blocks|key|760080|text|使用LinkedHashMap，当您需要按位置检索时，将值转换为ArrayList。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|760081|LinkedHashMap<String,String>+linkedHashMap+=+new+LinkedHashMap<String,String>();
/*+Populate+*/
linkedHashMap.put("key0","value0");
linkedHashMap.put("key1","value1");
linkedHashMap.put("key2","value2");
/*+Get+by+position+*/
int+pos+=+1;
String+value+=+(new+ArrayList<String>(linkedHashMap.values())).get(pos);|code-block|syntax|javascript|760082|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|$]]

Use a LinkedHashMap and when you need to retrieve by position, convert the values into an ArrayList.

<pre><code>LinkedHashMap&lt;String,String&gt; linkedHashMap = new LinkedHashMap&lt;String,String&gt;();
/* Populate */
linkedHashMap.put("key0","value0");
linkedHashMap.put("key1","value1");
linkedHashMap.put("key2","value2");
/* Get by position */
int pos = 1;
String value = (new ArrayList&lt;String&gt;(linkedHashMap.values())).get(pos);
</code></pre>

blocks|key|760248|text|如果希望保持将元素添加到映射中的顺序，请使用LinkedHashMap，而不是只使用HashMap。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|760249|下面是一种方法，它允许您通过地图中的索引来获取值：|760250|public+Object+getElementByIndex(LinkedHashMap+map,int+index){
++++return+map.get(+(map.keySet().toArray())[+index+]+);
}|code-block|syntax|javascript|760251|entityMap^0|M|D|16|7|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]|$9|R|A|S|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|T|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|U|8|@]|D|@]|E|$K|L]]|$1|M|3|-4|5|6|7|V|8|@]|D|@]|E|$]]]|N|$]]

If you want to maintain the order in which you added the elements to the map, use <code>LinkedHashMap</code> as opposed to just <code>HashMap</code>.

Here is an approach that will allow you to get a value by its index in the map:

<pre><code>public Object getElementByIndex(LinkedHashMap map,int index){
 return map.get( (map.keySet().toArray())[ index ] );
}
</code></pre>

blocks|key|760362|text|如果出于某种原因，您必须使用hashMap，您可以将keySet转换为数组，并对数组中的键进行索引，以获得映射中的值，如下所示：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|760363|Object[]+keys+=+map.keySet().toArray();|code-block|syntax|javascript|760364|然后您可以访问地图，如下所示：|760365|map.get(keys[i]);|760366|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 you, for some reason, have to stick with the hashMap, you can convert the keySet to an array and index the keys in the array to get the values in the map like so:

<pre><code>Object[] keys = map.keySet().toArray();
</code></pre>

You can then access the map like:

<pre><code>map.get(keys[i]);
</code></pre>

blocks|key|975805|text|使用LinkedHashMap并使用此函数。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|975806|private+LinkedHashMap<Integer,+String>+map+=+new+LinkedHashMap<Integer,+String>();|code-block|syntax|javascript|975807|像这样定义和。|975808|private+Entry+getEntry(int+id){
++++++++Iterator+iterator+=+map.entrySet().iterator();
++++++++int+n+=+0;
++++++++while(iterator.hasNext()){
++++++++++++Entry+entry+=+(Entry)+iterator.next();
++++++++++++if(n+==+id){
++++++++++++++++return+entry;
++++++++++++}
++++++++++++n+%2B%2B;
++++++++}
++++++++return+null;
++++}|975809|该函数可以返回所选条目。|975810|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|S|8|@]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

Use LinkedHashMap and use this function. 

<pre><code>private LinkedHashMap&lt;Integer, String&gt; map = new LinkedHashMap&lt;Integer, String&gt;();
</code></pre>

Define like this and.

<pre><code>private Entry getEntry(int id){
 Iterator iterator = map.entrySet().iterator();
 int n = 0;
 while(iterator.hasNext()){
 Entry entry = (Entry) iterator.next();
 if(n == id){
 return entry;
 }
 n ++;
 }
 return null;
 }
</code></pre>

The function can return the selected entry.

blocks|key|976047|text|默认情况下，java+LinkedHasMap不支持按位置获取值。所以我建议使用定制的IndexedLinkedHashMap|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|976048|public+class+IndexedLinkedHashMap<K,+V>+extends+LinkedHashMap<K,+V>+{

++++private+ArrayList<K>+keysList+=+new+ArrayList<>();

++++public+void+add(K+key,+V+val)+{
++++++++super.put(key,+val);
++++++++keysList.add(key);
++++}

