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I need to implement the following datastructure for my project. I have a relation of
const MyClass*
to
uint64_t
For every pointer I want to save a counter connected to it, which can be changed over time (in fact only incremented). This would be no problem, I could simply store it in a std::map. The problem is that I need fast access to the pointers which have the highest values.
That is why I came to the conclusion to use a boost::bimap. It is defined is follows for my project:
typedef boost::bimaps::bimap<
boost::bimaps::unordered_set_of< const MyClass* >,
boost::bimaps::multiset_of< uint64_t, std::greater<uint64_t> >
> MyBimap;
MyBimap bimap;
This would work fine, but am I right that I can not modify the uint64_t on pair which were inserted once? The documentation says that multiset_of is constant and therefore I cannot change a value of pair in the bimap.
What can I do? What would be the correct way to change the value of one key in this bimap? Or is there a simpler data structure possible for this problem?
Here's a simple hand-made solution.
Internally it keeps a map to store the counts indexed by object pointer, and a further multi-set of iterators, ordered by descending count of their pointees.
Whenever you modify a count, you must re-index. I have done this piecemeal, but you could do it as a batch update, depending on requirements.
Note that in c++17 there is a proposed splice operation for sets and maps, which would make the re-indexing extremely fast.
#include <map>
#include <set>
#include <vector>
struct MyClass { };
struct store
{
std::uint64_t add_value(MyClass* p, std::uint64_t count = 0)
{
add_index(_map.emplace(p, count).first);
return count;
}
std::uint64_t increment(MyClass* p)
{
auto it = _map.find(p);
if (it == std::end(_map)) {
// in this case, we'll create one - we could throw instead
return add_value(p, 1);
}
else {
remove_index(it);
++it->second;
add_index(it);
return it->second;
}
}
std::uint64_t query(MyClass* p) const {
auto it = _map.find(p);
if (it == std::end(_map)) {
// in this case, we'll create one - we could throw instead
return 0;
}
else {
return it->second;
}
}
std::vector<std::pair<MyClass*, std::uint64_t>> top_n(std::size_t n)
{
std::vector<std::pair<MyClass*, std::uint64_t>> result;
result.reserve(n);
for (auto idx = _value_index.begin(), idx_end = _value_index.end() ;
n && idx != idx_end ;
++idx, --n) {
result.emplace_back((*idx)->first, (*idx)->second);
}
return result;
}
private:
using map_type = std::map<MyClass*, std::uint64_t>;
struct by_count
{
bool operator()(map_type::const_iterator l, map_type::const_iterator r) const {
// note: greater than orders by descending count
return l->second > r->second;
}
};
using value_index_type = std::multiset<map_type::iterator, by_count>;
void add_index(map_type::iterator iter)
{
_value_index.emplace(iter->second, iter);
}
void remove_index(map_type::iterator iter)
{
for(auto range = _value_index.equal_range(iter);
range.first != range.second;
++range.first)
{
if (*range.first == iter) {
_value_index.erase(range.first);
return;
}
}
}
map_type _map;
value_index_type _value_index;
};
I have to process an std::vector either forwards or in reverse, depending upon a boolean flag. What's the most elegant way to accomplish this? Before needing to do it in reverse I had:
BOOST_FOREACH(const CType &foo, vec) {
...
}
However, now I have the horrendous-looking:
for (int i=undoing ? (vec.size()-1) : 0; undoing ? (i >= 0) : (i < vec.size()); undoing ? (i--) : (i++)) {
const CType &foo = vec[i];
...
}
Is there a better way?
I don't know that people would call it elegant, but there's:
auto do_it = [](const CType& elem)
{
...
