map <int, string> rollCallRegister;
map <int, string> :: iterator rollCallRegisterIter;
map <int, string> :: iterator temporaryRollCallRegisterIter;
rollCallRegisterIter = rollCallRegister.begin ();
tempRollCallRegisterIter = rollCallRegister.insert (rollCallRegisterIter, pair <int, string> (55, "swati"));
rollCallRegisterIter++;
tempRollCallRegisterIter = rollCallRegister.insert (rollCallRegisterIter, pair <int, string> (44, "shweta"));
rollCallRegisterIter++;
tempRollCallRegisterIter = rollCallRegister.insert (rollCallRegisterIter, pair <int, string> (33, "sindhu"));
// Displaying contents of this map.
cout << "\n\nrollCallRegister contains:\n";
for (rollCallRegisterIter = rollCallRegister.begin(); rollCallRegisterIter != rollCallRegister.end(); ++rollCallRegisterIter)
{
cout << (*rollCallRegisterIter).first << " => " << (*rollCallRegisterIter).second << endl;
}
Output:
rollCallRegister contains:
33 => sindhu
44 => shweta
55 => swati
I have incremented the iterator. Why is it still getting sorted? And if the position is supposed to be changed by the map on its own, then what's the purpose of providing an iterator?
Because std::map is a sorted associative container.
In a map, the key value is generally used to uniquely identify the element, while the mapped value is some sort of value associated to this key.
According to here position parameter is
the position of the first element to be compared for the insertion
operation. Notice that this does not force the new element to be in
that position within the map container (elements in a set always
follow a specific ordering), but this is actually an indication of a
possible insertion position in the container that, if set to the
element that precedes the actual location where the element is
inserted, makes for a very efficient insertion operation. iterator is
a member type, defined as a bidirectional iterator type.
So the purpose of this parameter is mainly slightly increasing the insertion speed by narrowing the range of elements.
You can use std::vector<std::pair<int,std::string>> if the order of insertion is important.
The interface is indeed slightly confusing, because it looks very much like std::vector<int>::insert (for example) and yet does not produce the same effect...
For associative containers, such as set, map and the new unordered_set and co, you completely relinquish the control over the order of the elements (as seen by iterating over the container). In exchange for this loss of control, you gain efficient look-up.
It would not make sense to suddenly give you control over the insertion, as it would let you break invariants of the container, and you would lose the efficient look-up that is the reason to use such containers in the first place.
And thus insert(It position, value_type&& value) does not insert at said position...
However this gives us some room for optimization: when inserting an element in an associative container, a look-up need to be performed to locate where to insert this element. By letting you specify a hint, you are given an opportunity to help the container speed up the process.
This can be illustrated for a simple example: suppose that you receive elements already sorted by way of some interface, it would be wasteful not to use this information!
template <typename Key, typename Value, typename InputStream>
void insert(std::map<Key, Value>& m, InputStream& s) {
typename std::map<Key, Value>::iterator it = m.begin();
for (; s; ++s) {
it = m.insert(it, *s).first;
}
}
Some of the items might not be well sorted, but it does not matter, if two consecutive items are in the right order, then we will gain, otherwise... we'll just perform as usual.
The map is always sorted, but you give a "hint" as to where the element may go as an optimisation.
The insertion is O(log N) but if you are able to successfully tell the container where it goes, it is constant time.
Thus if you are creating a large container of already-sorted values, then each value will get inserted at the end, although the tree will need rebalancing quite a few times.
As sad_man says, it's associative. If you set a value with an existing key, then you overwrite the previous value.
Now the iterators are necessary because you don't know what the keys are, usually.
Related
If I have a structure like
std::map<string, int> myMap;
myMap["banana"] = 1;
myMap["apple"] = 1;
myMap["orange"] = 1;
How can I access myMap[0]?
I know that the map sorts internally and I'm fine with this, I want to get a value in the map by index. I've tried myMap[0] but I get the error:
Error 1 error C2679: binary '[' : no operator found which takes a right-hand operand of type 'int' (or there is no acceptable conversion)
I realise I could do something like this:
string getKeyAtIndex (int index){
map<string, int>::const_iterator end = myMap.end();
int counter = 0;
for (map<string, int>::const_iterator it = myMap.begin(); it != end; ++it) {
counter++;
if (counter == index)
return it->first;
}
}
But surely this is hugely inefficient? Is there a better way?
