Given a vector such as this:
struct product {
float price;
float shipping;
};
vector<product> products;
how can I remove all the products from the vector apart from the one with the largest shipping to price ratio?
I tried keeping an iterator to the highest one found so far...
vector<product>::iterator it = products.begin();
vector<product>::iterator largest = products.begin();
while (it != products.end())
{
if (it->shipping / it->price > largest->shipping / largest->price)
{
products.erase(largest);
largest = it;
++it;
}
else
{
it = products.erase(it);
}
}
This is all well and good but it fails if the first element in the vector has the highest ratio (it gets deleted). I could get around the problem (I think) if largest was uninitialized and then checking for that in the if statement, but there is no real way of doing this from what I can tell (How to check if the iterator is initialized?).
Any suggestions?
vector<product> products;
//populate products
products.erase(
products.begin(),
std::max_element(
product.begin(),
producted.end()
)
);
products.resize(1u);
This assume you have a suitable operator< for your type, if not, make a comparison function and provide it as the third param to max_element.
EDIT:
This work also, in this case, instead of explicitly find the element and delete the element either side, it will sort to find 1 elemnt, then we can do one erase.
vector<product> products;
//populate products
std::nth_element(
products.begin(),
products.begin()+1,
products.end(),
std::greater<product>()
);
products.resize(1u);
Just rewrite the code to only make one single deletion:
std::vector<product>::iterator largest = products.begin();
for (std::vector<product>::iterator it = products.begin(); it != products.end(); ++it)
{
if (...) { largest = it; }
}
products.erase(it);
You could define largest as the first element and start iteration from the second element, just as followings:
bool operator < (const struct product& p1, const struct product& p2)
{
return p1.price/p1.shipping < p2.price/p2.shipping;
}
vector<product>::iterator largest = products.begin();
vector<product>::iterator it = products.begin();
++it;
while (it != products.end())
{
if (*largest < *it)
{
products.erase(largest);
largest = it;
++it;
}
else
{
it = products.erase(it);
}
}
But there is a bug here, after products.erase(largest) is called, it will be invalidated, so you'd better take the approach other suggested here.
Related
I am currently programming a little game for the console with an 2D map. 2 Elements of my game are: destroying fields and an enemy, which spreads in a random direction (its getting bigger). These two "entities" are saved in a structure which contains two vectors (X and Y). I am now trying to erase an element of "_Enemy"(<-private instance of the structure in a class, same as "_DestroyedFields") if you destroy the field where the enemy is.
I tried a lot of different variations to do so and whats giving me the error least is this method (I already searched the internet for a while now an couldn't find a answer to my question):
for (std::vector<int>::iterator itEX = _Enemys.X.begin(), itEY = _Enemys.Y.begin();
itEX != _Enemys.X.end() && itEY != _Enemys.Y.end();
++itEX, ++itEY) {
for (std::vector<int>::iterator itX = _DestroyedFields.X.begin(),
itY = _DestroyedFields.Y.begin();
itX != _DestroyedFields.X.end() && itY != _DestroyedFields.Y.end();
++itX, ++itY) {
if (*itY == *itEY && *itX == *itEX){
itEY = _Enemys.Y.erase(itEY);
itEX = _Enemys.X.erase(itEX);
}
}
}
PS: sorry if my english isn't the best, im german ^^
PSS: if you wanna watch over my whole code, you can find it on Github: https://github.com/Aemmel/ConsoleGame1
After erasing using iterator it, you cannot use it further as it is invalidated. You should use a result of a call to erase which is new, valid iterator.
for( it = v.begin(); it != v.end();)
{
//...
if(...)
{
it = v.erase( it);
}
else
{
++it;
}
...
}
I fixed the bug with first: making a "simple structure"(struct Entity{int X; intY} and then std::vector [insert name here]) and then with adding an break; if the condition is true.
for (Uint itE = 0; itE < _Enemys.size(); ++itE){
for (Uint it = 0; it<_DestroyedFields.size(); ++it){
if (_Enemys.at(itE).Y == _DestroyedFields.at(it).Y
&& _Enemys.at(itE).X == _DestroyedFields.at(it).X){
_Enemys.erase(_Enemys.begin()+itE);
break;
}
}
}
With struct Position {int x; int y;}; and some utility operators,
you may do one of the following: (https://ideone.com/0aiih0)
void filter(std::vector<Position>& positions, const std::vector<Position>& destroyedFields)
{
for (std::vector<Position>::iterator it = positions.begin(); it != positions.end(); ) {
if (std::find(destroyedFields.begin(), destroyedFields.end(), *it) != destroyedFields.end()) {
it = positions.erase(it);
} else {
++it;
}
}
}
Or, if input are sorted, you may use a 'difference':
std::vector<Position> filter2(const std::vector<Position>& positions, const std::vector<Position>& destroyedFields)
{
std::vector<Position> res;
std::set_difference(positions.begin(), positions.end(),
destroyedFields.begin(), destroyedFields.end(),
std::back_inserter(res));
return res;
}
I am trying to insert into this Vector (mCards) in alphabetical order. I can't use sort - it must be inserted in alphabetical order! Right now, I'm inserting in reverse. I can't figure out how to get this to insert in correct alphabetical order!
