Best way to extract a subvector from a vector?

vector<T>::const_iterator first = myVec.begin() + 100000;
vector<T>::const_iterator last = myVec.begin() + 101000;
vector<T> newVec(first, last);

It's an O(N) operation to construct the new vector, but there isn't really a better way.

Just use the vector constructor.

std::vector<int>   data();
// Load Z elements into data so that Z > Y > X

std::vector<int>   sub(&data[100000],&data[101000]);

This discussion is pretty old, but the simplest one isn't mentioned yet, with list-initialization:

 vector<int> subvector = {big_vector.begin() + 3, big_vector.end() - 2}; 

It requires c++11 or above.

Example usage:

#include <iostream>
#include <vector>
#include <algorithm>

using namespace std;

int main(){

    vector<int> big_vector = {5,12,4,6,7,8,9,9,31,1,1,5,76,78,8};
    vector<int> subvector = {big_vector.begin() + 3, big_vector.end() - 2};

    cout << "Big vector: ";
    for_each(big_vector.begin(), big_vector.end(),[](int number){cout << number << ";";});
    cout << endl << "Subvector: ";
    for_each(subvector.begin(), subvector.end(),[](int number){cout << number << ";";});
    cout << endl;
}

Result:

Big vector: 5;12;4;6;7;8;9;9;31;1;1;5;76;78;8;
Subvector: 6;7;8;9;9;31;1;1;5;76;

std::vector<T>(input_iterator, input_iterator), in your case foo = std::vector<T>(myVec.begin () + 100000, myVec.begin () + 150000);, see for example here

If both are not going to be modified (no adding/deleting items - modifying existing ones is fine as long as you pay heed to threading issues), you can simply pass around data.begin() + 100000 and data.begin() + 101000, and pretend that they are the begin() and end() of a smaller vector.

Or, since vector storage is guaranteed to be contiguous, you can simply pass around a 1000 item array:

T *arrayOfT = &data[0] + 100000;
size_t arrayOfTLength = 1000;

Both these techniques take constant time, but require that the length of data doesn't increase, triggering a reallocation.

You could just use insert

vector<type> myVec { n_elements };

vector<type> newVec;

newVec.insert(newVec.begin(), myVec.begin() + X, myVec.begin() + Y);

You can use STL copy with O(M) performance when M is the size of the subvector.

Suppose there are two vectors.

 vector<int> vect1{1, 2, 3, 4};
 vector<int> vect2;

Method 1. Using copy function. copy(first_iterator_index, last_iterator_index, back_inserter()) :- This function takes 3 arguments, firstly, the first iterator of old vector. Secondly, the last iterator of old vector and third is back_inserter function to insert values from back.

    // Copying vector by copy function
    copy(vect1.begin(), vect1.end(), back_inserter(vect2));

Method 2. By using Assign Function. assign(first_iterator_o, last_iterator_o). This method assigns the same values to new vector as old one. This takes 2 arguments, first iterator to old vector and last iterator to old vector.

    //Copying vector by assign function
    vect2.assign(vect1.begin(), vect1.end());

The only way to project a collection that is not linear time is to do so lazily, where the resulting "vector" is actually a subtype which delegates to the original collection. For example, Scala's List#subseq method create a sub-sequence in constant time. However, this only works if the collection is immutable and if the underlying language sports garbage collection.

Copy elements from one vector to another easily
In this example, I am using a vector of pairs to make it easy to understand
`

vector<pair<int, int> > v(n);

//we want half of elements in vector a and another half in vector b
vector<pair<lli, lli> > a(v.begin(),v.begin()+n/2);
vector<pair<lli, lli> > b(v.begin()+n/2, v.end());


//if v = [(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)]
//then a = [(1, 2), (2, 3)]
//and b = [(3, 4), (4, 5), (5, 6)]

//if v = [(1, 2), (2, 3), (3, 4), (4, 5), (5, 6), (6, 7)]
//then a = [(1, 2), (2, 3), (3, 4)]
//and b = [(4, 5), (5, 6), (6, 7)]

'
As you can see you can easily copy elements from one vector to another, if you want to copy elements from index 10 to 16 for example then we would use

vector<pair<int, int> > a(v.begin()+10, v.begin+16);

and if you want elements from index 10 to some index from end, then in that case

vector<pair<int, int> > a(v.begin()+10, v.end()-5);

hope this helps, just remember in the last case v.end()-5 > v.begin()+10

Yet another option: Useful for instance when moving between a thrust::device_vector and a thrust::host_vector, where you cannot use the constructor.

std::vector<T> newVector;
newVector.reserve(1000);
std::copy_n(&vec[100000], 1000, std::back_inserter(newVector));

Should also be complexity O(N)

You could combine this with top anwer code

vector<T>::const_iterator first = myVec.begin() + 100000;
vector<T>::const_iterator last = myVec.begin() + 101000;
std::copy(first, last, std::back_inserter(newVector));

vector::assign may be another solution

// note: size1 < src.size() && size2 < src.size()
std::vector<int> sub1(size1), sub2(size2);
sub1.assign(src.begin(), src.begin() + size1);
sub2.assign(src.begin(), src.begin() + size2);

Posting this late just for others..I bet the first coder is done by now. For simple datatypes no copy is needed, just revert to good old C code methods.

std::vector <int>   myVec;
int *p;
// Add some data here and set start, then
p=myVec.data()+start;

Then pass the pointer p and a len to anything needing a subvector.

notelen must be!! len < myVec.size()-start