Your map is not supposed to be accessed that way, it's indexed by keys not by positions. A map iterator is bidirectional, just like a list, so the function you are using is no more inefficient than accessing a list by position. If you want random access by position then use a vector or a deque.
Your function could be written with help from std::advance(iter, index) starting from begin():
auto it = myMap.begin();
std::advance(it, index);
return it->first;
There may be an implementation specific (non-portable) method to achieve your goal, but not one that is portable.
In general, the std::map is implemented as a type of binary tree, usually sorted by key. The definition of the first element differs depending on the ordering. Also, in your definition, is element[0] the node at the top of the tree or the left-most leaf node?
Many binary trees are implemented as linked lists. Most linked lists cannot be directly accessed like an array, because to find element 5, you have to follow the links. This is by definition.
You can resolve your issue by using both a std::vector and a std::map:
Allocate the object from dynamic memory.
Store the pointer, along with the key, into the std::map.
Store the pointer in the std::vector at the position you want it
at.
The std::map will allow an efficient method to access the object by key.
The std::vector will allow an efficient method to access the object by index.
Storing pointers allows for only one instance of the object instead of having to maintain multiple copies.
Well, actually you can't. The way you found is very unefficient, it have a computational complexity of O(n) (n operations worst case, where n is the number of elements in a map).
Accessing an item in a vector or in an array have complexity O(1) by comparison (constant computational complexity, a single operation).
Consider that map is internally implemented as a red black tree (or avl tree, it depends on the implementation) and every insert, delete and lookup operation are O(log n) worst case (it requires logarithm in base 2 operations to find an element in the tree), that is quite good.
A way you can deal with is to use a custom class that have inside both a vector and a map.
Insertion at the end of the class will be averaged O(1), lookup by name will be O(log n), lookup by index will be O(1) but in this case, removal operation will be O(n).
Previous answer (see comment): How about just myMap.begin();
You could implement a random-access map by using a vector backing-store, which is essentially a vector of pairs. You of course lose all the benefits of the standard library map at that point.
you can use some other map like containers .
keep a size fields can make binary search tree easy to random access .
here is my implementation ...
std style , random access iterator ...
size balanced tree ...
https://github.com/mm304321141/zzz_lib/blob/master/sbtree.h
and B+tree ...
https://github.com/mm304321141/zzz_lib/blob/master/bpptree.h
std::map is an ordered container, but it's iterators don't support random access, but rather bidirectional access. Therefore, you can only access the nth element by navigating all its prior elements. A shorter alternative to your example is using the standard iterator library:
std::pair<const std::string, int> &nth_element = *std::next(myMap.begin(), N);
This has linear complexity, which is not ideal if you plan to frequently access this way in large maps.
An alternative is to use an ordered container that supports random access. For example, boost::container::flat_map provides a member function nth which allows you exactly what you are looking for.
std::map<string,int>::iterator it = mymap.begin() + index;
Say I have a std::unordered_map<std::string, int> that represents a word and the number of times that word appeared in a book, and I want to be able to sort it by the value.
The problem is, I want the sorting to be stable, so that in case two items have equal value I want the one who got inserted first to the map to be first.
It is simple to implement it by adding addition field that will keep the time it got inserted. Then, create a comperator that uses both time and the value. Using simple std::sort will give me O(Nlog(N)) time complexity.
In my case, space is not an issue whenever time can be improved. I want to take advantage of it and do a bucket sorting. Which should give me O(N) time complexity. But when using bucket sorting, there is no comperator, when iterating the items in the map the order is not preserved.
How can I both make it stable and still keep the O(N) time complexity via bucket sorting or something else?
I guess that if I had some kind of hash map that preserves the order of insertion while iterating it, it would solve my issue.
Any other solutions with the same time complexity are acceptable.
Note - I already saw this and that and due to the fact that they are both from 2009 and that my case is more specific I think, I opened this question.