void Rolodex::Add(Card& card)
{
vector<Card>::iterator temp;
if (mCards.size() != 0)
{
for(vector<Card>::iterator it = mCards.begin(); it != mCards.end(); ++it)
{
Card& currentCard = *it;
temp = it;
int compareResult = currentCard.GetLastName().compare(card.GetLastName());
if (compareResult <= 0)
{
mIteratorNumber = it - mCards.begin();
mCards.insert(temp, card);
return;
}
}
}
else
{
mCards.push_back(card);
for(vector<Card>::iterator it = mCards.begin(); it != mCards.end(); ++it)
{
mIteratorNumber = it - mCards.begin();
}
}
}
If you wants a sorted container, you may instead look at std::map and std::set or their multi variant if you may have duplicated values.
To insert into a sorted vector, the right way to do it is to use std::upper_bound
myCards.insert( std::upper_bound( begin(myCards), end(myCards), card), card );
If the card do not have a valid operator<, use a predicate like this
auto it = std::upper_bound( begin(myCards), end(myCards), card,
[] ( Card const & a, Card const & b ) {
return a.GetLastName().compare(b.GetLastName()) < 0;
} );
myCards.insert( it, card );
I'm trying to create a pricing system for computers and I want to create two vectors, one stores item names, and the other one, prices for the item. My plan is to have two methods, "find_item",that finds the item name inside of a vector and returns its index, and "get_itemPrice", that takes the index from find_item and gets the price. My problem is coming up with a code that takes a string object inside of a vector and returns its index position.
You can simply use std::find. It will return an iterator to the first element equal to the one you are searching for, or to the end() if none is found. You can then use std::distance to get an index from that, if you really need it.
Like this:
vector<int> vect;
int i = 0;
for (vector<int>::iterator iter = vect.begin(); iter != vect.end(); ++iter)
{
if (*iter == vect)
break;
i++;
}
if (i == vect.size())
// not found
else
// i is your index
if you really would like to make an algorithm do it like this, but generally one should not try to beat those in the stl see the algorithm header
unsigned find_str( const vector<string>& vec, const string& key ){
unsigned count = 0;
vector<string>::const_iterator it;
for ( it = vec.begin() ; it < vec.end(); it++ ){
if( *it == key)
return count;
count++;
}
/*throw not found error here.*/
}
I've got code that looks like this:
for (std::list<item*>::iterator i=items.begin();i!=items.end();i++)
{
bool isActive = (*i)->update();
//if (!isActive)
// items.remove(*i);
//else
other_code_involving(*i);
}
items.remove_if(CheckItemNotActive);
I'd like remove inactive items immediately after update them, inorder to avoid walking the list again. But if I add the commented-out lines, I get an error when I get to i++: "List iterator not incrementable". I tried some alternates which didn't increment in the for statement, but I couldn't get anything to work.
What's the best way to remove items as you are walking a std::list?
You have to increment the iterator first (with i++) and then remove the previous element (e.g., by using the returned value from i++). You can change the code to a while loop like so:
std::list<item*>::iterator i = items.begin();
while (i != items.end())
{
bool isActive = (*i)->update();
if (!isActive)
{
items.erase(i++); // alternatively, i = items.erase(i);
}
else
{
other_code_involving(*i);
++i;
}
}
You want to do:
i= items.erase(i);
That will correctly update the iterator to point to the location after the iterator you removed.
You need to do the combination of Kristo's answer and MSN's:
// Note: Using the pre-increment operator is preferred for iterators because
// there can be a performance gain.
//
// Note: As long as you are iterating from beginning to end, without inserting
// along the way you can safely save end once; otherwise get it at the
// top of each loop.
std::list< item * >::iterator iter = items.begin();
std::list< item * >::iterator end = items.end();
while (iter != end)
{
item * pItem = *iter;
if (pItem->update() == true)
{
other_code_involving(pItem);
++iter;
}
else
{
// BTW, who is deleting pItem, a.k.a. (*iter)?
iter = items.erase(iter);
}
}
Of course, the most efficient and SuperCool® STL savy thing would be something like this:
// This implementation of update executes other_code_involving(Item *) if
// this instance needs updating.
