How do I erase an element from std::vector<> by index?

Question

I have a std::vector<int>, and I want to delete the nth element. How do I do that?

Example:

std::vector<int> vec;
vec.push_back(6);
vec.push_back(-17);
vec.push_back(12);
vec.erase(???);

• Dario: Consider using a std::deque which provides inserting and deleting at both ends.
• Owl: No, don't consider using deque just because you may want to delete an element, that's really poor advice. There's a whole load of reasons why you may want to use deque or vector. It is true that deleting an element from a vector can be costly - esp if the vector is large, but there's no reason to think that a deque would be any better than a vector from the code example you just posted.
• Michel Billaud: For example, if you have a graphical application where you display a "list" of things where you insert/remove things interactively, consider you run through the list 50-100 times each second to display them, and you add/remove things a few times each minute. So implementing the "list" as a vector is probably a better option in term of total efficiency.
• user11991978: I recommend std::vector.erase(...), which is also my preference - you can choose to delete either a single element or a range.

Answer 1

To delete a single element, you could do:

std::vector<int> vec;
vec.push_back(6);
vec.push_back(-17);
vec.push_back(12);
// Deletes the second element (vec[1])
vec.erase(std::next(vec.begin()));

Or, to delete more than one element at once:

// Deletes the second through third elements (vec[1], vec[2])
vec.erase(std::next(vec.begin(), 1), std::next(vec.begin(), 3));

• bobobobo: Note also binary operator+ is not necessarily defined for iterators on other container types, like list<T>::iterator (you cannot do list.begin() + 2 on an std::list, you have to use std::advance for that)
• Pierre: Thank you to all who have answered. What are we to think of a class design when such a simple operation as deleting an element, requires one to come to StackOverflow?
• Caleth: @Pierre because the numerical index of a particular element is not the primary model of access, iterator is. All the functions that look at elements of a container use that container's iterators. E.g. std::find_if
• Richard W: @Caleth yeah, but std::vector could still provide a method for this very common use-case. Everybody bashes Qt containers, but QList for instance has removeOne() which is just a no-brainer compared to the ugliness of std::vector.
• user686249: @Caleth Huh? Access by index (at() and operator[]) is one of the main distinctive properties of std::vector. It would only be consistent to also support inserting and erasing at an index.

Answer 2

The erase method on std::vector is overloaded, so it's probably clearer to call

vec.erase(vec.begin() + index);

when you only want to erase a single element.

• Anne Quinn: if there's only one element, index is 0, and so you get vec.begin() which is valid.
• droid192: I wish someone would have mentioned that vec.erase(0) does not work, but vec.erase(vec.begin()+0) (or without +0) does. Otherwise I get no matching function call, which is why I came here
• L. F.: @qrtLs vec.erase(0) may actually compile if 0 happens to be interpreted as the null pointer constant ...
• dixit_chandra: @qrtLs erase() function takes iterator type as the argument; as 0 is not an iterator, it would give compiler error as no matching function call.
• mschoenebeck: this should be the accepted answer as this is a solution based on index (as asked for)

Answer 3

template <typename T>
void remove(std::vector<T>& vec, std::size_t pos)
{
 std::vector<T>::iterator it = vec.begin();
 std::advance(it, pos);
 vec.erase(it);
}

• user1664047: Max, what makes that function better than: template <typename T> void remove(std::vector<T>& vec, size_t pos) { vec.erase(vec.begin + pos); } I'm not saying either is better, merely asking out of personal interest and to return the best result this question could get.
• Lily Ballard: @JoeyvG: Since a vector<T>::iterator is a random-access iterator, your version is fine and maybe a bit clearer. But the version that Max posted should work just fine if you change the container to another one that doesn't support random-access iterators
• Bim: This is imo the better answer, as it applies to other container formats too. You could also use std::next().
• BartoszKP: Much better approach since it doesn't rely on the internals of the container.
• Goblinhack: std::advance is only needed if you think this will not be a vector i.e. a list. But as you've specified that here, would operator+ not be simpler ? According to this stackoverflow.com/questions/1668088/… there is a possible gain in performance with operator+

Answer 4

The erase method will be used in two ways:

1. Erasing single element:

   vector.erase( vector.begin() + 3 ); // Deleting the fourth element
   

2. Erasing range of elements:

   vector.erase( vector.begin() + 3, vector.begin() + 5 ); // Deleting from fourth element to sixth element
   

• AlastairG: This is a duplicate answer almost 7 years after the accepted answer. Please don't do this.
• Nicholas Pipitone: @AlastairG This answer is a lot shorter and clearer than the original answer, though it could technically simply be an edit instead (Such an edit might be against the wishes of the original answer's OP though)
• Maciek Woźniak: I think we should add - it removes from 4th to 6th element excluded (6th element is not included/erased)

Answer 5

Erase an element with index :

vec.erase(vec.begin() + index);

Erase an element with value:

vec.erase(find(vec.begin(),vec.end(),value));

• Yunnosch: Please make more obvious the additional insight this answer provides in comparison to other existing older and upvoted answers.

