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python linspace in c++

Tags:

c++

python

numpy

im trying to write a similar version of python's numpy.linspace function.

double linspace(int a, int b, int c){
    double line[c];
    double delta =b-a/(c-1);
    for (int i=0; i<c; ++i){
            line[i]=0 + (i*delta);
    }
    return line;

with a and b being the first and the last component in the array, and c specifying the number of elements in the array. But when i compile this script it returns:

linspace.cpp: In function ‘double linspace(int, int, int)’:
linspace.cpp:11:9: error: cannot convert ‘double*’ to ‘double’ in return
  return line;
         ^

would anyone happen to knw how to solve this problem?

like image 749
user2804865 Avatar asked Dec 30 '25 03:12

user2804865


1 Answers

How about something like this:

#include <iostream>
#include <vector>

template<typename T>
std::vector<double> linspace(T start_in, T end_in, int num_in)
{

  std::vector<double> linspaced;

  double start = static_cast<double>(start_in);
  double end = static_cast<double>(end_in);
  double num = static_cast<double>(num_in);

  if (num == 0) { return linspaced; }
  if (num == 1) 
    {
      linspaced.push_back(start);
      return linspaced;
    }

  double delta = (end - start) / (num - 1);

  for(int i=0; i < num-1; ++i)
    {
      linspaced.push_back(start + delta * i);
    }
  linspaced.push_back(end); // I want to ensure that start and end
                            // are exactly the same as the input
  return linspaced;
}

void print_vector(std::vector<double> vec)
{
  std::cout << "size: " << vec.size() << std::endl;
  for (double d : vec)
    std::cout << d << " ";
  std::cout << std::endl;
}

int main()
{
  std::vector<double> vec_1 = linspace(1, 10, 3);
  print_vector(vec_1);

  std::vector<double> vec_2 = linspace(6.0, 23.4, 5);
  print_vector(vec_2);

  std::vector<double> vec_3 = linspace(0.0, 2.0, 1);
  print_vector(vec_3);

  std::vector<double> vec_4 = linspace(0.0, 2.0, 0);
  print_vector(vec_4);


  return 0;
}

C++ result:

size: 3
1 5.5 10 
size: 5
6 10.35 14.7 19.05 23.4 
size: 1
0 
size: 0

Numpy result:

In [14]: np.linspace(1, 10, 3)
Out[14]: array([  1. ,   5.5,  10. ])

In [15]: np.linspace(6, 23.4, 5)
Out[15]: array([  6.  ,  10.35,  14.7 ,  19.05,  23.4 ])

In [16]: np.linspace(0.0, 2.0, 1)
Out[16]: array([ 0.])

In [17]: np.linspace(0.0, 2.0, 0)
Out[17]: array([], dtype=float64)
like image 150
Akavall Avatar answered Dec 31 '25 17:12

Akavall