Here is a possible solution I came up with using an std::unordered_map and tracking the order of inserting using a std::vector.
Create a hash map with the string as key and count as value.
In addition, create a vector with iterators to that map type.
When counting elements, if the object is not yet in the map, add to both map and vector. Else, just increment the counter. The vector will preserve the order the elements got inserted to the map, and the insertion / update will still be in O(1) time complexity.
Apply bucket sort by iterating over the vector (instead of the map), this ensures the order is preserved and we'll get a stable sort. O(N)
Extract from the buckets to make a sorted array. O(N)
Implementation:
unordered_map<std::string, int> map;
std::vector<std::unordered_map<std::string,int>::iterator> order;
// Lets assume this is my string stream
std::vector<std::string> words = {"a","b","a" ... };
// Insert elements to map and the corresponding iterator to order
for (auto& word : words){
auto it = map.emplace(word,1);
if (!it.second){
it.first->second++;
}
else {
order.push_back(it.first);
}
max_count = std::max(max_count,it.first->second);
}
// Bucket Sorting
/* We are iterating over the vector and not the map
this ensures we are iterating by the order they got inserted */
std::vector<std::vector<std::string>> buckets(max_count);
for (auto o : order){
int count = o->second;
buckets[count-1].push_back(o->first);
}
std::vector<std::string> res;
for (auto it = buckets.rbegin(); it != buckets.rend(); ++it)
for (auto& str : *it)
res.push_back(str);
If I have a map with keys as strings. How can I know which string is in which place in the map? For example, 'cats' is the first entry and 'dogs' are the second entry. How will I know that dogs are the second entry in the map? Should I go through the map and search/check the order or is there any easier way?
You can find() the element, and then compute the distance between the element and the map's begin(). Since this is a map, the complexity of distance will be O(n) with the number of elements in the map. I doubt you'd see much performance improvement versus simply looping the map and counting, but I'd prefer to not write such hand-written loops.
But this smells like an XY Problem. Why would you need the position of an element in a map?
You can't do that anyway better, and here's two reasons for that:
a) std::map has a tree inside that, so initially leaves of tree don't have linear order, only a partial order (of course you can compare two keys, but you can't do that for a general binary tree - that's the reason for complication).
b) std::map has bidirectional iterator, so you can't have effective iterator arithmetics. You could do that like this:
typedef map<string, int> MyMapT;
MyMapT data;
// ...
auto it = data.find("dog");
if (it == data.end()) {
// don't have such a key
}
auto position = distance(data.begin(), data.end());
Unfortunately, that's not the best way to do that, because distance function works in O(n) time for bidirectional iterators. Again, that's because of tree structure inside the std::map.
If I have a structure like
std::map<string, int> myMap;
myMap["banana"] = 1;
myMap["apple"] = 1;
myMap["orange"] = 1;
How can I access myMap[0]?
I know that the map sorts internally and I'm fine with this, I want to get a value in the map by index. I've tried myMap[0] but I get the error:
Error 1 error C2679: binary '[' : no operator found which takes a right-hand operand of type 'int' (or there is no acceptable conversion)
I realise I could do something like this:
string getKeyAtIndex (int index){
map<string, int>::const_iterator end = myMap.end();
int counter = 0;
for (map<string, int>::const_iterator it = myMap.begin(); it != end; ++it) {
counter++;
if (counter == index)
return it->first;
}
}
But surely this is hugely inefficient? Is there a better way?
Your map is not supposed to be accessed that way, it's indexed by keys not by positions. A map iterator is bidirectional, just like a list, so the function you are using is no more inefficient than accessing a list by position. If you want random access by position then use a vector or a deque.
Your function could be written with help from std::advance(iter, index) starting from begin():
auto it = myMap.begin();
std::advance(it, index);
return it->first;
There may be an implementation specific (non-portable) method to achieve your goal, but not one that is portable.
In general, the std::map is implemented as a type of binary tree, usually sorted by key. The definition of the first element differs depending on the ordering. Also, in your definition, is element[0] the node at the top of the tree or the left-most leaf node?