};
if (iterate_forward) {
std::for_each(vec.begin(), vec.end(), do_it);
}
else {
std::for_each(vec.rbegin(), vec.rend(), do_it);
}
Add a template function that works with either the forward iterators or reverse iterators. Call the function using the appropriate iterator based on the value of undoing.
template <typename Iterator>
void doStuff(Iterator iter, Iterator end)
{
for ( ; iter != end; ++iter )
{
// Do stuff
}
}
if ( undoing )
{
doStuff(vec.rbegin(), vec.rend());
}
else
{
doStuff(vec.begin(), vec.end());
}
How about keeping the loop running as it is from 0 to vector.size, but reading the array in the direction you need.
int idx;
for (int i =0; i < vec.size(); i ++)
{
if (undoing) // assuming going forward
idx = i;
else // going backwards
idx = vec.size() - i - 1;
const CType &foo = vec[idx];
}
You may also use Boost.Range-based solution. It's similar to the one using STL algorithms, already proposed.
#include <boost/range/adaptor/reversed.hpp>
#include <boost/range/algorithm/for_each.hpp>
// In C++11 lambda expression can be used instead
struct my_fun
{
void operator()(const CType& elem) const
{
/*...*/
}
};
/*...*/
using namespace boost::adaptors;
if ( iterate_forward )
boost::for_each(my_vect, my_fun());
else
boost::for_each(my_vect | reversed, my_fun());
Well I think the question pretty much sums it up. I have a forward_list of unique items, and want to remove a single item from it:
std::forward_list<T> mylist;
// fill with stuff
mylist.remove_if([](T const& value)
{
return value == condition;
});
I mean, this method works fine but it's inefficient because it continues to search once the item is found and deleted. Is there a better way or do I need to do it manually?
If you only want to remove the first match, you can use std::adjacent_find followed by the member erase_after
#include <algorithm>
#include <cassert>
#include <forward_list>
#include <iostream>
#include <ios>
#include <iterator>
// returns an iterator before first element equal to value, or last if no such element is present
// pre-condition: before_first is incrementable and not equal to last
template<class FwdIt, class T>
FwdIt find_before(FwdIt before_first, FwdIt last, T const& value)
{
assert(before_first != last);
auto first = std::next(before_first);
if (first == last) return last;
if (*first == value) return before_first;
return std::adjacent_find(first, last, [&](auto const&, auto const& R) {
return R == value;
});
}
int main()
{
auto e = std::forward_list<int>{};
std::cout << std::boolalpha << (++e.before_begin() == end(e)) << "\n";
std::cout << (find_before(e.before_begin(), end(e), 0) == end(e)) << "\n";
auto s = std::forward_list<int>{ 0 };
std::cout << (find_before(s.before_begin(), end(s), 0) == s.before_begin()) << "\n";
auto d = std::forward_list<int>{ 0, 1 };
std::cout << (find_before(d.before_begin(), end(d), 0) == d.before_begin()) << "\n";
std::cout << (find_before(d.before_begin(), end(d), 1) == begin(d)) << "\n";
std::cout << (find_before(d.before_begin(), end(d), 2) == end(d)) << "\n";
// erase after
auto m = std::forward_list<int>{ 1, 2, 3, 4, 1, 3, 5 };
auto it = find_before(m.before_begin(), end(m), 3);
if (it != end(m))
m.erase_after(it);
std::copy(begin(m), end(m), std::ostream_iterator<int>(std::cout, ","));
}
Live Example
This will stop as soon as a match is found. Note that the adjacent_find takes a binary predicate, and by comparing only the second argument, we get an iterator before the element we want to remove, so that erase_after can actually remove it. Complexity is O(N) so you won't get it more efficient than this.
FWIW, here's another short version
template< typename T, class Allocator, class Predicate >
bool remove_first_if( std::forward_list< T, Allocator >& list, Predicate pred )
{
auto oit = list.before_begin(), it = std::next( oit );
while( it != list.end() ) {
if( pred( *it ) ) { list.erase_after( oit ); return true; }
oit = it++;
}
return false;
}
Going to have to roll your own...
template <typename Container, typename Predicate>
void remove_first_of(Container& container, Predicate p)
{
auto it = container.before_begin();
for (auto nit = std::next(it); ; it = nit, nit = std::next(it))
{
if (nit == container.end())
return;
if (p(*nit))
{
container.erase_after(it);
return;
}
}
}
A more complete example...
