blocks|key|3219106|text|基于文档，它非常简单，如：|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|3219107|Map<String,+Choice>+result+=
++++choices.stream().collect(Collectors.toMap(Choice::getName,
++++++++++++++++++++++++++++++++++++++++++++++Function.identity()));|code-block|syntax|javascript|3219108|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.oracle.com/javase/8/docs/api/java/util/stream/Collectors.html#toMap-java.util.function.Function-java.util.function.Function-^0|2|2|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]]]]

Based on <a href="http://docs.oracle.com/javase/8/docs/api/java/util/stream/Collectors.html#toMap-java.util.function.Function-java.util.function.Function-" rel="noreferrer"><code>Collectors</code> documentation</a> it's as simple as:

<pre><code>Map&lt;String, Choice&gt; result =
 choices.stream().collect(Collectors.toMap(Choice::getName,
 Function.identity()));
</code></pre>

blocks|key|3219157|text|如果您的密钥是，而不是，则应该将其转换为Map<String,+List<Choice>>而不是Map<String,+Choice>。|type|unstyled|depth|inlineStyleRanges|offset|length|style|BOLD|CODE|entityRanges|data|3219158|Map<String,+List<Choice>>+result+=
+choices.stream().collect(Collectors.groupingBy(Choice::getName));|code-block|syntax|javascript|3219159|entityMap^0|7|4|K|P|1C|J|0|0^^$0|@$1|2|3|4|5|6|7|N|8|@$9|O|A|P|B|C]|$9|Q|A|R|B|D]|$9|S|A|T|B|D]]|E|@]|F|$]]|$1|G|3|H|5|I|7|U|8|@]|E|@]|F|$J|K]]|$1|L|3|-4|5|6|7|V|8|@]|E|@]|F|$]]]|M|$]]

If your key is NOT guaranteed to be unique for all elements in the list, you should convert it to a <code>Map&lt;String, List&lt;Choice&gt;&gt;</code> instead of a <code>Map&lt;String, Choice&gt;</code>

<pre><code>Map&lt;String, List&lt;Choice&gt;&gt; result =
 choices.stream().collect(Collectors.groupingBy(Choice::getName));
</code></pre>

blocks|key|2137633|text|使用getName()作为键，使用Choice本身作为映射的值：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2137634|Map<String,+Choice>+result+=
++++choices.stream().collect(Collectors.toMap(Choice::getName,+c+->+c));|code-block|syntax|javascript|2137635|entityMap^0|2|9|H|6|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]|$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|R|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|S|8|@]|D|@]|E|$]]]|L|$]]

Use <code>getName()</code> as the key and <code>Choice</code> itself as the value of the map:

<pre><code>Map&lt;String, Choice&gt; result =
 choices.stream().collect(Collectors.toMap(Choice::getName, c -&gt; c));
</code></pre>

blocks|key|647435|text|列出的大多数答案都忽略了列表中有重复项的情况。在这种情况下，答案将抛出IllegalStateException。请参考下面的代码来处理列表重复的：|type|unstyled|depth|inlineStyleRanges|offset|length|style|BOLD|CODE|entityRanges|data|647436|public+Map<String,+Choice>+convertListToMap(List<Choice>+choices)+{
++++return+choices.stream()
++++++++.collect(Collectors.toMap(Choice::getName,+choice+->+choice,
++++++++++++(oldValue,+newValue)+->+newValue));
++}|code-block|syntax|javascript|647437|entityMap^0|C|7|1T|8|Z|L|0|0^^$0|@$1|2|3|4|5|6|7|N|8|@$9|O|A|P|B|C]|$9|Q|A|R|B|C]|$9|S|A|T|B|D]]|E|@]|F|$]]|$1|G|3|H|5|I|7|U|8|@]|E|@]|F|$J|K]]|$1|L|3|-4|5|6|7|V|8|@]|E|@]|F|$]]]|M|$]]

Most of the answers listed, miss a case when the list has duplicate items. In that case there answer will throw <code>IllegalStateException</code>. Refer the below code to handle list duplicates as well:

<pre><code>public Map&lt;String, Choice&gt; convertListToMap(List&lt;Choice&gt; choices) {
 return choices.stream()
 .collect(Collectors.toMap(Choice::getName, choice -&gt; choice,
 (oldValue, newValue) -&gt; newValue));
 }
</code></pre>

blocks|key|648514|text|这里还有一个，以防您不想使用Collectors.toMap()|type|unstyled|depth|inlineStyleRanges|entityRanges|data|648515|Map<String,+Choice>+result+=
+++choices.stream().collect(HashMap<String,+Choice>::new,+
+++++++++++++++++++++++++++(m,+c)+->+m.put(c.getName(),+c),
+++++++++++++++++++++++++++(m,+u)+->+{});|code-block|syntax|javascript|648516|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|$]]

Here's another one in case you don't want to use Collectors.toMap()

<pre><code>Map&lt;String, Choice&gt; result =
 choices.stream().collect(HashMap&lt;String, Choice&gt;::new, 
 (m, c) -&gt; m.put(c.getName(), c),
 (m, u) -&gt; {});
</code></pre>

blocks|key|3219265|text|还有一个简单的选择|type|blockquote|depth|inlineStyleRanges|entityRanges|data|3219266|Map<String,Choice>+map+=+new+HashMap<>();
choices.forEach(e->map.put(e.getName(),e));|code-block|syntax|javascript|3219267|unstyled|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|J|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|K|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|H|7|L|8|@]|9|@]|A|$]]]|I|$]]

<blockquote>
 One more option in simple way
</blockquote>

<pre><code>Map&lt;String,Choice&gt; map = new HashMap&lt;&gt;();
choices.forEach(e-&gt;map.put(e.getName(),e));
</code></pre>

blocks|key|3219375|text|例如，如果要将对象字段转换为：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3219376|示例对象：|3219377|class+Item{
++++++++private+String+code;
++++++++private+String+name;

++++++++public+Item(String+code,+String+name)+{
++++++++++++this.code+=+code;
++++++++++++this.name+=+name;
++++++++}

++++++++//getters+and+setters
++++}|code-block|syntax|javascript|3219378|并将操作转换列表为Map：|3219379|List<Item>+list+=+new+ArrayList<>();
list.add(new+Item("code1",+"name1"));
list.add(new+Item("code2",+"name2"));

Map<String,String>+map+=+list.stream()
+++++.collect(Collectors.toMap(Item::getCode,+Item::getName));|3219380|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|P|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|Q|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|R|8|@]|9|@]|A|$]]|$1|K|3|L|5|F|7|S|8|@]|9|@]|A|$G|H]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

For example, if you want convert object fields to map:

Example object:

<pre><code>class Item{
 private String code;
 private String name;

 public Item(String code, String name) {
 this.code = code;
 this.name = name;
 }

 //getters and setters
 }
</code></pre>

And operation convert List To Map:

<pre><code>List&lt;Item&gt; list = new ArrayList&lt;&gt;();
list.add(new Item("code1", "name1"));
list.add(new Item("code2", "name2"));

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

blocks|key|2137680|text|如果您不介意使用第三方库，AOL的独眼眼-反应库(公开我是贡献者)对所有JDK收藏类型都有扩展，包括列表和地图。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|2137681|ListX<Choices>+choices;
Map<String,+Choice>+map+=+choices.toMap(c->+c.getName(),c->c);|code-block|syntax|javascript|2137682|entityMap|0|LINK|mutability|MUTABLE|url|https://github.com/aol/cyclops-react|1|https://medium.com/@johnmcclean/extending-jdk-8-collections-8ae8d43dd75e#.j5lxtrqze|2|http://static.javadoc.io/com.aol.simplereact/cyclops-react/1.0.0-RC1/com/aol/cyclops/data/collections/extensions/standard/ListX.html|3|http://static.javadoc.io/com.aol.simplereact/cyclops-react/1.0.0-RC1/com/aol/cyclops/data/collections/extensions/standard/MapX.html^0|H|6|0|10|5|1|1E|2|2|1H|2|3|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@$A|X|B|Y|1|Z]|$A|10|B|11|1|12]|$A|13|B|14|1|15]|$A|16|B|17|1|18]]|C|$]]|$1|D|3|E|5|F|7|19|8|@]|9|@]|C|$G|H]]|$1|I|3|-4|5|6|7|1A|8|@]|9|@]|C|$]]]|J|$K|$5|L|M|N|C|$O|P]]|Q|$5|L|M|N|C|$O|R]]|S|$5|L|M|N|C|$O|T]]|U|$5|L|M|N|C|$O|V]]]]

