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

Question

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

std::vector<int> vec;

vec.push_back(6);
vec.push_back(-17);
vec.push_back(12);

vec.erase(???);

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));

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.

Answer 3

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

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

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));

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 i position, so instead we can access the pointer to the ith element by:

&vec[i]

So we can write:

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

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.

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).

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)

Answer 10

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/

Answer 11

I suggest to read this since I believe that is what are you looking for. https://en.wikipedia.org/wiki/Erase%E2%80%93remove_idiom

If you use for example

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

you will erase n -th element of vector but when you erase second element, all other elements of vector will be shifted and vector sized will be -1. This can be problem if you loop through vector since vector size() is decreasing. If you have problem like this provided link suggested to use existing algorithm in standard C++ library. and "remove" or "remove_if".

Hope that this helped

Answer 12

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

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

    if(condition)
        vec.erase(vec.begin() + i);

}

Answer 13

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 14

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 15

IF U ARE LOOKING FOR - .... vector vec ;

How to delete an element at particular position ? like vec[i][j] position

https://gist.github.com/kunal2013/86734b14e39160a3fbfe31278bfb4e1b

Answer 16

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).

Answer 17

How about this?

void squeeze(vector<int> &v)
{
    int j = 0;
    for (int i = 1; i < v.size(); i++)
        if (v[i] != v[j] && ++j != i)
            v[j] = v[i];
    v.resize(j + 1);
}

Answer 18

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);