如何从LinkedHashMap获取基于索引的值而不是键？ [重复]

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How to get position of key/value in LinkedHashMap using its key 4 answers
如何使用其关键4答案获取LinkedHashMap中键/值的位置

I have

我有

LinkedHashMap<String, List<String>> hMap;

I want to get List<String> by position not on key.

我想按位置而不是键来获取List 。

I don't want to use iterate.

我不想使用迭代。

Is there any other way to get Value based on index ?

有没有其他方法可以根据索引获得价值？

5 个解决方案

#1

You can't get the value of the Map based on index, Maps just don't work that way. A workaround would be to create a new list from your values and get the value based on index.

您无法根据索引获取Map的值，Maps只是不能以这种方式工作。解决方法是从您的值创建新列表并根据索引获取值。

LinkedHashMap<String, List<String>> hMap;
List<List<String>> l = new ArrayList<List<String>>(hMap.values());
l.get(0);

#2

public List<String> getByIndex(LinkedHashMap<String, List<String>> hMap, int index){
   return (List<String>) hMap.values().toArray()[index];
}

#3

you may want to consider either using another class to store your data, or write an extension to the linkedHashMap. something like

您可能想要考虑使用另一个类来存储数据，或者编写一个扩展到linkedHashMap。就像是

//this is pseudo code
public class IndexedLinkedHashMap<K,V> extends LinkedHashMap{

HashMap<int,K> index;
int curr = 0;

    @Override
    public void add(K key,V val){
        super.add(key,val);
        index.add(curr++, key);
    }

    public V getindexed(int i){
        return super.get(index.get(i));
    }

}

#4

As Kevin Bowersox stated, it's as simple as

正如Kevin Bowersox所说，它就像

List<String> result = (List<String>) hMap.values().toArray()[position];

But it should be noted that this will still iterate by using .toArray(). It's a simple statement and I'm not sure if there is one with better performance, but be aware that complexity is not log(n) (like indexed access in case of B*), but just n. Since LinkedHashMap is based on LinkedList, there is no way to randomly access elements, only in sequential order.

但应该注意的是，这仍将通过使用.toArray（）进行迭代。这是一个简单的陈述，我不确定是否有更好的性能，但要注意复杂性不是log（n）（如B *情况下的索引访问），但只是n。由于LinkedHashMap基于LinkedList，因此无法按顺序随机访问元素。

The cast to List is an unavoidable evil, since .toArray() follows the archaic concept of returning Object instead of a generic data type.

对List的强制转换是一个不可避免的恶魔，因为.toArray（）遵循返回Object而不是通用数据类型的古老概念。

While this might not be the main concept of a map, LinkedHashMap isn't just a map. it extends HashMap, and as an extending class, it's perfectly fine to bring additional methods that support the idiosyncracies of that class.

虽然这可能不是地图的主要概念，但LinkedHashMap不仅仅是一张地图。它扩展了HashMap，作为一个扩展类，它可以带来支持该类特性的其他方法。

#5

There is no direct DS in the standard Java Collections API to provide a indexed map. However, the following should let you achieve the result:

标准Java Collections API中没有直接DS来提供索引映射。但是，以下应该可以让您实现结果：

// An ordered map
Map<K, V> map = new LinkedHashMap<K, V>();
// To create indexed list, copy the references into an ArrayList (backed by an array)
List<Entry<K, V>> indexedList = new ArrayList<Map.Entry<K, V>>(map.entrySet());
// Get the i'th term
<Map.Entry<K,V>> entry = indexedList.get(index);
K key = entry.getKey();
V value = entry.getValue();

You might still want to retain the concerns of data persistence in the map separate from the retrieval.

您可能仍希望保留地图中与检索分开的数据持久性问题。

Update: Or use LinkedMap from Apache Commons.

更新：或使用Apache Commons中的LinkedMap。

#1