If you don't mind using 3rd party libraries, AOL's <a href="https://github.com/aol/cyclops-react">cyclops-react</a> lib (disclosure I am a contributor) has extensions for all <a href="https://medium.com/@johnmcclean/extending-jdk-8-collections-8ae8d43dd75e#.j5lxtrqze">JDK Collection</a> types, including <a href="http://static.javadoc.io/com.aol.simplereact/cyclops-react/1.0.0-RC1/com/aol/cyclops/data/collections/extensions/standard/ListX.html">List</a> and <a href="http://static.javadoc.io/com.aol.simplereact/cyclops-react/1.0.0-RC1/com/aol/cyclops/data/collections/extensions/standard/MapX.html">Map</a>.

<pre><code>ListX&lt;Choices&gt; choices;
Map&lt;String, Choice&gt; map = choices.toMap(c-&gt; c.getName(),c-&gt;c);
</code></pre>

blocks|key|648540|text|您可以使用IntStream创建索引流，然后将它们转换为Map：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|648541|Map<Integer,Item>+map+=+
IntStream.range(0,items.size())
+++++++++.boxed()
+++++++++.collect(Collectors.toMap+(i+->+i,+i+->+items.get(i)));|code-block|syntax|javascript|648542|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|$]]

You can create a Stream of the indices using an IntStream and then convert them to a Map :

<pre><code>Map&lt;Integer,Item&gt; map = 
IntStream.range(0,items.size())
 .boxed()
 .collect(Collectors.toMap (i -&gt; i, i -&gt; items.get(i)));
</code></pre>

blocks|key|647281|text|我试图这样做，并发现，使用上面的答案时，当使用Functions.identity()作为映射的键时，我在使用像this::localMethodName这样的本地方法时遇到了问题，因为输入问题。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|647282|在本例中，Functions.identity()实际上对输入做了一些操作，因此该方法只能通过返回Object并接受Object的参数来工作。|647283|为了解决这个问题，我放弃了Functions.identity()，转而使用了s->s。|647284|因此，在我的例子中，我的代码列出了目录中的所有目录，每个目录都使用目录名作为映射的键，然后使用目录名调用方法并返回一组项，如下所示：|647285|Map<String,+Collection<ItemType>>+items+=+Arrays.stream(itemFilesDir.listFiles(File::isDirectory))
.map(File::getName)
.collect(Collectors.toMap(s->s,+this::retrieveBrandItems));|code-block|syntax|javascript|647286|entityMap^0|N|K|1K|L|0|5|K|1D|6|1M|6|0|D|K|13|4|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]|$9|V|A|W|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|X|8|@$9|Y|A|Z|B|C]|$9|10|A|11|B|C]|$9|12|A|13|B|C]]|D|@]|E|$]]|$1|H|3|I|5|6|7|14|8|@$9|15|A|16|B|C]|$9|17|A|18|B|C]]|D|@]|E|$]]|$1|J|3|K|5|6|7|19|8|@]|D|@]|E|$]]|$1|L|3|M|5|N|7|1A|8|@]|D|@]|E|$O|P]]|$1|Q|3|-4|5|6|7|1B|8|@]|D|@]|E|$]]]|R|$]]

I was trying to do this and found that, using the answers above, when using <code>Functions.identity()</code> for the key to the Map, then I had issues with using a local method like <code>this::localMethodName</code> to actually work because of typing issues.

<code>Functions.identity()</code> actually does something to the typing in this case so the method would only work by returning <code>Object</code> and accepting a param of <code>Object</code>

To solve this, I ended up ditching <code>Functions.identity()</code> and using <code>s-&gt;s</code> instead.

