I wanted to use boost accumulators to calculate statistics of a variable that is a vector. Is there a simple way to do this. I think it\'s not possible to use the dumbest th
I've looked into your question a bit, and it seems to me that Boost.Accumulators already provides support for std::vector. Here is what I could find in a section of the user's guide :
Another example where the Numeric Operators Sub-Library is useful is when a type does not define the operator overloads required to use it for some statistical calculations. For instance,
std::vector<>does not overload any arithmetic operators, yet it may be useful to usestd::vector<>as a sample or variate type. The Numeric Operators Sub-Library defines the necessary operator overloads in theboost::numeric::operatorsnamespace, which is brought into scope by the Accumulators Framework with a using directive.
Indeed, after verification, the file boost/accumulators/numeric/functional/vector.hpp does contain the necessary operators for the 'naive' solution to work.
I believe you should try :
boost/accumulators/numeric/functional/vector.hpp before any other accumulators headerboost/accumulators/numeric/functional.hpp while defining BOOST_NUMERIC_FUNCTIONAL_STD_VECTOR_SUPPORTusing namespace boost::numeric::operators;.There's only one last detail left : execution will break at runtime because the initial accumulated value is default-constructed, and an assertion will occur when trying to add a vector of size n to an empty vector. For this, it seems you should initialize the accumulator with (where n is the number of elements in your vector) :
accumulator_set, stats > acc(std::vector(n));
I tried the following code, mean gives me a std::vector of size 2 :
int main()
{
accumulator_set, stats > acc(std::vector(2));
const std::vector v1 = boost::assign::list_of(1.)(2.);
const std::vector v2 = boost::assign::list_of(2.)(3.);
const std::vector v3 = boost::assign::list_of(3.)(4.);
acc(v1);
acc(v2);
acc(v3);
const std::vector &meanVector = mean(acc);
}
I believe this is what you wanted ?