fastest way to convert a std::vector to another std::vector

Question

What is the fastest way (if there is any other) to convert a std::vector from one datatype to another (with the idea to save space)? For example:

std::vector         


        

Answer 1

Copying element by element is not highly inefficient. std::vector provides constant access time to any of its elements, hence the operation will be O(n) overall. You will not notice it.

Answer 2

std::vector<bool> 

Stop.

A std::vector<bool> is... not. std::vector has a specialization for the use of the type bool, which causes certain changes in the vector. Namely, it stops acting like a std::vector.

There are certain things that the standard guarantees you can do with a std::vector. And vector<bool> violates those guarantees. So you should be very careful about using them.

Anyway, I'm going to pretend you said vector<int> instead of vector<bool>, as the latter really complicates things.

Copying element by element is highly inefficient in case of a really large vector.

Only if you do it wrong.

Vector casting of the type you want needs to be done carefully to be efficient.

If the the source T type is convertible to the destination T, then this is works just fine:

vector<Tnew> vec_new(vec_old.begin(), vec_old.end());

Decent implementations should recognize when they've been given random-access iterators and optimize the memory allocation and loop appropriately.

The biggest problem for non-convertible types you'll have for simple types is not doing this:

std::vector<int> newVec(oldVec.size());

That's bad. That will allocate a buffer of the proper size, but it will also fill it with data. Namely, default-constructed ints (int()).

Instead, you should do this:

std::vector<int> newVec;
newVec.reserve(oldVec.size());

This reserves capacity equal to the original vector, but it also ensures that no default construction takes place. You can now push_back to your hearts content, knowing that you will never cause reallocation in your new vector.

From there, you can just loop over each entry in the old vector, doing the conversion as needed.

Answer 3

You should be able to use assign like this:

vector<unsigned short> v;
//...
vector<bool> u;
//...
u.assign(v.begin(), v.end());

Answer 4

#ifdef VECTOR_H_TYPE1
#ifdef VECTOR_H_TYPE2
#ifdef VECTOR_H_CLASS
/* Other methods can be added as needed, provided they likewise carry out the same operations on both */

#include <vector>

using namespace std;

class VECTOR_H_CLASS {
public:
        vector<VECTOR_H_TYPE1> *firstVec;
        vector<VECTOR_H_TYPE2> *secondVec;

        VECTOR_H_CLASS(vector<VECTOR_H_TYPE1> &v1, vector<VECTOR_H_TYPE2> &v2) { firstVec = &v1; secondVec = &v2; }
        ~VECTOR_H_CLASS() {}

        void init() { // Use this to copy a full vector into an empty (or garbage) vector to equalize them
                secondVec->clear();
                for(vector<VECTOR_H_TYPE1>::iterator it = firstVec->begin(); it != firstVec->end(); it++) secondVec->push_back((VECTOR_H_TYPE2)*it);
        }

        void push_back(void *value) {
                firstVec->push_back((VECTOR_H_TYPE1)value);
                secondVec->push_back((VECTOR_H_TYPE2)value);
        }

        void pop_back() {
                firstVec->pop_back();
                secondVec->pop_back();
        }

        void clear() {
                firstVec->clear();
                secondVec->clear();
        }
};
#undef VECTOR_H_CLASS
#endif
#undef VECTOR_H_TYPE2
#endif
#undef VECTOR_H_TYPE1
#endif

Answer 5

There's no way to avoid the copy, since a std::vector<T> is a distinct type from std::vector<U>, and there's no way for them to share the memory. Other than that, it depends on how the data is mapped. If the mapping corresponds to an implicit conversion (e.g. unsigned short to bool), then simply creating a new vector using the begin and end iterators from the old will do the trick:

std::vector<bool> newV( oldV.begin(), oldV.end() );

If the mapping isn't just an implicit conversion (and this includes cases where you want to verify things; e.g. that the unsigned short does contain only 0 or 1), then it gets more complicated. The obvious solution would be to use std::transform:

std::vector<TargetType> newV;
newV.reserve( oldV.size() );    //  avoids unnecessary reallocations
std::transform( oldV.begin(), oldV.end(),
                std::back_inserter( newV ),
                TranformationObject() );

, where TranformationObject is a functional object which does the transformation, e.g.:

struct ToBool : public std::unary_function<unsigned short, bool>
{
    bool operator()( unsigned short original ) const
    {
        if ( original != 0 && original != 1 )
            throw Something();
        return original != 0;
    }
};

(Note that I'm just using this transformation function as an example. If the only thing which distinguishes the transformation function from an implicit conversion is the verification, it might be faster to verify all of the values in oldV first, using std::for_each, and then use the two iterator constructor above.)

Depending on the cost of default constructing the target type, it may be faster to create the new vector with the correct size, then overwrite it:

std::vector<TargetType> newV( oldV.size() );
std::transform( oldV.begin(), oldV.end(),
                newV.begin(),
                TranformationObject() );

Finally, another possibility would be to use a boost::transform_iterator. Something like:

std::vector<TargetType> newV(
    boost::make_transform_iterator( oldV.begin(), TranformationObject() ),
    boost::make_transform_iterator( oldV.end(), TranformationObject() ) );

In many ways, this is the solution I prefer; depending on how boost::transform_iterator has been implemented, it could also be the fastest.

Answer 6

The fastest way to do it is to not do it. For example, if you know in advance that your items only need a byte for storage, use a byte-size vector to begin with. You'll find it difficult to find a faster way than that :-)

If that's not possible, then just absorb the cost of the conversion. Even if it's a little slow (and that's by no means certain, see Nicol's excellent answer for details), it's still necessary. If it wasn't, you would just leave it in the larger-type vector.