I'm trying to get a single element from a vector and push it to the back of the vector then remove it so I won't have an empty section in memory. The erase-remove idiom can do this but it removes all instances of a particular value. I just want the first one removed.
I'm not too experienced with standard library algorithms and I can't find the appropriate methods (if any) to do this. Here is an example:
int main() {
std::vector<int> v{1, 2, 3, 3, 4};
remove_first(v, 3);
std::cout << v; // 1, 2, 3, 4
}
So how would I go about removing the first occurance of a 3 from this vector?
Find it first, then erase it:
auto it = std::find(v.begin(),v.end(),3);
// check that there actually is a 3 in our vector
if (it != v.end()) {
v.erase(it);
}
If you don't care about maintaining the ordering of the elements in the vector, you can avoid the copy of the "tail" of remaining elements on erase:
auto it = std::find(v.begin(), v.end(), 3);
if (it != v.end()) {
std::iter_swap(it, v.end() - 1);
v.erase(v.end() - 1);
}
Related
Given two vectors of copyable elements and a predicate over those items, what is an efficient and idiomatic method for:
Removing matching items from the first vector
Appending matching items to the second vector
The following snippet reflects my current thinking, but it does require two passes over the source vector.
vector<int> source{0, 1, 2, 3, 4, 5, 6, 7, 8, 9};
vector<int> target;
auto predicate = [](int n) { return n % 2 == 0; };
std::for_each(source.begin(), source.end(), [&predicate, &target](int n) {
if (predicate(n)) {
target.push_back(n);
}
});
auto it = std::remove_if(source.begin(), source.end(), predicate);
source.erase(it, source.end());
I'd use std::partition to partition sourceinto two parts.
auto itr = std::partition(source.begin(), source.end(), predicate);
target.insert(target.end(), itr, source.end());
source.resize(itr - source.begin());
If the ordering needs to remain the same then use stable_partition.
I agree with #acraig5075, that stable_partition, followed by copy and erase will do what you want.
However, if you don't want to shuffle elements around in the source container (probably a futile wish, IMHO), you can do it in a loop: (untested code)
auto itr = source.begin();
while (itr != source.end()) {
if (predicate(*itr)) {
target.push_back(*itr);
itr = source.erase(itr);
} else
++itr;
}
Note that this will "slide down" the elements of elements one at a time as they are moved into target, which is why I opined that avoiding this was probably foolish. (Note: If you're working with list, then it is less foolish)
Let's say I have std::vector<std::pair<int,Direction>>.
I am trying to use erase-remove_if idiom to remove pairs from the vector.
stopPoints.erase(std::remove_if(stopPoints.begin(),
stopPoints.end(),
[&](const stopPointPair stopPoint)-> bool { return stopPoint.first == 4; }));
I want to delete all pairs that have .first value set to 4.
In my example I have pairs:
- 4, Up
- 4, Down
- 2, Up
- 6, Up
However, after I execute erase-remove_if, I am left with:
- 2, Up
- 6, Up
- 6, Up
What am I doing wrong here?
The correct code is:
stopPoints.erase(std::remove_if(stopPoints.begin(),
stopPoints.end(),
[&](const stopPointPair stopPoint)-> bool
{ return stopPoint.first == 4; }),
stopPoints.end());
You need to remove the range starting from the iterator returned from std::remove_if to the end of the vector, not only a single element.
"Why?"
std::remove_if swaps elements inside the vector in order to put all elements that do not match the predicate towards the beginning of the container. This means that if the predicate (body of the lambda function) returns true, then that element will be placed at the end of the vector.
remove_if then **returns an iterator which points to the first element which matches the predicate **. In other words, an iterator to the first element to be removed.
std::vector::erase erases the range starting from the returned iterator to the end of the vector, such that all elements that match the predicate are removed.
More information: Erase-remove idiom (Wikipedia).
The method std::vector::erase has two overloads:
iterator erase( const_iterator pos );
iterator erase( const_iterator first, const_iterator last );
The first one only remove the element at pos while the second one remove the range [first, last).
