDo(list,+new+Closure()+{
++++++++@Override
++++++++public+void+execute(Object+input)+{
++++++++++++Item+item+=+(Item)+input;
++++++++++++put(i.getKey(),+item);
++++++++}
++++});
}};|code-block|syntax|javascript|1096920|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Without java-8, you'll be able to do this in one line Commons collections, and the Closure class

<pre class="lang-java prettyprint-override"><code>List&lt;Item&gt; list;
@SuppressWarnings("unchecked")
Map&lt;Key, Item&gt; map = new HashMap&lt;Key, Item&gt;&gt;(){{
 CollectionUtils.forAllDo(list, new Closure() {
 @Override
 public void execute(Object input) {
 Item item = (Item) input;
 put(i.getKey(), item);
 }
 });
}};
</code></pre>

blocks|key|1096950|text|将对象的List<?>转换为Map<k,+v>的Java8示例|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1096951|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+:+"+%2B+result1);

//example+2
Map<Integer,+String>+result2+=+list.stream().collect(
++++Collectors.toMap(x+->+x.getId(),+x+->+x.getName()));|code-block|syntax|javascript|1096952|代码复制自：|1096953|https://www.mkyong.com/java8/java-8-convert-list-to-map/|1096954|entityMap|0|LINK|mutability|MUTABLE|url^0|4|7|E|9|0|0|0|0|1K|0|0^^$0|@$1|2|3|4|5|6|7|V|8|@$9|W|A|X|B|C]|$9|Y|A|Z|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|10|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|11|8|@]|D|@]|E|$]]|$1|M|3|N|5|6|7|12|8|@]|D|@$9|13|A|14|1|15]]|E|$]]|$1|O|3|-4|5|6|7|16|8|@]|D|@]|E|$]]]|P|$Q|$5|R|S|T|E|$U|N]]]]

A Java 8 example to convert a <code>List&lt;?&gt;</code> of objects into a <code>Map&lt;k, v&gt;</code>:

<pre><code>List&lt;Hosting&gt; list = new ArrayList&lt;&gt;();
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&lt;Integer, String&gt; result1 = list.stream().collect(
 Collectors.toMap(Hosting::getId, Hosting::getName));

System.out.println("Result 1 : " + result1);

//example 2
Map&lt;Integer, String&gt; result2 = list.stream().collect(
 Collectors.toMap(x -&gt; x.getId(), x -&gt; x.getName()));
</code></pre>

Code copied from: 
<a href="https://www.mkyong.com/java8/java-8-convert-list-to-map/" rel="nofollow noreferrer">https://www.mkyong.com/java8/java-8-convert-list-to-map/</a>

blocks|key|1096986|text|您可以利用Java+8的streams+API。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1096987|public+class+ListToMap+{

++public+static+void+main(String[]+args)+{
++++List<User>+items+=+Arrays.asList(new+User("One"),+new+User("Two"),+new+User("Three"));

++++Map<String,+User>+map+=+createHashMap(items);
++++for(String+key+:+map.keySet())+{
++++++System.out.println(key+%2B"+:+"%2Bmap.get(key));
++++}
++}

++public+static+Map<String,+User>+createHashMap(List<User>+items)+{
++++Map<String,+User>+map+=+items.stream().collect(Collectors.toMap(User::getId,+Function.identity()));
++++return+map;
++}
}|code-block|syntax|javascript|1096988|更多详情请访问：http://codecramp.com/java-8-streams-api-convert-list-map/|offset|length|1096989|entityMap|0|LINK|mutability|MUTABLE|url|http://codecramp.com/java-8-streams-api-convert-list-map/^0|0|0|8|1L|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|U|8|@]|9|@$I|V|J|W|1|X]]|A|$]]|$1|K|3|-4|5|6|7|Y|8|@]|9|@]|A|$]]]|L|$M|$5|N|O|P|A|$Q|R]]]]

You can leverage the streams API of Java 8.