There is nothing in the standard library which would be directly applicable. Actually, there is. See #TemplateRex's answer for that.
You can also write this yourself (especially if you want to combine the search with the erasure), something like this:
template <class T, class Allocator, class Predicate>
bool remove_first_if(std::forward_list<T, Allocator> &list, Predicate pred)
{
auto itErase = list.before_begin();
auto itFind = list.begin();
const auto itEnd = list.end();
while (itFind != itEnd) {
if (pred(*itFind)) {
list.erase_after(itErase);
return true;
} else {
++itErase;
++itFind;
}
}
return false;
}
This kind of stuff used to be a standard exercise when I learned programming way back in the early '80s. It might be interesting to to recall the solution, and compare that with what one can do in C++. Actually that was in Algol 68, but I won't impose that on you and give the translation into C. Given
typedef ... T;
typedef struct node *link;
struct node { link next; T data; };
one could write, realising that one needs to pass the address of the list head pointer if is to be possible to unlink the first node:
void search_and_destroy(link *p_addr, T y)
{
while (*p_addr!=NULL && (*p_addr)->data!=y)
p_addr = &(*p_addr)->next;
if (*p_addr!=NULL)
{
link old = *p_addr;
*p_addr = old->next; /* unlink node */
free(old); /* and free memory */
}
}
There are a lot of occurrences of *p_addr there; it is the last one, where it is the LHS of an assignment, that is the reason one needs the address of a pointer here in the first place. Note that in spite of the apparent complication, the statement p_addr = &(*p_addr)->next; is just replacing a pointer by the value it points to, and then adding an offset (which is 0 here).
One could introduce an auxiliary pointer value to lighten the code a bit up, as follows
void search_and_destroy(link *p_addr, T y)
{
link p=*p_addr;
while (p!=NULL && p->data!=y)
p=*(p_addr = &p->next);
if (p!=NULL)
{
*p_addr = p->next;
free(p);
}
}
but that is fundamentally the same code: any decent compiler should realise that the pointer value *p_addr is used multiple times in succession in the first example, and keep it in a register.
Now with std::forward_list<T>, we are not allowed access to the pointers that link the nodes, and get those awkward "iterators pointing one node before the real action" instead. Our solution becomes
void search_and_destroy(std::forward_list<T> list, T y)
{
std::forward_list<T>::iterator it = list.before_begin();
const std::forward_list<T>::iterator NIL = list.end();
while (std::next(it)!=NIL && *std::next(it)!=y)
++it;
if (std::next(it)!=NIL)
list.erase_after(it);
}
Again we could keep a second iterator variable to hold std::next(it) without having to spell it out each time (not forgetting to refresh its value when we increment it) and arrive at essentially the answer by Daniel Frey. (We could instead try to make that variable a pointer of type *T equal to &*std::next(it) instead, which suffices for the use we make of it, but it would actually be a bit of a hassle to ensure it becomes the null pointer when std::next(it)==NIL, as the standard will not let us take &*NIL).
I cannot help feel that since the old days the solution to this problem has not become more elegant.
I have a list of tuple, the list was sorted based on the first element of the tuple but the second and last elements are in random order. Now I want to find all tuples with the first element within a range, i.e. return all tuples for (tuple.first>-X and tuple.first<X). Among all these returning tuples, I need to find the maximum and minimum value in the second element of the tuples. How can a STL algorithm implement this?
ListType::iterator itrFirst = std::find_if( ls.begin(), ls.end(), boost::bind( &TupleType::get< 0 >, _1 ) >= rangeStart );
ListType::iterator itrLast = std::find_if( itrFirst, ls.end(), boost::bind( &TupleType::get< 0 >, _1 ) > rangeEnd );
for( ;itrFirst != itrLast; ++itrFirst ) // Print keys for elements in range
std::cout << itrFirst->get<0>() << std::endl;
I presume boost:: can be replaced with std:: if you have a recent compiler ( I don't ).
