问scipy.statsβ函数的替代逆计算

EN

#include <boost/python.hpp>
#include <boost/math/distributions/beta.hpp>

using namespace boost::python;

class betainvClass {
    public: double betainv(double p, double a, double b);
};

double betainvClass::betainv(double p, double a, double b) { 
    return boost::math::ibeta_inv(a, b, p);
}

// Expose classes and methods to Python
BOOST_PYTHON_MODULE(betainv) {
    class_<betainvClass> ("create_betainv_instance")
        .def("betainv", &betainvClass::betainv)
    ;
}

TARGET = betainv
PYTHON_INC = /usr/local/Cellar/python/2.7.5/Frameworks/Python.framework/Versions/2.7/include/python2.7
PYTHON_LIB = /usr/local/Cellar/python/2.7.5/Frameworks/Python.framework/Versions/2.7/lib/python2.7/config
BOOST_INC = /usr/local/include
BOOST_LIB = /usr/local/lib

$(TARGET).so: $(TARGET).o
    g++ -shared -Wl \
    $(TARGET).o -L$(BOOST_LIB) -lboost_python \
    -L$(PYTHON_LIB) -lpython2.7 \
    -o $(TARGET).so

$(TARGET).o: $(TARGET).cpp
    g++ -I$(PYTHON_INC) -I$(BOOST_INC) -c $(TARGET).cpp

clean:
    rm -f *.o *.a *.so *~ core

import betainv

beta = betainv.create_betainv_instance()
print "0.25, 0.0342, 170 -> " + str(beta.betainv(0.25, 0.0342, 170))
print "0.25, 0.0342, 171 -> " + str(beta.betainv(0.25, 0.0342, 171))
print "0.25, 0.0342, 172 -> " + str(beta.betainv(0.25, 0.0342, 172))

required time scipy = 295.145019531
required time c++   = 7.68383789062