<pre><code>public class ListToMap {

 public static void main(String[] args) {
 List&lt;User&gt; items = Arrays.asList(new User("One"), new User("Two"), new User("Three"));

 Map&lt;String, User&gt; map = createHashMap(items);
 for(String key : map.keySet()) {
 System.out.println(key +" : "+map.get(key));
 }
 }

 public static Map&lt;String, User&gt; createHashMap(List&lt;User&gt; items) {
 Map&lt;String, User&gt; map = items.stream().collect(Collectors.toMap(User::getId, Function.identity()));
 return map;
 }
}
</code></pre>

For more details visit: <a href="http://codecramp.com/java-8-streams-api-convert-list-map/" rel="nofollow noreferrer">http://codecramp.com/java-8-streams-api-convert-list-map/</a>

blocks|key|882106|text|简短而甜蜜。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|882107|使用Java+8，您可以执行以下操作：|882108|Map<Key,+Value>+result=+results
+++++++++++++++++++++++.stream()
+++++++++++++++++++++++.collect(Collectors.toMap(Value::getName,Function.identity()));|code-block|syntax|javascript|882109|Value可以是您使用的任何对象。|offset|length|style|CODE|882110|entityMap^0|0|0|0|0|5|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|R|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|S|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|T|8|@$K|U|L|V|M|N]]|9|@]|A|$]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

Short and sweet.
Using Java 8 you can do following :
<pre><code>Map&lt;Key, Value&gt; result= results
 .stream()
 .collect(Collectors.toMap(Value::getName,Function.identity()));
</code></pre>
<code>Value</code> can be any object you use.

blocks|key|882895|text|Apache+Commons+MapUtils.populateMap|type|unstyled|depth|inlineStyleRanges|entityRanges|data|882896|如果您不使用Java8，并且出于某种原因不想使用显式循环，那么可以尝试使用Apache+Commons中的MapUtils.populateMap。|offset|length|style|CODE|882897|MapUtils.populateMap|882898|假设您有一个Pair列表。|882899|List<ImmutablePair<String,+String>>+pairs+=+ImmutableList.of(
++++new+ImmutablePair<>("A",+"aaa"),
++++new+ImmutablePair<>("B",+"bbb")
);|code-block|syntax|javascript|882900|现在，您需要一个指向Pair对象的Pair键的映射。|882901|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);|882902|给出输出：|882903|{A=(A,aaa),+B=(B,bbb)}|882904|也就是说，for循环可能更容易理解。(下面给出了相同的输出)：|882905|Map<String,+Pair<String,+String>>+map+=+new+HashMap<>();
for+(Pair<String,+String>+pair+:+pairs)+{
++map.put(pair.getKey(),+pair);
}
System.out.println(map);|882906|entityMap|0|LINK|mutability|MUTABLE|url|https://commons.apache.org/proper/commons-collections/apidocs/org/apache/commons/collections4/MapUtils.html#populateMap(java.util.Map,%2520java.lang.Iterable,%2520org.apache.commons.collections4.Transformer)^0|0|1H|K|0|0|K|0|0|6|4|0|0|A|4|H|4|0|0|0|0|5|3|0|0^^$0|@$1|2|3|4|5|6|7|1A|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|1B|8|@$D|1C|E|1D|F|G]]|9|@]|A|$]]|$1|H|3|I|5|6|7|1E|8|@]|9|@$D|1F|E|1G|1|1H]]|A|$]]|$1|J|3|K|5|6|7|1I|8|@$D|1J|E|1K|F|G]]|9|@]|A|$]]|$1|L|3|M|5|N|7|1L|8|@]|9|@]|A|$O|P]]|$1|Q|3|R|5|6|7|1M|8|@$D|1N|E|1O|F|G]|$D|1P|E|1Q|F|G]]|9|@]|A|$]]|$1|S|3|T|5|N|7|1R|8|@]|9|@]|A|$O|P]]|$1|U|3|V|5|6|7|1S|8|@]|9|@]|A|$]]|$1|W|3|X|5|N|7|1T|8|@]|9|@]|A|$O|P]]|$1|Y|3|Z|5|6|7|1U|8|@$D|1V|E|1W|F|G]]|9|@]|A|$]]|$1|10|3|11|5|N|7|1X|8|@]|9|@]|A|$O|P]]|$1|12|3|-4|5|6|7|1Y|8|@]|9|@]|A|$]]]|13|$14|$5|15|16|17|A|$18|19]]]]