Answer 6

Actually, the erase function works for two profiles:

• Removing a single element

  iterator erase (iterator position);
  

• Removing a range of elements

  iterator erase (iterator first, iterator last);
  

Since std::vec.begin() marks the start of container and if we want to delete the ith element in our vector, we can use:

vec.erase(vec.begin() + index);

If you look closely, vec.begin() is just a pointer to the starting position of our vector and adding the value of i to it increments the pointer to the position i. So, we instead can access the pointer to the ith element with &vec[i]:

vec.erase(&vec[i]); // To delete the ith element

• CuriousGeorge: -1 The last line does not compile (at least in VS2017). The code assumes that vector::iterator is implicitly constructible from a raw pointer, which is not required by the standard.
• Nishant Singh: This is true especially for debug iterators

Answer 7

If you have an unordered vector you can take advantage of the fact that it's unordered and use something I saw from Dan Higgins at CPPCON

template< typename TContainer >
static bool EraseFromUnorderedByIndex( TContainer& inContainer, size_t inIndex )
{
 if ( inIndex < inContainer.size() )
 {
 if ( inIndex != inContainer.size() - 1 )
 inContainer[inIndex] = inContainer.back();
 inContainer.pop_back();
 return true;
 }
 return false;
}

Since the list order doesn't matter, just take the last element in the list and copy it over the top of the item you want to remove, then pop and delete the last item.

• theferrit32: I think this is the best answer if the vector is unordered. It does not rely on the assumption that iterator + index will actually give you back the iterator position at that index, which is not true of all iterable containers. It is also constant complexity instead of linear by taking advantage of the back pointer.
• Will Crawford: This totally needs to be added to the std lib as unordered_remove and unordered_remove_if ... unless it has been and I missed it, which is happening more and more often these days :)
• Carsten S: If would suggest to use move-assignment or swap instead of a copy-assignment.
• keith: std::remove reorders the container so that all the elements to be removed are at the end, no need to do it manually like this if you are using C++ 17.
• Paul Du Bois: @keith how does std::remove help? cppreference claims that even in C++17, all remove overloads require a predicate, and none take an index.

Answer 8

It may seem obvious to some people, but to elaborate on the above answers:

If you are doing removal of std::vector elements using erase in a loop over the whole vector, you should process your vector in reverse order, that is to say using

for (int i = v.size() - 1; i >= 0; i--)

instead of (the classical)

for (int i = 0; i < v.size(); i++)

The reason is that indices are affected by erase so if you remove the 4-th element, then the former 5-th element is now the new 4-th element, and it won't be processed by your loop if you're doing i++.

Below is a simple example illustrating this where I want to remove all the odds element of an int vector;

#include <iostream>
#include <vector>
using namespace std;
void printVector(const vector<int> &v)
{
 for (size_t i = 0; i < v.size(); i++)
 {
 cout << v[i] << " ";
 }
 cout << endl;
}
int main()
{ 
 vector<int> v1, v2;
 for (int i = 0; i < 10; i++)
 {
 v1.push_back(i);
 v2.push_back(i);
 }
 // print v1
 cout << "v1: " << endl;
 printVector(v1);
 
 cout << endl;
 
 // print v2
 cout << "v2: " << endl;
 printVector(v2);
 
 // Erase all odd elements
 cout << "--- Erase odd elements ---" << endl;
 
 // loop with decreasing indices
 cout << "Process v2 with decreasing indices: " << endl;
 for (int i = v2.size() - 1; i >= 0; i--)
 {
 if (v2[i] % 2 != 0)
 {
 cout << "# ";
 v2.erase(v2.begin() + i);
 }
 else
 {
 cout << v2[i] << " ";
 }
 }
 cout << endl;
 cout << endl;
 
 // loop with increasing indices
 cout << "Process v1 with increasing indices: " << endl;
 for (int i = 0; i < v1.size(); i++)
 {
 if (v1[i] % 2 != 0)
 {
 cout << "# ";
 v1.erase(v1.begin() + i);
 }
 else
 {
 cout << v1[i] << " ";
 }
 }
 
 
 return 0;
}

Output:

v1:
0 1 2 3 4 5 6 7 8 9
v2:
0 1 2 3 4 5 6 7 8 9
--- Erase odd elements ---
Process v2 with decreasing indices:
# 8 # 6 # 4 # 2 # 0
Process v1 with increasing indices:
0 # # # # #

Note that on the second version with increasing indices, even numbers are not displayed as they are skipped because of i++

Note also that processing the vector in reverse order, you CAN'T use unsigned types for indices (for (uint8_t i = v.size() -1; ... won't work). This because when i equals 0, i-- will overflow and be equal to 255 for uint8_t for example (so the loop won't stop as i will still be >= 0, and probably out of bounds of the vector).

• Wormer: But if one uses iterators instead of index i to iterate then he can take advantage of the fact that erase returns the iterator to the next element after the one being removed. Even so it is advisable to use the remove-erase idiom to minimize the number of moves of the remaining (non-removed) elements -- first pop everything to remove to the end of the container, then remove them all at once in one operation.