Since you forget the last iterator in your call, the first version is chosen by overload resolution, and you only remove the first pair shifted to the end by std::remove_if. You need to do this:
stopPoints.erase(std::remove_if(stopPoints.begin(),
stopPoints.end(),
[&](const stopPointPair stopPoint)-> bool {
return stopPoint.first == 4;
}),
stopPoints.end());
The erase-remove idiom works as follow. Let say you have a vector {2, 4, 3, 6, 4} and you want to remove the 4:
std::vector<int> vec{2, 4, 3, 6, 4};
auto it = std::remove(vec.begin(), vec.end(), 4);
Will transform the vector into {2, 3, 6, A, B} by putting the "removed" values at the end (the values A and B at the end are unspecified (as if the value were moved), which is why you got 6 in your example) and return an iterator to A (the first of the "removed" value).
If you do:
vec.erase(it)
...the first overload of std::vector::erase is chosen and you only remove the value at it, which is the A and get {2, 3, 6, B}.
By adding the second argument:
vec.erase(it, vec.end())
...the second overload is chosen, and you erase value between it and vec.end(), so both A and B are erased.
I know that at the time this question was asked there was no C++20, so just adding answer for completeness & up-to-date answer for this question,
C++20 has now much cleaner & less verbose pattern, using std::erase_if.
See generic code example:
#include <vector>
int main()
{
std::vector<char> cnt(10);
std::iota(cnt.begin(), cnt.end(), '0');
auto erased = std::erase_if(cnt, [](char x) { return (x - '0') % 2 == 0; });
std::cout << erased << " even numbers were erased.\n";
}
Specific question code snippet:
std::erase_if(stopPoints, [&](const stopPointPair stopPoint)-> bool { return stopPoint.first == 4; });
see complete details here:
https://en.cppreference.com/w/cpp/container/vector/erase2
Let's say I have std::vector<std::pair<int,Direction>>.
I am trying to use erase-remove_if idiom to remove pairs from the vector.
stopPoints.erase(std::remove_if(stopPoints.begin(),
stopPoints.end(),
[&](const stopPointPair stopPoint)-> bool { return stopPoint.first == 4; }));
I want to delete all pairs that have .first value set to 4.
In my example I have pairs:
- 4, Up
- 4, Down
- 2, Up
- 6, Up
However, after I execute erase-remove_if, I am left with:
- 2, Up
- 6, Up
- 6, Up
What am I doing wrong here?
The correct code is:
stopPoints.erase(std::remove_if(stopPoints.begin(),
stopPoints.end(),
[&](const stopPointPair stopPoint)-> bool
{ return stopPoint.first == 4; }),
stopPoints.end());
You need to remove the range starting from the iterator returned from std::remove_if to the end of the vector, not only a single element.
"Why?"
std::remove_if swaps elements inside the vector in order to put all elements that do not match the predicate towards the beginning of the container. This means that if the predicate (body of the lambda function) returns true, then that element will be placed at the end of the vector.
remove_if then **returns an iterator which points to the first element which matches the predicate **. In other words, an iterator to the first element to be removed.
std::vector::erase erases the range starting from the returned iterator to the end of the vector, such that all elements that match the predicate are removed.
More information: Erase-remove idiom (Wikipedia).
The method std::vector::erase has two overloads:
iterator erase( const_iterator pos );
iterator erase( const_iterator first, const_iterator last );
The first one only remove the element at pos while the second one remove the range [first, last).
Since you forget the last iterator in your call, the first version is chosen by overload resolution, and you only remove the first pair shifted to the end by std::remove_if. You need to do this:
stopPoints.erase(std::remove_if(stopPoints.begin(),
stopPoints.end(),
[&](const stopPointPair stopPoint)-> bool {
return stopPoint.first == 4;
}),
stopPoints.end());
The erase-remove idiom works as follow. Let say you have a vector {2, 4, 3, 6, 4} and you want to remove the 4:
std::vector<int> vec{2, 4, 3, 6, 4};
auto it = std::remove(vec.begin(), vec.end(), 4);
Will transform the vector into {2, 3, 6, A, B} by putting the "removed" values at the end (the values A and B at the end are unspecified (as if the value were moved), which is why you got 6 in your example) and return an iterator to A (the first of the "removed" value).