So my code, in my case to list all directories inside a directory, and for each one use the name of the directory as the key to the map and then call a method with the directory name and return a collection of items, looks like:

<pre><code>Map&lt;String, Collection&lt;ItemType&gt;&gt; items = Arrays.stream(itemFilesDir.listFiles(File::isDirectory))
.map(File::getName)
.collect(Collectors.toMap(s-&gt;s, this::retrieveBrandItems));
</code></pre>

blocks|key|3219484|text|我将编写如何使用泛型、和反向控制将列表转换为映射。只是通用方法！|type|unstyled|depth|inlineStyleRanges|offset|length|style|BOLD|entityRanges|data|3219485|，可能是，我们有整数的列表，或对象列表。因此，问题是:映射的键应该是什么？|3219486|创建接口|3219487|public+interface+KeyFinder<K,+E>+{
++++K+getKey(E+e);
}|code-block|syntax|javascript|3219488|现在使用控制反转：|3219489|++static+<K,+E>+Map<K,+E>+listToMap(List<E>+list,+KeyFinder<K,+E>+finder)+{
++++++++return++list.stream().collect(Collectors.toMap(e+->+finder.getKey(e)+,+e+->+e));
++++}|3219490|例如，如果我们有书籍的对象，这个类将选择映射的键。|3219491|public+class+BookKeyFinder+implements+KeyFinder<Long,+Book>+{
++++@Override
++++public+Long+getKey(Book+e)+{
++++++++return+e.getPrice()
++++}
}|3219492|entityMap^0|8|3|C|4|R|5|0|0|4|B|3|F|M|0|0|4|0|0|0|9|0|0|0|P|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@$9|Z|A|10|B|C]|$9|11|A|12|B|C]|$9|13|A|14|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|15|8|@$9|16|A|17|B|C]|$9|18|A|19|B|C]|$9|1A|A|1B|B|C]]|D|@]|E|$]]|$1|H|3|I|5|6|7|1C|8|@$9|1D|A|1E|B|C]]|D|@]|E|$]]|$1|J|3|K|5|L|7|1F|8|@]|D|@]|E|$M|N]]|$1|O|3|P|5|6|7|1G|8|@$9|1H|A|1I|B|C]]|D|@]|E|$]]|$1|Q|3|R|5|L|7|1J|8|@]|D|@]|E|$M|N]]|$1|S|3|T|5|6|7|1K|8|@$9|1L|A|1M|B|C]]|D|@]|E|$]]|$1|U|3|V|5|L|7|1N|8|@]|D|@]|E|$M|N]]|$1|W|3|-4|5|6|7|1O|8|@]|D|@]|E|$]]]|X|$]]

I will write how to convert list to map using generics and inversion of control. Just universal method!

Maybe we have list of Integers or list of objects. So the question is the following: what should be key of the map?

create interface

<pre><code>public interface KeyFinder&lt;K, E&gt; {
 K getKey(E e);
}
</code></pre>

now using inversion of control:

<pre><code> static &lt;K, E&gt; Map&lt;K, E&gt; listToMap(List&lt;E&gt; list, KeyFinder&lt;K, E&gt; finder) {
 return list.stream().collect(Collectors.toMap(e -&gt; finder.getKey(e) , e -&gt; e));
 }
</code></pre>

For example, if we have objects of book , this class is to choose key for the map 

<pre><code>public class BookKeyFinder implements KeyFinder&lt;Long, Book&gt; {
 @Override
 public Long getKey(Book e) {
 return e.getPrice()
 }
}
</code></pre>

blocks|key|648474|text|我使用这个语法|type|unstyled|depth|inlineStyleRanges|entityRanges|data|648475|Map<Integer,+List<Choice>>+choiceMap+=+
choices.stream().collect(Collectors.groupingBy(choice+->+choice.getName()));|code-block|syntax|javascript|648476|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|$]]

I use this syntax

<pre><code>Map&lt;Integer, List&lt;Choice&gt;&gt; choiceMap = 
choices.stream().collect(Collectors.groupingBy(choice -&gt; choice.getName()));
</code></pre>

blocks|key|647384|text|可以使用流来完成这一任务。为了消除显式使用Collectors的需要，可以静态地导入toMap+(正如有效Java第三版所建议的那样)。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|647385|import+static+java.util.stream.Collectors.toMap;

private+static+Map<String,+Choice>+nameMap(List<Choice>+choices)+{
++++return+choices.stream().collect(toMap(Choice::getName,+it+->+it));
}|code-block|syntax|javascript|647386|entityMap^0|L|A|16|5|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]|$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|R|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|S|8|@]|D|@]|E|$]]]|L|$]]

It's possible to use streams to do this. To remove the need to explicitly use <code>Collectors</code>, it's possible to import <code>toMap</code> statically (as recommended by Effective Java, third edition).