//
// This method returns true if this still needs future updates.
//
bool Item::update(void)
{
if (m_needsUpdates == true)
{
m_needsUpdates = other_code_involving(this);
}
return (m_needsUpdates);
}
// This call does everything the previous loop did!!! (Including the fact
// that it isn't deleting the items that are erased!)
items.remove_if(std::not1(std::mem_fun(&Item::update)));
I have sumup it, here is the three method with example:
1. using while loop
list<int> lst{4, 1, 2, 3, 5};
auto it = lst.begin();
while (it != lst.end()){
if((*it % 2) == 1){
it = lst.erase(it);// erase and go to next
} else{
++it; // go to next
}
}
for(auto it:lst)cout<<it<<" ";
cout<<endl; //4 2
2. using remove_if member funtion in list:
list<int> lst{4, 1, 2, 3, 5};
lst.remove_if([](int a){return a % 2 == 1;});
for(auto it:lst)cout<<it<<" ";
cout<<endl; //4 2
3. using std::remove_if funtion combining with erase member function:
list<int> lst{4, 1, 2, 3, 5};
lst.erase(std::remove_if(lst.begin(), lst.end(), [](int a){
return a % 2 == 1;
}), lst.end());
for(auto it:lst)cout<<it<<" ";
cout<<endl; //4 2
4. using for loop , should note update the iterator:
list<int> lst{4, 1, 2, 3, 5};
for(auto it = lst.begin(); it != lst.end();++it){
if ((*it % 2) == 1){
it = lst.erase(it); erase and go to next(erase will return the next iterator)
--it; // as it will be add again in for, so we go back one step
}
}
for(auto it:lst)cout<<it<<" ";
cout<<endl; //4 2
Use std::remove_if algorithm.
Edit:
Work with collections should be like:
prepare collection.
process collection.
Life will be easier if you won't mix this steps.
std::remove_if. or list::remove_if ( if you know that you work with list and not with the TCollection )
std::for_each
The alternative for loop version to Kristo's answer.
You lose some efficiency, you go backwards and then forward again when deleting but in exchange for the extra iterator increment you can have the iterator declared in the loop scope and the code looking a bit cleaner. What to choose depends on priorities of the moment.
The answer was totally out of time, I know...
typedef std::list<item*>::iterator item_iterator;
for(item_iterator i = items.begin(); i != items.end(); ++i)
{
bool isActive = (*i)->update();
if (!isActive)
{
items.erase(i--);
}
else
{
other_code_involving(*i);
}
}
Here's an example using a for loop that iterates the list and increments or revalidates the iterator in the event of an item being removed during traversal of the list.
for(auto i = items.begin(); i != items.end();)
{
if(bool isActive = (*i)->update())
{
other_code_involving(*i);
++i;
}
else
{
i = items.erase(i);
}
}
items.remove_if(CheckItemNotActive);
Removal invalidates only the iterators that point to the elements that are removed.
So in this case after removing *i , i is invalidated and you cannot do increment on it.
What you can do is first save the iterator of element that is to be removed , then increment the iterator and then remove the saved one.
If you think of the std::list like a queue, then you can dequeue and enqueue all the items that you want to keep, but only dequeue (and not enqueue) the item you want to remove. Here's an example where I want to remove 5 from a list containing the numbers 1-10...
std::list<int> myList;
int size = myList.size(); // The size needs to be saved to iterate through the whole thing
for (int i = 0; i < size; ++i)
{
int val = myList.back()
myList.pop_back() // dequeue
if (val != 5)
{
myList.push_front(val) // enqueue if not 5
}
}
myList will now only have numbers 1-4 and 6-10.
Iterating backwards avoids the effect of erasing an element on the remaining elements to be traversed:
typedef list<item*> list_t;
for ( list_t::iterator it = items.end() ; it != items.begin() ; ) {
--it;
bool remove = <determine whether to remove>
if ( remove ) {
items.erase( it );
}
}
PS: see this, e.g., regarding backward iteration.
PS2: I did not thoroughly tested if it handles well erasing elements at the ends.