Since it is already sorted, you can use equal_range to get a pair of iterators that delimit the range of "interesting" tuples:
It const begin = std::lower_bound(list.begin(), list.end(),
[X](Tuple const& t) {
return t.first > -X;
});
It const end = std::upper_bound(begin, list.end(),
[X](Tuple const& t) {
return t.first < X;
});
std::pair<It,It> const range = std::make_range(begin, end);
Then, you can simply iterate over this range, and register the minimum and maximum values that you see:
int min = INT_MAX, max = INT_MIN;
for (Tuple const& t: range) {
if (t.second < min) { min = t.second; }
if (t.second > max) { max = t.second; }
}
// min and max are correctly set
So... it's not a single STL algorithm.
Note: std::min_element and std::max_element do exist, but that would mean looping twice over the range, it's certainly feasible though.
Tuple const& min = *std::min_element(range.first, range.second,
[](Tuple const& left, Tuple const& right) {
return left.second < right.second;
});
Tuple const& max = *std::max_element(range.first, range.second,
[](Tuple const& left, Tuple const& right) {
return left.second < right.second;
});
// Or as noted by Vitali, slightly more efficient:
auto const minmax = std::minmax_element(range.first, range.second,
[](Tuple const& left, Tuple const& right) {
return left.second < right.second;
});
Tuple const& min = *minmax.first;
Tuple const& max = *minmax.second;
Note that it gives a tuple, and not the .second member.
I'm trying to check if a given key is in a map and somewhat can't do it:
typedef map<string,string>::iterator mi;
map<string, string> m;
m.insert(make_pair("f","++--"));
pair<mi,mi> p = m.equal_range("f");//I'm not sure if equal_range does what I want
cout << p.first;//I'm getting error here
so how can I print what is in p?
Use map::find and map::end:
if (m.find("f") == m.end()) {
// not found
} else {
// found
}
To check if a particular key in the map exists, use the count member function in one of the following ways:
m.count(key) > 0
m.count(key) == 1
m.count(key) != 0
The documentation for map::find says: "Another member function, map::count, can be used to just check whether a particular key exists."
The documentation for map::count says: "Because all elements in a map container are unique, the function can only return 1 (if the element is found) or zero (otherwise)."
To retrieve a value from the map via a key that you know to exist, use map::at:
value = m.at(key)
Unlike map::operator[], map::at will not create a new key in the map if the specified key does not exist.
C++20 gives us std::map::contains to do that.
#include <iostream>
#include <string>
#include <map>
int main()
{
std::map<int, std::string> example = {{1, "One"}, {2, "Two"},
{3, "Three"}, {42, "Don\'t Panic!!!"}};
if(example.contains(42)) {
std::cout << "Found\n";
} else {
std::cout << "Not found\n";
}
}
You can use .find():
map<string,string>::iterator i = m.find("f");
if (i == m.end()) { /* Not found */ }
else { /* Found, i->first is f, i->second is ++-- */ }
C++17 simplified this a bit more with an If statement with initializer.
This way you can have your cake and eat it too.
if ( auto it{ m.find( "key" ) }; it != std::end( m ) )
{
// Use `structured binding` to get the key
// and value.
const auto&[ key, value ] { *it };
// Grab either the key or value stored in the pair.
// The key is stored in the 'first' variable and
// the 'value' is stored in the second.
const auto& mkey{ it->first };
const auto& mvalue{ it->second };
// That or just grab the entire pair pointed
// to by the iterator.
const auto& pair{ *it };
}
else
{
// Key was not found..
}
m.find == m.end() // not found
If you want to use other API, then find go for m.count(c)>0
if (m.count("f")>0)
cout << " is an element of m.\n";
else
cout << " is not an element of m.\n";
I think you want map::find. If m.find("f") is equal to m.end(), then the key was not found. Otherwise, find returns an iterator pointing at the element found.