<pre><code>import static java.util.stream.Collectors.toMap;

private static Map&lt;String, Choice&gt; nameMap(List&lt;Choice&gt; choices) {
 return choices.stream().collect(toMap(Choice::getName, it -&gt; it));
}
</code></pre>

blocks|key|647472|text|另一种可能只出现在评论中：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|647473|Map<String,+Choice>+result+=
choices.stream().collect(Collectors.toMap(c+->+c.getName(),+c+->+c)));|code-block|syntax|javascript|647474|如果要使用子对象的参数作为键，则非常有用：|647475|Map<String,+Choice>+result+=
choices.stream().collect(Collectors.toMap(c+->+c.getUser().getName(),+c+->+c)));|647476|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|$]]

Another possibility only present in comments yet:
<pre><code>Map&lt;String, Choice&gt; result =
choices.stream().collect(Collectors.toMap(c -&gt; c.getName(), c -&gt; c)));
</code></pre>
Useful if you want to use a parameter of a sub-object as Key:
<pre><code>Map&lt;String, Choice&gt; result =
choices.stream().collect(Collectors.toMap(c -&gt; c.getUser().getName(), c -&gt; c)));
</code></pre>

blocks|key|648494|text|Map<String,+Set<String>>+collect+=+Arrays.asList(Locale.getAvailableLocales()).stream().collect(Collectors
++++++++++++++++.toMap(l+->+l.getDisplayCountry(),+l+->+Collections.singleton(l.getDisplayLanguage())));|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|648495|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>Map&lt;String, Set&lt;String&gt;&gt; collect = Arrays.asList(Locale.getAvailableLocales()).stream().collect(Collectors
 .toMap(l -&gt; l.getDisplayCountry(), l -&gt; Collections.singleton(l.getDisplayLanguage())));
</code></pre>

blocks|key|648566|text|这可以通过两种方式来完成。让人成为我们将要用来演示它的类。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|648567|public+class+Person+{

++++private+String+name;
++++private+int+age;

++++public+String+getAge()+{
++++++++return+age;
++++}
}|code-block|syntax|javascript|648568|让Person成为要转换为地图的人的列表|offset|length|style|BOLD|648569|1.在列表中使用简单foreach和Lambda表达式|648570|Map<Integer,List<Person>>+mapPersons+=+new+HashMap<>();
persons.forEach(p->mapPersons.put(p.getAge(),p));|648571|2.在给定列表上定义的流上使用收集器。|648572|+Map<Integer,List<Person>>+mapPersons+=+
+++++++++++persons.stream().collect(Collectors.groupingBy(Person::getAge));|648573|entityMap^0|0|0|1|6|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|X|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Y|8|@$I|Z|J|10|K|L]]|9|@]|A|$]]|$1|M|3|N|5|6|7|11|8|@]|9|@]|A|$]]|$1|O|3|P|5|D|7|12|8|@]|9|@]|A|$E|F]]|$1|Q|3|R|5|6|7|13|8|@]|9|@]|A|$]]|$1|S|3|T|5|D|7|14|8|@]|9|@]|A|$E|F]]|$1|U|3|-4|5|6|7|15|8|@]|9|@]|A|$]]]|V|$]]

This can be done in 2 ways. Let person be the class we are going to use to demonstrate it.

<pre><code>public class Person {

 private String name;
 private int age;

 public String getAge() {
 return age;
 }
}
</code></pre>

Let persons be the list of Persons to be converted to the map

1.Using Simple foreach and a Lambda Expression on the List

<pre><code>Map&lt;Integer,List&lt;Person&gt;&gt; mapPersons = new HashMap&lt;&gt;();
persons.forEach(p-&gt;mapPersons.put(p.getAge(),p));
</code></pre>

2.Using Collectors on Stream defined on the given List.