++++public+void+update(K+key,+V+val)+{
++++++++super.put(key,+val);
++++}

++++public+void+removeItemByKey(K+key)+{
++++++++super.remove(key);
++++++++keysList.remove(key);
++++}

++++public+void+removeItemByIndex(int+index)+{
++++++++super.remove(keysList.get(index));
++++++++keysList.remove(index);
++++}

++++public+V+getItemByIndex(int+i)+{
++++++++return+(V)+super.get(keysList.get(i));
++++}

++++public+int+getIndexByKey(K+key)+{
++++++++return+keysList.indexOf(key);
++++}
}|code-block|syntax|javascript|976049|然后，您可以像这样使用这个自定义的LinkedHasMap|976050|IndexedLinkedHashMap<String,UserModel>+indexedLinkedHashMap=new+IndexedLinkedHashMap<>();|976051|添加价值的步骤|976052|indexedLinkedHashMap.add("key1",UserModel);|976053|按索引执行getValue|976054|indexedLinkedHashMap.getItemByIndex(position);|976055|entityMap^0|17|K|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@$9|Z|A|10|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|11|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|12|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|13|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|14|8|@]|D|@]|E|$]]|$1|Q|3|R|5|H|7|15|8|@]|D|@]|E|$I|J]]|$1|S|3|T|5|6|7|16|8|@]|D|@]|E|$]]|$1|U|3|V|5|H|7|17|8|@]|D|@]|E|$I|J]]|$1|W|3|-4|5|6|7|18|8|@]|D|@]|E|$]]]|X|$]]

By default, java LinkedHasMap does not support for getting value by position. So I suggest go with customized <code>IndexedLinkedHashMap</code>

<pre><code>public class IndexedLinkedHashMap&lt;K, V&gt; extends LinkedHashMap&lt;K, V&gt; {

 private ArrayList&lt;K&gt; keysList = new ArrayList&lt;&gt;();

 public void add(K key, V val) {
 super.put(key, val);
 keysList.add(key);
 }

 public void update(K key, V val) {
 super.put(key, val);
 }

 public void removeItemByKey(K key) {
 super.remove(key);
 keysList.remove(key);
 }

 public void removeItemByIndex(int index) {
 super.remove(keysList.get(index));
 keysList.remove(index);
 }

 public V getItemByIndex(int i) {
 return (V) super.get(keysList.get(i));
 }

 public int getIndexByKey(K key) {
 return keysList.indexOf(key);
 }
}
</code></pre>

Then you can use this customized LinkedHasMap as

<pre><code>IndexedLinkedHashMap&lt;String,UserModel&gt; indexedLinkedHashMap=new IndexedLinkedHashMap&lt;&gt;();
</code></pre>

TO add Values 

<pre><code>indexedLinkedHashMap.add("key1",UserModel);
</code></pre>

To getValue by index

<pre><code>indexedLinkedHashMap.getItemByIndex(position);
</code></pre>

blocks|key|760007|text|我假设“位置”指的是将元素插入到HashMap中的顺序。在这种情况下，您需要使用LinkedHashMap。但是，LinkedHashMap没有提供访问器方法；您需要编写一个类似这样的方法|type|unstyled|depth|inlineStyleRanges|entityRanges|data|760008|public+Object+getElementAt(LinkedHashMap+map,+int+index)+{
++++for+(Map.Entry+entry+:+map.entrySet())+{
++++++++if+(index--+==+0)+{
++++++++++++return+entry.value();
++++++++}
++++}
++++return+null;
}|code-block|syntax|javascript|760009|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'm assuming by 'position' you're referring to the order in which you've inserted the elements into the HashMap. In that case you want to be using a LinkedHashMap. The LinkedHashMap doesn't offer an accessor method however; you will need to write one like

<pre><code>public Object getElementAt(LinkedHashMap map, int index) {
 for (Map.Entry entry : map.entrySet()) {
 if (index-- == 0) {
 return entry.value();
 }
 }
 return null;
}
</code></pre>

blocks|key|975932|text|另一种可行的方法是将映射值转换为数组，然后在索引处检索元素。使用以下方法在LinkedHashMap中对100,000个对象进行100,000个元素的索引搜索的测试运行产生了以下结果：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|975933|//My+answer:
public+Particle+getElementByIndex(LinkedHashMap<Point,+Particle>+map,int+index){
++++return+map.values().toArray(new+Particle[map.values().size()])[index];
}+//68+965+ms