The error is because p.first is an iterator, which doesn't work for stream insertion. Change your last line to cout << (p.first)->first;. p is a pair of iterators, p.first is an iterator, p.first->first is the key string.
A map can only ever have one element for a given key, so equal_range isn't very useful. It's defined for map, because it's defined for all associative containers, but it's a lot more interesting for multimap.
template <typename T, typename Key>
bool key_exists(const T& container, const Key& key)
{
return (container.find(key) != std::end(container));
}
Of course if you wanted to get fancier you could always template out a function that also took a found function and a not found function, something like this:
template <typename T, typename Key, typename FoundFunction, typename NotFoundFunction>
void find_and_execute(const T& container, const Key& key, FoundFunction found_function, NotFoundFunction not_found_function)
{
auto& it = container.find(key);
if (it != std::end(container))
{
found_function(key, it->second);
}
else
{
not_found_function(key);
}
}
And use it like this:
std::map<int, int> some_map;
find_and_execute(some_map, 1,
[](int key, int value){ std::cout << "key " << key << " found, value: " << value << std::endl; },
[](int key){ std::cout << "key " << key << " not found" << std::endl; });
The downside to this is coming up with a good name, "find_and_execute" is awkward and I can't come up with anything better off the top of my head...
map<string, string> m;
check key exist or not, and return number of occurs(0/1 in map):
int num = m.count("f");
if (num>0) {
//found
} else {
// not found
}
check key exist or not, and return iterator:
map<string,string>::iterator mi = m.find("f");
if(mi != m.end()) {
//found
//do something to mi.
} else {
// not found
}
in your question, the error caused by bad operator<< overload, because p.first is map<string, string>, you can not print it out. try this:
if(p.first != p.second) {
cout << p.first->first << " " << p.first->second << endl;
}
Be careful in comparing the find result with the the end like for map 'm' as all answer have
done above
map::iterator i = m.find("f");
if (i == m.end())
{
}
else
{
}
you should not try and perform any operation such as printing the key or value with iterator i if its equal to m.end() else it will lead to segmentation fault.
Comparing the code of std::map::find and std::map::count, I'd say the first may yield some performance advantage:
const_iterator find(const key_type& _Keyval) const
{ // find an element in nonmutable sequence that matches _Keyval
const_iterator _Where = lower_bound(_Keyval); // Here one looks only for lower bound
return (_Where == end()
|| _DEBUG_LT_PRED(this->_Getcomp(),
_Keyval, this->_Key(_Where._Mynode()))
? end() : _Where);
}
size_type count(const key_type& _Keyval) const
{ // count all elements that match _Keyval
_Paircc _Ans = equal_range(_Keyval); // Here both lower and upper bounds are to be found, which is presumably slower.
size_type _Num = 0;
_Distance(_Ans.first, _Ans.second, _Num);
return (_Num);
}
I know this question already has some good answers but I think my solution is worth of sharing.
It works for both std::map and std::vector<std::pair<T, U>> and is available from C++11.
template <typename ForwardIterator, typename Key>
bool contains_key(ForwardIterator first, ForwardIterator last, Key const key) {
using ValueType = typename std::iterator_traits<ForwardIterator>::value_type;
auto search_result = std::find_if(
first, last,
[&key](ValueType const& item) {
return item.first == key;
}
);
if (search_result == last) {
return false;
} else {
return true;
}
}
map <int , char>::iterator itr;
for(itr = MyMap.begin() ; itr!= MyMap.end() ; itr++)
{
if (itr->second == 'c')
{
cout<<itr->first<<endl;
}
}
If you want to compare pair of map you can use this method:
typedef map<double, double> TestMap;
TestMap testMap;
pair<map<double,double>::iterator,bool> controlMapValues;
controlMapValues= testMap.insert(std::pair<double,double>(x,y));
if (controlMapValues.second == false )
{
TestMap::iterator it;
it = testMap.find(x);
if (it->second == y)
{
cout<<"Given value is already exist in Map"<<endl;
}
}
This is a useful technique.