<pre><code> Map&lt;Integer,List&lt;Person&gt;&gt; mapPersons = 
 persons.stream().collect(Collectors.groupingBy(Person::getAge));
</code></pre>

blocks|key|648529|text|以下是StreamEx的解决方案|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|648530|StreamEx.of(choices).toMap(Choice::getName,+c+->+c);|code-block|syntax|javascript|648531|entityMap|0|LINK|mutability|MUTABLE|url|https://github.com/amaembo/streamex^0|3|8|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]]]]

Here is solution by <a href="https://github.com/amaembo/streamex" rel="nofollow noreferrer">StreamEx</a>

<pre><code>StreamEx.of(choices).toMap(Choice::getName, c -&gt; c);
</code></pre>

blocks|key|2137860|text|Map<String,Choice>+map=list.stream().collect(Collectors.toMap(Choice::getName,+s->s));|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|2137861|甚至为我服务，|unstyled|2137862|Map<String,Choice>+map=++list1.stream().collect(()->+new+HashMap<String,Choice>(),+
++++++++++++(r,s)+->+r.put(s.getString(),s),(r,s)+->+r.putAll(s));|2137863|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|L|8|@]|9|@]|A|$]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$B|C]]|$1|I|3|-4|5|F|7|N|8|@]|9|@]|A|$]]]|J|$]]

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

Even serves this purpose for me,

<pre><code>Map&lt;String,Choice&gt; map= list1.stream().collect(()-&gt; new HashMap&lt;String,Choice&gt;(), 
 (r,s) -&gt; r.put(s.getString(),s),(r,s) -&gt; r.putAll(s));
</code></pre>

blocks|key|648639|text|如果必须重写同一键名的每个新值：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|648640|public+Map+<+String,+Choice+>+convertListToMap(List+<+Choice+>+choices)+{
++++return+choices.stream()
++++++++.collect(Collectors.toMap(Choice::getName,
++++++++++++Function.identity(),
++++++++++++(oldValue,+newValue)+-+>+newValue));
}|code-block|syntax|javascript|648641|如果必须将所有选项分组到一个名称的列表中：|648642|public+Map+<+String,+Choice+>+convertListToMap(List+<+Choice+>+choices)+{
++++return+choices.stream().collect(Collectors.groupingBy(Choice::getName));
}|648643|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|$]]

If every new value for the same key name has to be overridden:

<pre class="lang-java prettyprint-override"><code>public Map &lt; String, Choice &gt; convertListToMap(List &lt; Choice &gt; choices) {
 return choices.stream()
 .collect(Collectors.toMap(Choice::getName,
 Function.identity(),
 (oldValue, newValue) - &gt; newValue));
}
</code></pre>

If all choices have to be grouped in a list for a name:

<pre class="lang-java prettyprint-override"><code>public Map &lt; String, Choice &gt; convertListToMap(List &lt; Choice &gt; choices) {
 return choices.stream().collect(Collectors.groupingBy(Choice::getName));
}
</code></pre>

blocks|key|647466|text|List<V>+choices;+//+your+list
Map<K,V>+result+=+choices.stream().collect(Collectors.toMap(choice::getKey(),choice));
//assuming+class+"V"+has+a+method+to+get+the+key,+this+method+must+handle+case+of+duplicates+too+and+provide+a+unique+key.|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|647467|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>List&lt;V&gt; choices; // your list
Map&lt;K,V&gt; result = choices.stream().collect(Collectors.toMap(choice::getKey(),choice));
//assuming class "V" has a method to get the key, this method must handle case of duplicates too and provide a unique key.
</code></pre>

blocks|key|648670|text|作为guava的替代方案，可以使用科特林-斯图利布|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|648671|private+Map<String,+Choice>+nameMap(List<Choice>+choices)+{
++++return+CollectionsKt.associateBy(choices,+Choice::getName);
}|code-block|syntax|javascript|648672|entityMap|0|LINK|mutability|MUTABLE|url|https://search.maven.org/artifact/org.jetbrains.kotlin/kotlin-stdlib^0|2|5|H|8|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]]|D|@$9|V|A|W|1|X]]|E|$]]|$1|F|3|G|5|H|7|Y|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Z|8|@]|D|@]|E|$]]]|L|$M|$5|N|O|P|E|$Q|R]]]]

As an alternative to <code>guava</code> one can use <a href="https://search.maven.org/artifact/org.jetbrains.kotlin/kotlin-stdlib" rel="nofollow noreferrer">kotlin-stdlib</a>

