Re: index of std::vector with condition

From:

Stuart Golodetz <sgolodetz@NdOiSaPlA.pMiPpLeExA.ScEom>

Newsgroups:

comp.lang.c++

Date:

Tue, 13 Jul 2010 00:53:04 +0100

Message-ID:

<laGdna_eDoR9N6bRnZ2dnUVZ8rydnZ2d@pipex.net>

Philipp Kraus wrote:

Hello,

is there a fast way to create a index with condition of a std::vector?

For illustrating:

I have a vector mydata with some data and I would like to create a
vector with unique elements

std::vector<T> l_unique( mydata );

std::sort( l_unique.begin(), l_unique.end() );

l_unique.erase( std::unique( l_unique.begin(), l_unique.end()),
l_unique.end() );

in the next step I will do this

std::vector< std::vector<std::size:t> > index;

for(std::size_t i=0; i < l_unique.size(); ++i) {

      std::vector<std::size_t> y;



      for(std::size_t n=0; n < mydata.size(); ++n)

if (mydata[n] == l_unique[i])

y.push_back(n);

index.push_back(y);

}

Can I do this two for loops more efficient? I would like to get the
index position in my data vector from a unique set of the elements
within the vector.

Thanks for help

Phil

Anything wrong with this sort of approach?

int arr[] = {3,2,2,3,3,1,1,5,6};
std::vector<int> v(&arr[0], &arr[sizeof(arr)/sizeof(int)]);

std::map<int,std::vector<size_t> > index;
for(size_t i=0, size=v.size(); i<size; ++i)
{
index[v[i]].push_back(i);
}

Regards,
Stu