Best way to extract a subvector from a vector?

Question

Suppose I have a std::vector (let's call it myVec) of size N. What's the simplest way to construct a new vector consisting of a copy of elements X through Y, where 0 <= X <= Y <= N-1? For example, myVec [100000] through myVec [100999] in a vector of size 150000.

If this cannot be done efficiently with a vector, is there another STL datatype that I should use instead?

Answer 1

These days, we use spans! So you would write:

#include <span>

...
auto start_pos = 100000;
auto length = 1000;
auto span_of_myvec = std::span(myvec);
auto my_subspan = span_of_myvec.subspan(start_pos, length);

to get a span of 1000 elements of the same type as myvec's. Or a more terse form:

auto my_subspan = std::span(myvec).subspan(1000000, 1000);

(but I don't like this as much, since the meaning of each numeric argument is not entirely clear; and it gets worse if the length and start_pos are of the same order of magnitude.)

Anyway, remember that this is not a copy, it's just a view of the data in the vector, so be careful. If you want an actual copy, you could do:

std::vector<T> new_vec(my_subspan.cbegin(), my_subspan.cend());

Notes:

• If you use a version of C++ older than C++20, you can use gsl::span - from the Guidelines Support Library. For more information, see: http://www.modernescpp.com/index.php/c-core-guideline-the-guidelines-support-library and https://github.com/martinmoene/gsl-lite for a concrete implementation
• For more information about spans, see: What is a "span" and when should I use one?
• std::vector has a gazillion constructors, it's super-easy to fall into one you didn't intend to use, so be careful.

Answer 2

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;

Note: the extent of the range is defined to take the second iterator as exclusive. e.g. in the above example the last element in the subvector is 76, which is 3 positions before big_vector.end()

Answer 3

You can use a combination of views::drop() and views::take():

#include <vector>
#include <string>
#include <ranges>
#include <format>
using namespace std;
using namespace std::ranges;

int main(){
    const vector<int> v = views::iota(0,10) | ranges::to<vector>();

    const vector<int> sub = v | views::drop(3) | views::take(4) | ranges::to<vector>() ;

    const string buffer = sub |
                views::transform([](int i){return format("{}",i);}) | 
                views::join_with(',') | 
                ranges::to<string>();

    printf("%s", buffer.c_str());
    return 0;
}

Answer 4

Just to add another recent (C++14 onwards) option to the table, for a similar situation we created a subvector class (see igormcoelho/view_wrapper project). The subvector creates a constant time O(1) range view to a vector class, that can also generate a copy with method as_copy:

using view_wrapper::subvector;
// creates vector with 150000 elements
std::vector<int> myVec(150000, 0);
// get range interval [100000, 100999) in O(1) constant time
subvector<int> v(myVec, 100000, 100999); 
// make copy in O(M) time, where M=100999-100000
std::vector<int> v2 = v.as_copy()

The coolest thing on subvector is that, as a range, one can change it and this will have an effect on the original vector:

// add 1 in the end of the subvector v and in the middle of myVec
v.push_back(1);
// size of myVec is now one more
assert(myVec.size() == 150001);

Note that a subvector push_back operation will in fact occur as an emplace in the middle of the original vector, so it may cost O(N), and not O(1) amortized cost of typical push_back.

Finally, if your compiler supports c++20 you can easy convert subvector into a std::span with method as_span():

std::span sp = v.as_span();

Answer 5

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);

Answer 6

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());

Answer 7

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

Answer 8

You could just use insert

vector<type> myVec { n_elements };

vector<type> newVec;

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

Answer 9

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));

Answer 10

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

Answer 11

Maybe the array_view/span in the GSL library is a good option.

Here is also a single file implementation: array_view.

Answer 12

You didn't mention what type std::vector<...> myVec is, but if it's a simple type or struct/class that doesn't include pointers, and you want the best efficiency, then you can do a direct memory copy (which I think will be faster than the other answers provided). Here is a general example for std::vector<type> myVec where type in this case is int:

typedef int type; //choose your custom type/struct/class
int iFirst = 100000; //first index to copy
int iLast = 101000; //last index + 1
int iLen = iLast - iFirst;
std::vector<type> newVec;
newVec.resize(iLen); //pre-allocate the space needed to write the data directly
memcpy(&newVec[0], &myVec[iFirst], iLen*sizeof(type)); //write directly to destination buffer from source buffer

Answer 13

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

Answer 14

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.

Answer 15

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]);

Answer 16

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.

Answer 17

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.

Answer 18

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