You can write
std::list<item*>::iterator i = items.begin();
while (i != items.end())
{
bool isActive = (*i)->update();
if (!isActive) {
i = items.erase(i);
} else {
other_code_involving(*i);
i++;
}
}
You can write equivalent code with std::list::remove_if, which is less verbose and more explicit
items.remove_if([] (item*i) {
bool isActive = (*i)->update();
if (!isActive)
return true;
other_code_involving(*i);
return false;
});
The std::vector::erase std::remove_if idiom should be used when items is a vector instead of a list to keep compexity at O(n) - or in case you write generic code and items might be a container with no effective way to erase single items (like a vector)
items.erase(std::remove_if(begin(items), end(items), [] (item*i) {
bool isActive = (*i)->update();
if (!isActive)
return true;
other_code_involving(*i);
return false;
}));
do while loop, it's flexable and fast and easy to read and write.
auto textRegion = m_pdfTextRegions.begin();
while(textRegion != m_pdfTextRegions.end())
{
if ((*textRegion)->glyphs.empty())
{
m_pdfTextRegions.erase(textRegion);
textRegion = m_pdfTextRegions.begin();
}
else
textRegion++;
}
I'd like to share my method. This method also allows the insertion of the element to the back of the list during iteration
#include <iostream>
#include <list>
int main(int argc, char **argv) {
std::list<int> d;
for (int i = 0; i < 12; ++i) {
d.push_back(i);
}
auto it = d.begin();
int nelem = d.size(); // number of current elements
for (int ielem = 0; ielem < nelem; ++ielem) {
auto &i = *it;
if (i % 2 == 0) {
it = d.erase(it);
} else {
if (i % 3 == 0) {
d.push_back(3*i);
}
++it;
}
}
for (auto i : d) {
std::cout << i << ", ";
}
std::cout << std::endl;
// result should be: 1, 3, 5, 7, 9, 11, 9, 27,
return 0;
}
I think you have a bug there, I code this way:
for (std::list<CAudioChannel *>::iterator itAudioChannel = audioChannels.begin();
itAudioChannel != audioChannels.end(); )
{
CAudioChannel *audioChannel = *itAudioChannel;
std::list<CAudioChannel *>::iterator itCurrentAudioChannel = itAudioChannel;
itAudioChannel++;
if (audioChannel->destroyMe)
{
audioChannels.erase(itCurrentAudioChannel);
delete audioChannel;
continue;
}
audioChannel->Mix(outBuffer, numSamples);
}
Given a sorted vector with a number of values, as in the following example:
std::vector<double> f;
f.pushback(10);
f.pushback(100);
f.pushback(1000);
f.pushback(10000);
I'm looking for the most elegant way to retrieve for any double d the two values that are immediately adjacent to it. For example, given the value "45", I'd like this to return "10" and "100".
I was looking at lower_bound and upper_bound, but they don't do what I want. Can you help?
EDIT: I've decided to post my own anser, as it is somewhat a composite of all the helpful answers that I got in this thread. I've voted up those answers which I thought were most helpful.
Thanks everyone,
Dave
You can grab both values (if they exist) in one call with equal_range(). It returns a std::pair of iterators, with first being the first location and second being the last location in which you could insert the value passed without violating ordering. To strictly meet your criteria, you'd have to decrement the iterator in first, after verifying that it wasn't equal to the vector's begin().
You can use STL's lower_bound to get want you want in a few lines of code. lower_bound uses binary search under the hood, so your runtime is O(log n).
double val = 45;
double lower, upper;
std::vector<double>::iterator it;
it = lower_bound(f.begin(), f.end(), val);
if (it == f.begin()) upper = *it; // no smaller value than val in vector
else if (it == f.end()) lower = *(it-1); // no bigger value than val in vector
else {
lower = *(it-1);
upper = *it;
}
You could simply use a binary search, which will run in O(log(n)).
Here is a Lua snippet (I don't have time to do it in C++, sorry) which does what you want, except for limit conditions (that you did not define anyway) :
function search(value, list, first, last)
if not first then first = 1; last = #list end
if last - first < 2 then
return list[first], list[last]
end
local median = math.ceil(first + (last - first)/2)
if list[median] > value then
return search(value, list, first, median)
else
return search(value, list, median, last)
end
end
local list = {1,10,100,1000}
print(search(arg[1] + 0, list))
It takes the value to search from the command line :
$ lua search.lua 10 # didn't know what to do in this case
10 100
$ lua search.lua 101
100 1000
$ lua search.lua 99
10 100
I'm going to post my own anser, and vote anyone up that helped me to reach it, since this is what I'll use in the end, and you've all helped me reach this conclusion. Comments are welcome.
std::pair<value_type, value_type> GetDivisions(const value_type& from) const
{
if (m_divisions.empty())
throw 0; // Can't help you if we're empty.