If you do:
vec.erase(it)
...the first overload of std::vector::erase is chosen and you only remove the value at it, which is the A and get {2, 3, 6, B}.
By adding the second argument:
vec.erase(it, vec.end())
...the second overload is chosen, and you erase value between it and vec.end(), so both A and B are erased.
I know that at the time this question was asked there was no C++20, so just adding answer for completeness & up-to-date answer for this question,
C++20 has now much cleaner & less verbose pattern, using std::erase_if.
See generic code example:
#include <vector>
int main()
{
std::vector<char> cnt(10);
std::iota(cnt.begin(), cnt.end(), '0');
auto erased = std::erase_if(cnt, [](char x) { return (x - '0') % 2 == 0; });
std::cout << erased << " even numbers were erased.\n";
}
Specific question code snippet:
std::erase_if(stopPoints, [&](const stopPointPair stopPoint)-> bool { return stopPoint.first == 4; });
see complete details here:
https://en.cppreference.com/w/cpp/container/vector/erase2
I'm trying to get a single element from a vector and push it to the back of the vector then remove it so I won't have an empty section in memory. The erase-remove idiom can do this but it removes all instances of a particular value. I just want the first one removed.
I'm not too experienced with standard library algorithms and I can't find the appropriate methods (if any) to do this. Here is an example:
int main() {
std::vector<int> v{1, 2, 3, 3, 4};
remove_first(v, 3);
std::cout << v; // 1, 2, 3, 4
}
So how would I go about removing the first occurance of a 3 from this vector?
Find it first, then erase it:
auto it = std::find(v.begin(),v.end(),3);
// check that there actually is a 3 in our vector
if (it != v.end()) {
v.erase(it);
}
If you don't care about maintaining the ordering of the elements in the vector, you can avoid the copy of the "tail" of remaining elements on erase:
auto it = std::find(v.begin(), v.end(), 3);
if (it != v.end()) {
std::iter_swap(it, v.end() - 1);
v.erase(v.end() - 1);
}
I was looking at the API documentation for stl vector, and noticed there was no method on the vector class that allowed the removal of an element with a certain value. This seems like a common operation, and it seems odd that there's no built in way to do this.
std::remove does not actually erase elements from the container: it moves the elements to be removed to the end of the container, and returns the new end iterator which can be passed to container_type::erase to do the actual removal of the extra elements that are now at the end of the container:
std::vector<int> vec;
// .. put in some values ..
int int_to_remove = n;
vec.erase(std::remove(vec.begin(), vec.end(), int_to_remove), vec.end());
If you want to remove an item, the following will be a bit more efficient.
std::vector<int> v;
auto it = std::find(v.begin(), v.end(), 5);
if(it != v.end())
v.erase(it);
or you may avoid overhead of moving the items if the order does not matter to you:
std::vector<int> v;
auto it = std::find(v.begin(), v.end(), 5);
if (it != v.end()) {
using std::swap;
// swap the one to be removed with the last element
// and remove the item at the end of the container
// to prevent moving all items after '5' by one
swap(*it, v.back());
v.pop_back();
}
Use the global method std::remove with the begin and end iterator, and then use std::vector.erase to actually remove the elements.
Documentation links
std::remove http://www.cppreference.com/cppalgorithm/remove.html
std::vector.erase http://www.cppreference.com/cppvector/erase.html
std::vector<int> v;
v.push_back(1);
v.push_back(2);
//Vector should contain the elements 1, 2
//Find new end iterator
std::vector<int>::iterator newEnd = std::remove(v.begin(), v.end(), 1);
//Erase the "removed" elements.
v.erase(newEnd, v.end());
//Vector should now only contain 2
Thanks to Jim Buck for pointing out my error.
From c++20:
A non-member function introduced std::erase, which takes the vector and value to be removed as inputs.
ex:
std::vector<int> v = {90,80,70,60,50};
std::erase(v,50);
The other answers cover how to do this well, but I thought I'd also point out that it's not really odd that this isn't in the vector API: it's inefficient, linear search through the vector for the value, followed by a bunch of copying to remove it.
If you're doing this operation intensively, it can be worth considering std::set instead for this reason.
If you have an unsorted vector, then you can simply swap with the last vector element then resize().