Many binary trees are implemented as linked lists. Most linked lists cannot be directly accessed like an array, because to find element 5, you have to follow the links. This is by definition.
You can resolve your issue by using both a std::vector and a std::map:
Allocate the object from dynamic memory.
Store the pointer, along with the key, into the std::map.
Store the pointer in the std::vector at the position you want it
at.
The std::map will allow an efficient method to access the object by key.
The std::vector will allow an efficient method to access the object by index.
Storing pointers allows for only one instance of the object instead of having to maintain multiple copies.
Well, actually you can't. The way you found is very unefficient, it have a computational complexity of O(n) (n operations worst case, where n is the number of elements in a map).
Accessing an item in a vector or in an array have complexity O(1) by comparison (constant computational complexity, a single operation).
Consider that map is internally implemented as a red black tree (or avl tree, it depends on the implementation) and every insert, delete and lookup operation are O(log n) worst case (it requires logarithm in base 2 operations to find an element in the tree), that is quite good.
A way you can deal with is to use a custom class that have inside both a vector and a map.
Insertion at the end of the class will be averaged O(1), lookup by name will be O(log n), lookup by index will be O(1) but in this case, removal operation will be O(n).
Previous answer (see comment): How about just myMap.begin();
You could implement a random-access map by using a vector backing-store, which is essentially a vector of pairs. You of course lose all the benefits of the standard library map at that point.
you can use some other map like containers .
keep a size fields can make binary search tree easy to random access .
here is my implementation ...
std style , random access iterator ...
size balanced tree ...
https://github.com/mm304321141/zzz_lib/blob/master/sbtree.h
and B+tree ...
https://github.com/mm304321141/zzz_lib/blob/master/bpptree.h
std::map is an ordered container, but it's iterators don't support random access, but rather bidirectional access. Therefore, you can only access the nth element by navigating all its prior elements. A shorter alternative to your example is using the standard iterator library:
std::pair<const std::string, int> &nth_element = *std::next(myMap.begin(), N);
This has linear complexity, which is not ideal if you plan to frequently access this way in large maps.
An alternative is to use an ordered container that supports random access. For example, boost::container::flat_map provides a member function nth which allows you exactly what you are looking for.
std::map<string,int>::iterator it = mymap.begin() + index;
At the moment my solution is to iterate through the map to solve this.
I see there is a upper_bound method which can make this loop faster, but is there a quicker or more succinct way?
The end:
m.rbegin();
Maps (and sets) are sorted, so the first element is the smallest, and the last element is the largest. By default maps use std::less, but you can switch the comparer and this would of course change the position of the largest element. (For example, using std::greater would place it at begin().
Keep in mind rbegin returns an iterator. To get the actual key, use m.rbegin()->first. You might wrap it up into a function for clarity, though I'm not sure if it's worth it:
template <typename T>
inline const typename T::key_type& last_key(const T& pMap)
{
return pMap.rbegin()->first;
}
typedef std::map</* types */> map_type;
map_type myMap;
// populate
map_type::key_type k = last_key(myMap);
The entries in a std::map are sorted, so for a std::map m (assuming m.empty() is false), you can get the biggest key easily: (--m.end())->first
As std::map is assosiative array one can easily find biggest or smallest key very easily. By defualt compare function is less(<) operator so biggest key will be last element in map. Similarly if someone has different requirement anyone can modify compare function while declaring map.
std::map< key, Value, compare< key,Value > >
By default compare=std::less
Since you're not using unordered_map, your keys should be in order. Depending upon what you want to do with an iterator, you have two options:
If you want a forwards-iterator then you can use std::prev(myMap.end()). Note that --myMap.end() isn't guaranteed to work in all scenarios, so I'd usually avoid it.
If you want to iterate in reverse then use myMap.rbegin()
Since the map is just an AVL tree then, it's sorted -in an ascending order-. So, the element with largest key is the last element and you can obtain it using one of the following two methods:
1.
largestElement = (myMap.rbegin())-> first; // rbegin(): returns an iterator pointing to the last element
largestElement = (--myMap.end())->first; // end(): returns an iterator pointing to the theortical element following the last element