CODEWITHSUNDEEP
c++visual-c++vector
Hum
Hum

Reputation: 474

extracting the elements from specified indexes in c++ vector

I have two questions, Any help will be highly appriciated.

I have a matrix A ={ 0 1 0 0 1 1 0 0}. Now I found the locations of 0's indexes and saved in the vector B={0 2 3 6 7}.

  1. How Can I extract the elemnents indexed by vector B in A to a new vector, without damaging original vector A? i.e. I want to get C= {0 0 0 0 0} which is the data from A, indexed by B.

  2. How can I erase the elements in A indexed by B?

I tried something like this for question no. 2,but did not suceed. 

///// erasing the elements in particular locations

sort (B.begin(), B.end());

for(int i=A.size() - 1; i >= 0; i--){

   A.erase(A.begin() + B[i]);
}

Upvotes: 3

Views: 7548

Answers (4)

AMA
AMA

Reputation: 4214

Here are some std-syle generic utilities (standard c++98).

1. Extract elements at indices

/// Extract elements from vector at indices specified by range
template <typename ForwardIt, typename IndicesForwardIt>
inline std::vector<typename std::iterator_traits<ForwardIt>::value_type>
extract_at(
    ForwardIt first,
    IndicesForwardIt indices_first,
    IndicesForwardIt indices_last)
{
    typedef std::vector<typename std::iterator_traits<ForwardIt>::value_type>
        vector_type;
    vector_type extracted;
    extracted.reserve(static_cast<typename vector_type::size_type>(
        std::distance(indices_first, indices_last)));
    for(; indices_first != indices_last; ++indices_first)
        extracted.push_back(*(first + *indices_first));
    return extracted;
}

/// Extract elements from collection specified by collection of indices
template <typename TVector, typename TIndicesVector>
inline TVector extract_at(const TVector& data, const TIndicesVector& indices)
{
    return extract_at(data.begin(), indices.begin(), indices.end());
}

2. Remove elements at indices

//! Remove one element with given index from the range [first; last)
template <typename ForwardIt>
inline ForwardIt remove_at(ForwardIt first, ForwardIt last, const size_t index)
{
    std::advance(first, index);
    for(ForwardIt it = first + 1; it != last; ++it, ++first)
        *first = *it;
    return first;
}

/*!
 * Remove elements in the range [first; last) with indices from the sorted
 * range [indices_first, indices_last)
 */
template <typename ForwardIt, typename SortedIndicesForwardIt>
inline ForwardIt remove_at(
    ForwardIt first,
    ForwardIt last,
    SortedIndicesForwardIt indices_first,
    SortedIndicesForwardIt indices_last)
{
    typedef typename std::vector<bool> flags;
    // flag elements to keep
    flags is_keep(
        static_cast<flags::size_type>(std::distance(first, last)), true);
    for(; indices_first != indices_last; ++indices_first)
        is_keep[static_cast<flags::size_type>(*indices_first)] = false;
    // move kept elements to beginning
    ForwardIt result = first;
    for(flags::const_iterator it = is_keep.begin(); first != last; ++first, ++it)
        if(*it) // keep element
            *result++ = *first; //in c++11 and later use: *result++ = std::move(*first);
    return result;
}

Usage (erase-remove idiom):

std::vector<int> vec{0, 1, 2, 3, 4, 5, 6, 7, 8, 9};
std::vector<int> ii{6, 3, 1};
std::sort(ii.begin(), ii.end());
vec.erase(remove_at(vec.begin(), vec.end(), ii.begin(), ii.end()), vec.end());

Upvotes: 0

Baldrickk
Baldrickk

Reputation: 4409

Q1:
If you are happy making a new vector, PaulMcKenxie's answer is what you want:

std::vector<int> C;
for (size_t i = 0; i < B.size(); ++i )
   C.push_back(A[B[i]]);

Q2:
Otherwise, you need to remove each instance not indexed by B. This is relatively complex, as by removing an entry the way you did, you force realocation that can/(will?) invalidate your iterators / pointers to the data.

Probably the best solution (simple and efficient) to this is to create a temporary vector C above, and then swapping the reduced data in.

void delete_indexes(std::vector<int> &data, std::vector<int> &indexes)
{
    std::vector<int> temp;
    for (size_t i = 0; i < indexes.size(); ++i )
    {
       temp.push_back(data[indexes[i]]);
    }
    data.swap(temp);  //does the work
}

int main()
{
    //do stuff
    delete_indexes(A,B);
}

The swap option is fast (just swaps instead of removing and writing) and the temp vector (with your original data) is disposed of when it goes out of scope.

Edit:
This answer could also be what you are looking for, assuming you have a function for generating each element of B that you can apply (even if it is A[i] == 1 (code edited to suit):

for(auto it = A.begin(); it != A.end();)
{
  if (criteria_for_B(*it))
    it = A.erase(it);
  else
    ++it;
}

Upvotes: 2

hogren
hogren

Reputation: 103

For me, I use the erase function but with a counter to décrement the iterator :

#include <iostream>
#include <vector>
using namespace std;

int main(){
  vector<int> A;
  A.push_back(0);
  A.push_back(1);
  A.push_back(0);
  A.push_back(0);
  A.push_back(1);
  A.push_back(1);
  A.push_back(0);
  A.push_back(0);

  vector<int> B;
  B.push_back(0);
  B.push_back(2);
  B.push_back(3);
  B.push_back(6);
  B.push_back(7);

  for(int i=0; i<A.size(); i++){
    cout << A[i] << "-";
  }
  cout << endl;

  vector<int> C = A;
  int ii=0;
  for(int i=0; i<B.size(); i++){
    C.erase(C.begin() -ii + B[i] );
    ii++;
  }
  for(int i=0; i<C.size(); i++){
    cout << C[i] << "-";
  }

}

You can use a third vector as me or just directly modify A.

I hope that will help you !

Upvotes: 0

PaulMcKenzie
PaulMcKenzie

Reputation: 35440

1.How Can I extarct the elemnents indexed by vector B in A, without damaging original vector A? i.e. I want to get C= {0 0 0 0 0} which is the data from A, indexed by B.

std::vector<int> C;
for (size_t i = 0; i < B.size(); ++i )
   C.push_back(A[B[i]]);

Of course, we're assuming that B does not have entries that will go out-of-bounds of the A vector.

Upvotes: 2

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