blocks|key|270748|text|地图中没有索引这回事。映射不作为“对”序列存储(至少不是必须的；实际上在大多数实现中也不是这样)。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|270749|然而，不管实现是什么，std::map都不会对具有索引的容器进行建模。|270750|根据您提出这个问题的目的，"index“可以是迭代器(如其他人所建议的)，也可以是键本身。|270751|然而，你问这个问题听起来很奇怪。如果你能给我们提供更多的细节，我们也许能为你的问题指出一个更好的解决方案。|270752|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|J|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|K|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|L|8|@]|9|@]|A|$]]|$1|F|3|G|5|6|7|M|8|@]|9|@]|A|$]]|$1|H|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|I|$]]

There is no such thing as an index in a map. Maps are not stored (not necessarly, at least; and indeed they are not in most implementations) as a sequence of "pairs".

Regardless of the implementation, however, std::map does not model a container having an index.

Depending on what you are asking this question for, the "index" can be an iterator (as suggested by others) or the key itself.

However, it sounds strange you asked this question. If you could give us a bit more details we would probably be able to point you to a better solution to your problem.

blocks|key|54273|text|std::map实际上并没有索引，相反，它有一个键/值对的迭代器。这类似于索引，因为它表示集合中排序的位置，但它不是数字。要获取键/值对的迭代器，请使用find方法|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|54274|std::map<string,+int>::iterator+it+=+myMap.find("myKey");|code-block|syntax|javascript|54275|entityMap^0|0|8|24|4|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|$]]

A <code>std::map</code> doesn't really have an index, instead it has an iterator for a key / value pair. This is similar to an index in that it represents a position of sorts in the collection but it is not numeric. To get the iterator of a key / value pair use the <code>find</code> method 

<pre><code>std::map&lt;string, int&gt;::iterator it = myMap.find("myKey");
</code></pre>

blocks|key|54327|text|映射的语义不包括索引。要理解这一点，您可以注意到Map通常是以树的形式实现的。因此，其中的元素没有索引(尝试以一种自然的方式为树定义索引)。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|54328|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|$]]

The semantic of a map does not include indexes. To understand that, you can note that Maps are typically implemented as trees. Therefore, elements in it do not have an index (try to define an index in a natural way for a tree).

blocks|key|54387|text|Well+-+map将密钥和数据保持为一对，因此您可以通过将map的迭代器解引用到pair或直接引用到pair的第一个元素来提取密钥。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|54388|std::map<string,+int>+myMap;
std::map<string,+int>::iterator+it;

for(it=myMap.begin();it!=myMap.end();it%2B%2B)
{
++++std::cout<<it->first<<std::endl;
}|code-block|syntax|javascript|54389|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|$]]

Well - map is keeping the key and the data as a pair 
so you can extract key by dereferecing the map's iterator into pair or directly into pair's first element.

<pre><code>std::map&lt;string, int&gt; myMap;
std::map&lt;string, int&gt;::iterator it;

for(it=myMap.begin();it!=myMap.end();it++)
{
 std::cout&lt;&lt;it-&gt;first&lt;&lt;std::endl;
}
</code></pre>

blocks|key|270926|text|我来到这里寻找这个答案，但我发现这个距离函数需要2个迭代器并返回一个索引|type|unstyled|depth|inlineStyleRanges|entityRanges|data|270927|cout+<<+distance(mymap.begin(),mymap.find("198765432"));|code-block|syntax|javascript|270928|希望这能有所帮助:D|270929|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|$]]

I came here seeking for this answer but i found this
distance function takes 2 iterators and returns an index

<pre><code>cout &lt;&lt; distance(mymap.begin(),mymap.find("198765432"));
</code></pre>

hope this helps :D

blocks|key|270981|text|Map是一种键值数据结构，其内部数据以树结构形式存在。有上述的O(n)解。“distance(mymap.begin()，mymap.find("198765432"))”不会给你带来正确的答案。为了满足您的需求，您必须为O+log(n)竞争操作构建自己的段树型数据结构。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|270982|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|$]]

Map is a key-value data structure which internally data in a tree structure. There are O(n) solution stated above.
" distance(mymap.begin(),mymap.find("198765432")) " will not bring you the correct answer.
For your requirement, you have to build your own segment tree type data structure for O log(n) competitive operations.