Answer 9

If you work with large vectors (size> 100,000) and want to delete lots of elements, I would recommend to do something like this:

int main(int argc, char** argv) {
 vector <int> vec;
 vector <int> vec2;
 for (int i = 0; i < 20000000; i++){
 vec.push_back(i);}
 for (int i = 0; i < vec.size(); i++)
 {
 if(vec.at(i) %3 != 0)
 vec2.push_back(i);
 }
 vec = vec2;
 cout << vec.size() << endl;
}

The code takes every number in vec that can't be divided by 3 and copies it to vec2. Afterwards it copies vec2 in vec. It is pretty fast. To process 20,000,000 elements this algorithm only takes 0.8 sec!

I did the same thing with the erase-method, and it takes lots and lots of time:

Erase-Version (10k elements) : 0.04 sec
Erase-Version (100k elements) : 0.6 sec
Erase-Version (1000k elements): 56 sec
Erase-Version (10000k elements): ...still calculating (>30 min)

• Regis Portalez: how does this answer the question?
• Roddy: Interesting, but not relevant to the question!
• user202729: Won't a in-place algorithm faster?
• RiaD: that's std::remove_if (+erase)

Answer 10

I suggest you read about the erase–remove idiom.
For example:

vec.erase(vec.begin() + 1, vec.begin() + 3);

With this, you will erase the nth element of vec but before you erase the second element, all the other elements of vec will be shifted and the vector size will be reduced by 1. This can be a problem as you may be looping through vec while its size() is decreasing.

If you have problem like this the provided link suggests to use remove and remove_if.

• Jan Schultke: This idiom has been obsoleted by std::erase_if.

Answer 11

To delete an element use the following way:

// declaring and assigning array1 
std:vector<int> array1 {0,2,3,4};
// erasing the value in the array
array1.erase(array1.begin()+n);

For a more broad overview you can visit: http://www.cplusplus.com/reference/vector/vector/erase/

• L. F.: Consider using cppreference. See this, this, etc.

Answer 12

if you need to erase an element inside of a for-loop, do the following:

int i = 0;
while(i < vec.size()){
 if(condition)
 vec.erase(vec.begin() + i);
 else
 ++i;
}

Answer 13

You need to use the Standard Template Library's std::vector::erase function.

Example: Deleting an element from a vector (using index)

// Deleting the eleventh element from vector vec
vec.erase( vec.begin() + 10 ); 

Explanation of the above code

std::vector<T,Allocator>::erase Usage:

iterator erase (iterator position); // until C++11

iterator erase( const_iterator pos ); // since C++11 and until C++20

constexpr iterator erase( const_iterator pos ); // since C++20

Here there is a single parameter, position which is an iterator pointing to a single element to be removed from the vector. Member types iterator and const_iterator are random access iterator types that point to elements.

How it works

erase function does the following:

• It removes from the vector either a single element (position) or a range of elements ([first, last)).

• It reduces the container size by the number of elements removed, which are destroyed.

Note: The iterator pos must be valid and dereferenceable. Thus the end() iterator (which is valid, but is not dereferenceable) cannot be used as a value for pos.

Return value and Complexity

The return value is an iterator pointing to the new location of the element that followed the last element that was erased by the function call. This is the container end of the operation that erased the last element in the sequence.

Member type iterator is a random access iterator type that points to elements.

Here, the time complexity is linear on the number of elements erased (destructions) plus the number of elements after the last element is deleted (moving).

• Usukhbayar Batbayar: deleting Nth index example is super helpful. Thank you.

Answer 14

The previous answers assume that you always have a signed index. Sadly, std::vector uses size_type for indexing, and difference_type for iterator arithmetic, so they don't work together if you have "-Wconversion" and friends enabled. This is another way to answer the question, while being able to handle both signed and unsigned:

To remove:

template<class T, class I, class = typename std::enable_if<std::is_integral<I>::value>::type>
void remove(std::vector<T> &v, I index)
{
 const auto &iter = v.cbegin() + gsl::narrow_cast<typename std::vector<T>::difference_type>(index);
 v.erase(iter);
}

To take:

template<class T, class I, class = typename std::enable_if<std::is_integral<I>::value>::type>
T take(std::vector<T> &v, I index)
{
 const auto &iter = v.cbegin() + gsl::narrow_cast<typename std::vector<T>::difference_type>(index);
 auto val = *iter;
 v.erase(iter);
 return val;
}

Answer 15

here is one more way to do this if you want to delete a element by finding this with its value in vector,you just need to do this on vector.

vector<int> ar(n);
ar.erase(remove(ar.begin(), ar.end()), (place your value here from vector array));

it will remove your value from here. thanks

Answer 16

the fastest way (for programming contests by time complexity() = constant)

can erase 100M item in 1 second;

 vector<int> it = (vector<int>::iterator) &vec[pos];
 vec.erase(it);

and most readable way : vec.erase(vec.begin() + pos);

• Marshall Clow: This is very non-portable; it will work with libstdc++, but not libc++, and not with MSVC. vector<int>::iterator is not necessarily the same as int *
• Jonathan Wakely: It's disgusting, I think I'll change libstdc++ to stop it from working.