With an ordered container, you'll be best off with std::vector::erase(). Note that there is a std::remove() defined in <algorithm>, but that doesn't actually do the erasing. (Read the documentation carefully).
*
C++ community has heard your request :)
*
C++ 20 provides an easy way of doing it now.
It gets as simple as :
#include <vector>
...
vector<int> cnt{5, 0, 2, 8, 0, 7};
std::erase(cnt, 0);
You should check out std::erase and std::erase_if.
Not only will it remove all elements of the value (here '0'), it will do it in O(n) time complexity. Which is the very best you can get.
If your compiler does not support C++ 20, you should use erase-remove idiom:
#include <algorithm>
...
vec.erase(std::remove(vec.begin(), vec.end(), 0), vec.end());
See also std::remove_if to be able to use a predicate...
Here's the example from the link above:
vector<int> V;
V.push_back(1);
V.push_back(4);
V.push_back(2);
V.push_back(8);
V.push_back(5);
V.push_back(7);
copy(V.begin(), V.end(), ostream_iterator<int>(cout, " "));
// The output is "1 4 2 8 5 7"
vector<int>::iterator new_end =
remove_if(V.begin(), V.end(),
compose1(bind2nd(equal_to<int>(), 0),
bind2nd(modulus<int>(), 2)));
V.erase(new_end, V.end()); [1]
copy(V.begin(), V.end(), ostream_iterator<int>(cout, " "));
// The output is "1 5 7".
A shorter solution (which doesn't force you to repeat the vector name 4 times) would be to use Boost:
#include <boost/range/algorithm_ext/erase.hpp>
// ...
boost::remove_erase(vec, int_to_remove);
See http://www.boost.org/doc/libs/1_64_0/libs/range/doc/html/range/reference/algorithms/new/remove_erase.html
Two ways are there by which you can use to erase an item particularly.
lets take a vector
std :: vector < int > v;
v.push_back(10);
v.push_back(20);
v.push_back(30);
v.push_back(40);
v.push_back(40);
v.push_back(50);
1) Non efficient way : Although it seems to be quite efficient but it's not because erase function delets the elements and shifts all the elements towards left by 1.
so its complexity will be O(n^2)
std :: vector < int > :: iterator itr = v.begin();
int value = 40;
while ( itr != v.end() )
{
if(*itr == value)
{
v.erase(itr);
}
else
++itr;
}
2) Efficient way ( RECOMMENDED ) : It is also known as ERASE - REMOVE idioms .
std::remove transforms the given range into a range with all the elements that compare not equal to given element shifted to the start of the container.
So, actually don't remove the matched elements.
It just shifted the non matched to starting and gives an iterator to new valid end.
It just requires O(n) complexity.
output of the remove algorithm is :
10 20 30 50 40 50
as return type of remove is iterator to the new end of that range.
template <class ForwardIterator, class T>
ForwardIterator remove (ForwardIterator first, ForwardIterator last, const T& val);
Now use vector’s erase function to delete elements from the new end to old end of the vector. It requires O(1) time.
v.erase ( std :: remove (v.begin() , v.end() , element ) , v.end () );
so this method work in O(n)
Similar to the erase remove idiom, for vector one could use resize and remove and use iterator distance computation:
std::vector<int> vec;
// .. put in some values ..
int int_to_remove = n;
vec.resize(std::remove(vec.begin(), vec.end(), int_to_remove) - vec.begin());
Tested here.
If you want to do it without any extra includes:
vector<IComponent*> myComponents; //assume it has items in it already.
void RemoveComponent(IComponent* componentToRemove)
{
IComponent* juggler;
if (componentToRemove != NULL)
{
for (int currComponentIndex = 0; currComponentIndex < myComponents.size(); currComponentIndex++)
{
if (componentToRemove == myComponents[currComponentIndex])
{
//Since we don't care about order, swap with the last element, then delete it.
juggler = myComponents[currComponentIndex];
myComponents[currComponentIndex] = myComponents[myComponents.size() - 1];
myComponents[myComponents.size() - 1] = juggler;
//Remove it from memory and let the vector know too.
myComponents.pop_back();
delete juggler;
}
}
}
}