<pre><code>private Map&lt;String, Choice&gt; nameMap(List&lt;Choice&gt; choices) {
 return CollectionsKt.associateBy(choices, Choice::getName);
}
</code></pre>

blocks|key|2137980|text|List<Integer>+listA+=+new+ArrayList<>();
++++listA.add(1);
++++listA.add(5);
++++listA.add(3);
++++listA.add(4);+++++++


++++System.out.println(listA.stream().collect(Collectors.toMap(x+->x,+x->x)));|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|2137981|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>List&lt;Integer&gt; listA = new ArrayList&lt;&gt;();
 listA.add(1);
 listA.add(5);
 listA.add(3);
 listA.add(4); 


 System.out.println(listA.stream().collect(Collectors.toMap(x -&gt;x, x-&gt;x)));
</code></pre>

blocks|key|648657|text|String+array[]+=+{"ASDFASDFASDF","AA",+"BBB",+"CCCC",+"DD",+"EEDDDAD"};
++++List<String>+list+=+Arrays.asList(array);
++++Map<Integer,+String>+map+=+list.stream()
++++++++++++.collect(Collectors.toMap(s+->+s.length(),+s+->+s,+(x,+y)+->+{
++++++++++++++++System.out.println("Dublicate+key"+%2B+x);
++++++++++++++++return+x;
++++++++++++},()->+new+TreeMap<>((s1,s2)->s2.compareTo(s1))));
++++System.out.println(map);|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|648658|双倍键AA|unstyled|648659|{12=ASDFASDFASDF,+7=EEDDDAD,+4=CCCC,+3=BBB,+2=AA}|648660|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|L|8|@]|9|@]|A|$]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$B|C]]|$1|I|3|-4|5|F|7|N|8|@]|9|@]|A|$]]]|J|$]]

<pre><code>String array[] = {&quot;ASDFASDFASDF&quot;,&quot;AA&quot;, &quot;BBB&quot;, &quot;CCCC&quot;, &quot;DD&quot;, &quot;EEDDDAD&quot;};
 List&lt;String&gt; list = Arrays.asList(array);
 Map&lt;Integer, String&gt; map = list.stream()
 .collect(Collectors.toMap(s -&gt; s.length(), s -&gt; s, (x, y) -&gt; {
 System.out.println(&quot;Dublicate key&quot; + x);
 return x;
 },()-&gt; new TreeMap&lt;&gt;((s1,s2)-&gt;s2.compareTo(s1))));
 System.out.println(map);
</code></pre>
Dublicate key AA
<pre><code>{12=ASDFASDFASDF, 7=EEDDDAD, 4=CCCC, 3=BBB, 2=AA}
</code></pre>

I want to translate a List of objects into a Map using Java 8's streams and lambdas.

This is how I would write it in Java 7 and below.

<pre><code>private Map&lt;String, Choice&gt; nameMap(List&lt;Choice&gt; choices) {
 final Map&lt;String, Choice&gt; hashMap = new HashMap&lt;&gt;();
 for (final Choice choice : choices) {
 hashMap.put(choice.getName(), choice);
 }
 return hashMap;
}
</code></pre>

I can accomplish this easily using Java 8 and Guava but I would like to know how to do this without Guava.

In Guava:

<pre><code>private Map&lt;String, Choice&gt; nameMap(List&lt;Choice&gt; choices) {
 return Maps.uniqueIndex(choices, new Function&lt;Choice, String&gt;() {

 @Override
 public String apply(final Choice input) {
 return input.getName();
 }
 });
}
</code></pre>

And Guava with Java 8 lambdas.

<pre><code>private Map&lt;String, Choice&gt; nameMap(List&lt;Choice&gt; choices) {
 return Maps.uniqueIndex(choices, Choice::getName);
}
</code></pre>

Java 8 List<V> into Map<K, V>

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我希望使用Java 8的流和lambdas将对象列表转换为Map。这就是我用Java 7和更低版本编写它的方式。private Map<String, Choice> nameMap(List<Choice> choices) { final Map<String, Choice> hashMap = new HashMap<>(); for (final Choice

问Java 8 List<V> V> Map<K，V>
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