blocks|key|54494|text|一个用例:如果你想知道在一个向量上有多少项是小的或者相等的。约束条件:i<=+j，有多少vi小于等于主播)。让我们将其插入到映射或集合中。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|54495|vector<int>+v={1,+4,+2,+3};
set<int>+s;
s+=+{1};+//+1's+position+is+1+(one+based)
s+=+{1,4};+//4's+positon+is+2
s+=+{1,+2,+4}+;//2's+position+is+2
s+=+{1+,+2,+3,+4};+//3's+positon+is+3|code-block|syntax|javascript|54496|这似乎是标准的:距离将需要O(n)时间。我可以使用set.lower_bound()和倒数到set.begin()来达到同样的效果。有没有比O(n)更好的解决方案，比如使用额外的数据结构？|54497|好吧，再想一想，这里有一个存储索引(基于1)的解决方案来解决这个特定的问题。然而，它可能不能解决在完成的映射中获得正确的项目索引的问题。|54498|vector<int>+arr={1+,+1+,+2,+4,+2};
multimap<int,+int>+track;+
for(auto+a:arr)
{
++++auto+it+=+track.insert(make_pair(a,+1));+//first+item+is+1
++++if(it!=track.begin())
++++{
++++++++--it;
++++++++int+prev=it->second;
++++++++it%2B%2B;
++++++++it->second%2B=prev;
++++}++
++++cout<<a<<','<<it->second-1<<endl;+++++
}|54499|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]]|$1|G|3|H|5|6|7|Q|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|R|8|@]|9|@]|A|$]]|$1|K|3|L|5|D|7|S|8|@]|9|@]|A|$E|F]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

A use case: if you want to know how many items are smaller or equal as you progress on a vector. Constraint : i &lt; = j, how many v[i]'s are smaller or equal to v[j]). let's insert it into a map or set.

<pre><code>vector&lt;int&gt; v={1, 4, 2, 3};
set&lt;int&gt; s;
s = {1}; // 1's position is 1 (one based)
s = {1,4}; //4's positon is 2
s = {1, 2, 4} ;//2's position is 2
s = {1 , 2, 3, 4}; //3's positon is 3
</code></pre>

it seems std:distance would need a O(n) time.
I could achieve same affect using set.lower_bound() and counting backward till set.begin(). Does anyone have a better solution than requiring O(n) , perhaps using additional data structures? 

OK, on a second thought here is a solution to store index (1 based) for this specific problem. However it may not solve the problem for get the correct index of items in the finished map.

<pre><code>vector&lt;int&gt; arr={1 , 1 , 2, 4, 2};
multimap&lt;int, int&gt; track; 
for(auto a:arr)
{
 auto it = track.insert(make_pair(a, 1)); //first item is 1
 if(it!=track.begin())
 {
 --it;
 int prev=it-&gt;second;
 it++;
 it-&gt;second+=prev;
 } 
 cout&lt;&lt;a&lt;&lt;','&lt;&lt;it-&gt;second-1&lt;&lt;endl; 
}
</code></pre>

blocks|key|54530|text|使用|type|unstyled|depth|inlineStyleRanges|entityRanges|data|54531|int+k+=+distance(mymap.begin(),+mymap.find(mykey));|code-block|syntax|javascript|54532|它将为您提供key元素的索引。|54533|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|$]]

Use
<pre><code>int k = distance(mymap.begin(), mymap.find(mykey));
</code></pre>
It will give you the index of the key element.

I have a <code>std::map</code> called <code>myMap</code> in my C++ application, and I want to get an element using either <code>myMap.find(key)</code> or <code>myMap[key]</code>. However, I would also like to get the index of that element in the map.

<pre><code>std::map&lt;string, int&gt; myMap;
// Populate myMap with a bunch of items...
myElement = myMap["myKey"];
// Now I need to get the index of myElement in myMap
</code></pre>

Is there a clean way to do that?

Thank you.

Get index of element in C++ map

我的C++应用程序中有一个名为myMap的std::map，我想使用myMap.find(key)或myMap[key]获取一个元素。但是，我还希望获得映射中该元素的索引。std::map<string, int> myMap;// Populate myMap with a bunch of items...myEl...

问获取C++映射中元素的索引
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