Normal(Gaussian) Distribution Function in C++

问题

I need to know a way to have Gaussian Distribution of 50 numbers. I know of the Boost library which generates random numbers. In my case i don't need random, i need the normal distribution of 50 numbers. Any way to have that? thanks.

回答1:

I think the OP was asking for a random number generator, in which the random numbers are not uniformly distributed (as is typical e.g. rand() in C) but are Gaussian distributed.

This short routine adapted from "Numerical Recipes in C" (Press et al, 1992) may be of use:

double grand() {

  double r,v1,v2,fac;

  r=2;
  while (r>=1) {
    v1=(2*((double)rand()/(double)RAND_MAX)-1);
    v2=(2*((double)rand()/(double)RAND_MAX)-1);
    r=v1*v1+v2*v2;
  }
  fac=sqrt(-2*log(r)/r);

  return(v2*fac);

}

...ensure the relevant #includes are present for the math functions and rand, and that srand(time(NULL)) or similar has been called to appropriately seed the C rand() RNG.

回答2:

As of C++11 there is a normal (gaussian) distribution available in the standard library:

http://www.cplusplus.com/reference/random/normal_distribution/

The mean value and standard deviation are passed as arguments when creating it. The link above provides a good example:

// normal_distribution
#include <iostream>
#include <random>

int main()
{
  const int nrolls=10000;  // number of experiments
  const int nstars=100;    // maximum number of stars to distribute

  std::default_random_engine generator;
  std::normal_distribution<double> distribution(5.0,2.0);

  int p[10]={};

  for (int i=0; i<nrolls; ++i) {
    double number = distribution(generator);
    if ((number>=0.0)&&(number<10.0)) ++p[int(number)];
  }

  std::cout << "normal_distribution (5.0,2.0):" << std::endl;

  for (int i=0; i<10; ++i) {
    std::cout << i << "-" << (i+1) << ": ";
    std::cout << std::string(p[i]*nstars/nrolls,'*') << std::endl;
  }

  return 0;
}