//Syd+Lambert's+answer:
public+Particle+getElementByIndex(LinkedHashMap<Point,+Particle>+map,int+index){
++++return+map.get(+(map.keySet().toArray())[+index+]+);
}+//80+700+ms|code-block|syntax|javascript|975934|总而言之，从LinkedHashMap中按索引检索元素似乎是一项相当繁重的操作。|975935|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

Another working approach is transforming map values into an array and then retrieve element at index. Test run of 100 000 element by index searches in LinkedHashMap of 100 000 objects using following approaches led to following results: 

<pre><code>//My answer:
public Particle getElementByIndex(LinkedHashMap&lt;Point, Particle&gt; map,int index){
 return map.values().toArray(new Particle[map.values().size()])[index];
} //68 965 ms

//Syd Lambert's answer:
public Particle getElementByIndex(LinkedHashMap&lt;Point, Particle&gt; map,int index){
 return map.get( (map.keySet().toArray())[ index ] );
} //80 700 ms
</code></pre>

All in all retrieving element by index from LinkedHashMap seems to be pretty heavy operation.

blocks|key|759945|text|HashMap+-和底层的数据结构-哈希表，没有位置的概念。与LinkedList或向量不同，输入键被转换为存储值的“存储桶”。这些存储桶不是以在HashMap接口之外有意义的方式排序的，因此，放入HashMap中的项的顺序与其他数据结构的顺序不同|type|unstyled|depth|inlineStyleRanges|entityRanges|data|759946|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

HashMap - and the underlying data structure - hash tables, do not have a notion of position. Unlike a LinkedList or Vector, the input key is transformed to a 'bucket' where the value is stored. These buckets are not ordered in a way that makes sense outside the HashMap interface and as such, the items you put into the HashMap are not in order in the sense that you would expect with the other data structures

blocks|key|975596|text|HashMap没有位置的概念，因此无法通过位置获取对象。贴图中的对象是通过关键点设置和获取的。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|975597|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

HashMap has no concept of position so there is no way to get an object by position. Objects in Maps are set and get by keys.

blocks|key|976009|text|您可以使用以下代码来获取密钥：String+[]++keys+=+(String[])+item.keySet().toArray(new+String[0]);|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|976010|并使用以下项的键获取插入到HashMap中的对象或列表：item.get(keys[position]);|976011|entityMap^0|F|1U|0|S|P|0^^$0|@$1|2|3|4|5|6|7|J|8|@$9|K|A|L|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|H|3|-4|5|6|7|P|8|@]|D|@]|E|$]]]|I|$]]

you can use below code to get key : 
<code>String [] keys = (String[]) item.keySet().toArray(new String[0]);</code>

and get object or list that insert in HashMap with key of this item like this :
<code>item.get(keys[position]);</code>

blocks|key|760060|text|HashMaps不允许按位置访问，它只知道散列码，并且如果它可以计算键的散列码，它就可以检索值。TreeMaps有排序的概念。Linkedhas地图保留了它们进入地图的顺序。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|760061|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

HashMaps don't allow access by position, it only knows about the hash code and and it can retrieve the value if it can calculate the hash code of the key. TreeMaps have a notion of ordering. Linkedhas maps preserve the order in which they entered the map.

blocks|key|760273|text|你可以尝试实现这样的东西，看一下：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|760274|Map<String,+Integer>+map+=+new+LinkedHashMap<String,+Integer>();
map.put("juan",+2);
map.put("pedro",+3);
map.put("pablo",+5);
map.put("iphoncio",9)

List<String>+indexes+=+new+ArrayList<String>(map.keySet());+//+<==+Parse

System.out.println(indexes.indexOf("juan"));+++++//+==>+0
System.out.println(indexes.indexOf("iphoncio"));++++++//+==>+3|code-block|syntax|javascript|760275|我希望这对你有用。|760276|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

You can try to implement something like that, look at: 

<pre><code>Map&lt;String, Integer&gt; map = new LinkedHashMap&lt;String, Integer&gt;();
map.put("juan", 2);
map.put("pedro", 3);
map.put("pablo", 5);
map.put("iphoncio",9)

List&lt;String&gt; indexes = new ArrayList&lt;String&gt;(map.keySet()); // &lt;== Parse

System.out.println(indexes.indexOf("juan")); // ==&gt; 0
System.out.println(indexes.indexOf("iphoncio")); // ==&gt; 3
</code></pre>

I hope this works for you.

How to retrieve an element from HashMap by its position, is it possible at all?

Is it possible to get element from HashMap by its position?

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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如何根据元素的位置从HashMap中检索元素，这是可能的吗？

问是否可以根据元素的位置从HashMap获取元素？
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问是否可以根据元素的位置从HashMap获取元素？
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