std::vector<value_type>::const_iterator it =
std::lower_bound(m_divisions.begin(), m_divisions.end(), from);
if (it == m_divisions.end())
return std::make_pair(m_divisions.back(), m_divisions.back());
else if (it == m_divisions.begin())
return std::make_pair(m_divisions.front(), m_divisions.front());
else
return std::make_pair(*(it - 1), *(it));
}
What if (in your case) d is less than the first element or more than the last? And how to deal with negative values? By the way: guaranteeing that your "d" lives between the first and the last value of your vector you can do like that:
// Your initializations
std::vector<double>::const_iterator sit = f.begin();
double upper, lower;
Here is the rest:
while ( *sit < d ) // if the element is still less than your d
++sit; // increase your iterator
upper = *sit; // here you get the upper value
lower = *(--sit); // and here your lower
Elegant enough? :/
You could do a search in your vector for your value (which would tell you where your value would be if it were in the vector) and then return the value before and after that location. So searching for 45 would tell you it should be at index=1 and then you would return 0 and 1 (depending on your implementation of the search, you'll either get the index of the smaller value or the index of the larger value, but this is easy to check with a couple boundary conditions). This should be able to run in O(log n) where n is the number of elements in your vector.
I would write something like this, didn't test if this compiles, but you get the idea:
template <typename Iterator>
std::pair<Iterator, Iterator> find_best_pair(Iterator first, Iterator last, const typename Iterator::value_type & val)
{
std::pair<Iterator, Iterator> result(last, last);
typename Iterator::difference_type size = std::distance(first, last);
if (size == 2)
{
// if the container is of size 2, the answer is the two elements
result.first = first;
result.first = first;
++result.first;
}
else
{
// must be of at lease size 3
if (size > 2)
{
Iterator second = first;
++second;
Iterator prev_last = last;
--prev_last;
Iterator it(std::lower_bound(second, last, val));
if (it != last)
{
result.first = it;
result.second = it;
if (it != prev_last)
{
// if this is not the previous last
// then the answer is (it, it + 1)
++result.second;
}
else
{
// if this the previous last
// then the answer is (it - 1, it)
--result.first;
}
}
}
}
return result;
}
I wrote up this little function, which seems to fit the more general case you wanted. I haven't tested it totally, but I did write a little test code (included).
#include <algorithm>
#include <iostream>
#include <vector>
template <class RandomAccessIt, class Container, class T>
std::pair<RandomAccessIt, RandomAccessIt> bracket_range(RandomAccessIt begin, RandomAccessIt end, Container& c, T val)
{
typename Container::iterator first;
typename Container::iterator second;
first = std::find(begin, end, val);
//Find the first value after this by iteration
second = first;
if (first == begin){ // Found the first element, so set this to end to indicate no lower values
first = end;
}
else if (first != end && first != begin) --first; //Set this to the first value before the found one, if the value was found
while (second != end && *second == val) ++second;
return std::make_pair(first,second);
}
int main(int argc, _TCHAR* argv[])
{
std::vector<int> values;
std::pair<std::vector<int>::iterator, std::vector<int>::iterator> vals;
for (int i = 1; i < 9; ++i) values.push_back(i);
for (int i = 0; i < 10; ++i){
vals = bracket_range(values.begin(), values.end(),values, i);
if (vals.first == values.end() && vals.second == values.end()){ // Not found at all
std::cout << i << " is not in the container." << std::endl;
}
else if (vals.first == values.end()){ // No value lower
std::cout << i << ": " << "None Lower," << *(vals.second) << std::endl;
}
else if (vals.second == values.end()) { // No value higher
std::cout << i << ": " << *(vals.first) << ", None Higher" << std::endl;
}
else{
std::cout << i << ": " << *(vals.first) << "," << *(vals.second) << std::endl;
}
}
return 0;
}
Based on the code that tunnuz posted, here you have some improvements regarding bound checking:
template<typename T>
void find_enclosing_values(const std::vector<T> &vec, const T &value, T &lower, T &upper, const T &invalid_value)
{
std::vector<T>::const_iterator it = vec.begin();
while (it != vec.end() && *it < value)
++it;
if(it != vec.end())
upper = *it;
else
upper = invalid_value;
if(it == vec.begin())
lower = invalid_value;
else
lower = *(--it);
}
Example of usage:
std::vector<int> v;
v.push_back(3);
v.push_back(7);
v.push_back(10);
int lower, upper;
find_enclosing_values(v, 4, lower, upper, -1);
std::cout<<"lower "<<lower<<" upper "<<upper<<std::endl;
If you have the ability to use some other data structure (not a vector), I'd suggest a B-tree. If you data is unchanging, I believe you can retrieve the result in constant time (logarithmic time at the worst).