CODEWITHSUNDEEP
c++vector
Wawel100
Wawel100

Reputation: 1281

Adding vectors of doubles of differing sizes in C++

I have a number of vector containers of varying size each containing doubles. I would like to add the elements of each vector to create a single vector of doubles. This simple example will example what I'm talking about:

Consider two vectors A with three elements 3.0 2.0 1.0 and B with two elements 2.0 1.0. I would like to add both vectors starting at the last element and working backwards. This would give an array C with entries 3.0 4.0 2.0.

What would be the most elegant/efficent way of doing this?

Thanks!

Upvotes: 3

Views: 1298

Answers (3)

andand
andand

Reputation: 17497

Copy the larger vector into C and then add (+=) the elements of the smaller against the associated elements of C.

Somthing like:

std::vector<double> add(const std::vector<double>& a,
                        const std::vector<double>& b)
{
    std::vector<double> c( (a.size() > b.size()) ? a : b );
    const std::vector<double>& aux = (a.size() > b.size() ? b : a);
    size_t diff = c.size() - aux.size();

    for (size_t i = diff; i < c.size(); ++i)
        c[i] += aux[i-diff];

    return c;
}

Edit Based on comment below objecting to use of [] vs iterators.

Personally I find iterators to be excessively verbose for something like this, but if you prefer them, then you could try somehting like the following:

std::vector<double> add(const std::vector<double>& a, 
                        const std::vector<double>& b)
{
    std::vector<double> c( (a.size() > b.size()) ? a : b);
    std::vector<double>::reverse_iterator c_i;

    const std::vector<double>& aux = (a.size() > b.size()) ? b : a;
    std::vector<double>::const_reverse_iterator aux_i;

    for (c_i=c.rbegin(), aux_i=aux.rbegin(); aux_i!=aux.rend(); ++c_i, ++aux_i)
        *c_i += *aux_i;

    return c;
}

I didn't test or compile either of these, but I think you get the idea.

Upvotes: 3

no one important
no one important

Reputation: 94

Once you know you have one vector that is larger than the other

std::vector<double> new_vector = bigger_vector; // Copy the largest
std::transform(smaller_vector.rbegin(), smaller_vector.rend(), // iterate over the complete smaller vector 
    bigger_vector.rbegin(), // 2nd input is the corresponding entries of the larger vector  
    new_vector.rbegin(),    // Output is the new vector 
    std::plus<double>());   // Add em

This is nice because you don't have to do any loop indentation, and works on any sequence container that supports reverse iterators.

Upvotes: 4

Seth Johnson
Seth Johnson

Reputation: 15192

Try this with iterators:

#include <vector>

void add(
        std::vector<double>& result,
        const std::vector<double>& a,
        const std::vector<double>& b)
{
    std::vector<double>::const_reverse_iterator sit;
    std::vector<double>::const_reverse_iterator send;

    // copy the larger vector
    if (a.size() > b.size() ) {
        result = a;
        sit  = b.rbegin();
        send = b.rend();
    }
    else {
        result = b;
        sit  = a.rbegin();
        send = a.rend();
    }

    // add the smaller one, starting from the back
    for (std::vector<double>::reverse_iterator it = result.rbegin();
            sit != send;
            ++it, ++sit)
    {
        *it += *sit;
    }
}

Upvotes: 4

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