filling a boost vector or matrix

Question

Is there a single-expression way to assign a scalar to all elements of a boost matrix or vector? I'm trying to find a more compact way of representing:

boost::numeric::ublas::c_vector<float, N> v;
for (size_t i=0; i<N; i++) {
    v[i] = myScalar;
 }

The following do not work:

boost::numeric::ublas::c_vector<float, N> 
   v(myScalar, myScalar, ...and so on..., myScalar);

boost::numeric::ublas::c_vector<float, N> v;
v = myScalar;

Thuythuya

You should also tag this "C++". – Greenling

Answer 1

Because the vector models a standard random access container you should be able to use the standard STL algorithms. Something like:

c_vector<float,N> vec;
std::fill_n(vec.begin(),N,0.0f);

or

std::fill(vec.begin(),vec.end(),0.0f);

It probably also is compatible with Boost.Assign but you'd have to check.

Holiday

The STL algorithms do seem to work. Thanks. Boost.Assign doesn't seem to work for me, but I think it's because I'm using a c_vector (const-sized vector) instead of a vector (dynamically sized vector), so push_back doesn't work. – Thuythuya

This doesn't answer how you would fill a boost::numeric::ublas::matrix – Ballesteros

Answer 2

I have started using boost::assign for cases that I want to statically assign specific values (examples lifted from link above).

#include <boost/assign/std/vector.hpp>
using namespace boost::assign; // bring 'operator+()' into scope

{
  vector<int> values;
  values += 1,2,3,4,5,6,7,8,9;
}

You can also use boost::assign for maps.

#include <boost/assign/list_inserter.hpp>
#include <string>
using boost::assign;

std::map<std::string, int> months;
insert( months )
        ( "january",   31 )( "february", 28 )
        ( "march",     31 )( "april",    30 )
        ( "may",       31 )( "june",     30 )
        ( "july",      31 )( "august",   31 )
        ( "september", 30 )( "october",  31 )
        ( "november",  30 )( "december", 31 );

You can allow do direct assignment with list_of() and map_list_of()

#include <boost/assign/list_of.hpp> // for 'list_of()'
#include <list>
#include <stack>
#include <string>
#include <map>
using namespace std;
using namespace boost::assign; // bring 'list_of()' into scope

{
    const list<int> primes = list_of(2)(3)(5)(7)(11);
    const stack<string> names = list_of( "Mr. Foo" )( "Mr. Bar")
                                       ( "Mrs. FooBar" ).to_adapter();

    map<int,int> next = map_list_of(1,2)(2,3)(3,4)(4,5)(5,6);

    // or we can use 'list_of()' by specifying what type
    // the list consists of
    next = list_of< pair<int,int> >(6,7)(7,8)(8,9);

}

There are also functions for repeat(), repeat_fun(), and range() which allows you to add repeating values or ranges of values.

Ludicrous

The first example seems to be for std::vector (for which it works), not boost::numeric::ublas::vector (for which it does not work). – Urethra

Answer 3

The recommended way looks like this:

boost::numeric::ublas::c_vector<float, N> v;
v = boost::numeric::ublas::zero_vector<float>(N);
v = boost::numeric::ublas::scalar_vector<float>(N, value);

The same is for matrix types:

boost::numeric::ublas::matrix<float> m(4,4);
m = boost::numeric::ublas::identity_matrix<float>(4,4);
m = boost::numeric::ublas::scalar_matrix<float>(4,4);
m = boost::numeric::ublas::zero_matrix<float>(4,4);

Beniamino

How do I get out the value type if I don't know it? – Badajoz

Answer 4

Have you tried this?

ublas::c_vector v = ublas::scalar_vector(N, myScalar);

Norman

Answer 5

Been a while since I used C++. Does the following work?

for (size_t i = 0; i < N; v[i++] = myScalar) ;

Tabatha

That'll work, though it is a full statement as opposed to an expression. – Thuythuya

True, but it is more compact way which is what you wanted to find based on the post. – Tabatha

-1 sorry. You must use "v[i++]" -- "v[++i]" will skip initialisation of v[0] and overwrite the memory past the end of the vector. – Gunmaker

Ack. Sorry! I did acknowledge that I had to use the correct unary operator but for some reason I kept thinking ++i is the one that increments i after evaluating. Mostly since everyone prefer i++ and the "i += 1" behaviour seems more logical. Fixed now in any case.. And I guess I should thank you as well, I like 600 rep more than 601! – Tabatha