<h2>Apache Commons MapUtils.populateMap</h2>
If you don't use Java 8 and you don't want to use a explicit loop for some reason, try <code>MapUtils.populateMap</code> from Apache Commons.
<a href="https://commons.apache.org/proper/commons-collections/apidocs/org/apache/commons/collections4/MapUtils.html#populateMap(java.util.Map,%20java.lang.Iterable,%20org.apache.commons.collections4.Transformer)" rel="nofollow noreferrer">MapUtils.populateMap</a>
Say you have a list of <code>Pair</code>s.
<pre><code>List&lt;ImmutablePair&lt;String, String&gt;&gt; pairs = ImmutableList.of(
 new ImmutablePair&lt;&gt;(&quot;A&quot;, &quot;aaa&quot;),
 new ImmutablePair&lt;&gt;(&quot;B&quot;, &quot;bbb&quot;)
);
</code></pre>
And you now want a Map of the <code>Pair</code>'s key to the <code>Pair</code> object.
<pre><code>Map&lt;String, Pair&lt;String, String&gt;&gt; map = new HashMap&lt;&gt;();
MapUtils.populateMap(map, pairs, new Transformer&lt;Pair&lt;String, String&gt;, String&gt;() {

 @Override
 public String transform(Pair&lt;String, String&gt; input) {
 return input.getKey();
 }
});

System.out.println(map);
</code></pre>
gives output:
<pre><code>{A=(A,aaa), B=(B,bbb)}
</code></pre>
That being said, a <code>for</code> loop is maybe easier to understand. (This below gives the same output):
<pre><code>Map&lt;String, Pair&lt;String, String&gt;&gt; map = new HashMap&lt;&gt;();
for (Pair&lt;String, String&gt; pair : pairs) {
 map.put(pair.getKey(), pair);
}
System.out.println(map);
</code></pre>

blocks|key|1097777|text|如前所述，在java-8中，我们有一个由Collectors提供的简明解决方案：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1097778|++list.stream().collect(
+++++++++groupingBy(Item::getKey)
++++++++)|code-block|syntax|javascript|1097779|此外，您还可以嵌套多个组，将另一个groupingBy方法作为第二个参数进行传递：|1097780|++list.stream().collect(
+++++++++groupingBy(Item::getKey,+groupingBy(Item::getOtherKey))
++++++++)|1097781|这样，我们就有了多层次的地图，就像这样：Map<key,+Map<key,+List<Item>>>|offset|length|style|CODE|1097782|entityMap^0|0|0|0|0|K|U|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|U|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|W|8|@$M|X|N|Y|O|P]]|9|@]|A|$]]|$1|Q|3|-4|5|6|7|Z|8|@]|9|@]|A|$]]]|R|$]]

like already said, in java-8 we have the concise solution by Collectors:

<pre><code> list.stream().collect(
 groupingBy(Item::getKey)
 )
</code></pre>

and also, you can nest multiple group passing an other groupingBy method as second parameter:

<pre><code> list.stream().collect(
 groupingBy(Item::getKey, groupingBy(Item::getOtherKey))
 )
</code></pre>

In this way, we'll have multi level map, like this: <code>Map&lt;key, Map&lt;key, List&lt;Item&gt;&gt;&gt;</code>

blocks|key|882984|text|使用java-8流|type|unstyled|depth|inlineStyleRanges|entityRanges|data|882985|Map<Integer,+String>+map+=+results.stream().collect(Collectors.toMap(e+->+((Integer)+e[0]),+e+->+(String)+e[1]));|code-block|syntax|javascript|882986|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Using java-8 streams
<pre class="lang-java prettyprint-override"><code>Map&lt;Integer, String&gt; map = results.stream().collect(Collectors.toMap(e -&gt; ((Integer) e[0]), e -&gt; (String) e[1]));
</code></pre>

Recently I have conversation with a colleague about what would be the optimal way to convert <code>List</code> to <code>Map</code> in Java and if there any specific benefits of doing so.
I want to know optimal conversion approach and would really appreciate if any one can guide me.
Is this good approach:
<pre><code>List&lt;Object[]&gt; results;
Map&lt;Integer, String&gt; resultsMap = new HashMap&lt;Integer, String&gt;();
for (Object[] o : results) {
 resultsMap.put((Integer) o[0], (String) o[1]);
}
</code></pre>

How to convert List to Map?

Java

最近，我与一位同事讨论了在Java语言中将List转换为Map的最佳方法，以及这样做是否有任何特定的好处。我想知道最优转换方法，如果有人能指导我，我将不胜感激。这是一种好方法吗：List<Object[]> results;Map<Integer, String> resultsMap = new HashMap<In...

问如何将List转换为Map？
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