How to derive equation from Numpy's polyfit?

Question

Given an array of x and y values, the following code will calculate a regression curve for these data points.

# calculate polynomial
z = np.polyfit(x, y, 5)
         


        

Answer 1

If you want to show the equation, you can use sympy to output latex:

from sympy import S, symbols, printing
from matplotlib import pyplot as plt
import numpy as np

x=np.linspace(0,1,100)
y=np.sin(2 * np.pi * x)

p = np.polyfit(x, y, 5)
f = np.poly1d(p)

# calculate new x's and y's
x_new = np.linspace(x[0], x[-1], 50)
y_new = f(x_new)

x = symbols("x")
poly = sum(S("{:6.2f}".format(v))*x**i for i, v in enumerate(p[::-1]))
eq_latex = printing.latex(poly)

plt.plot(x_new, y_new, label="${}$".format(eq_latex))
plt.legend(fontsize="small")
plt.show()

the result:

Answer 2

Construct a simple example:

In [94]: x=np.linspace(0,1,100)
In [95]: y=2*x**3-3*x**2+x-1

In [96]: z=np.polyfit(x,y,3)
In [97]: z
Out[97]: array([ 2., -3.,  1., -1.])

The z coefficients correspond to the [2,-3,1,-1] I used to construct y.

In [98]: f=np.poly1d(z)
In [99]: f
Out[99]: poly1d([ 2., -3.,  1., -1.])

The str, or print, string for f is a representation of this polynomial equation. But it's the z coeff that defines the equation.

In [100]: print(f)
   3     2
2 x - 3 x + 1 x - 1
In [101]: str(f)
Out[101]: '   3     2\n2 x - 3 x + 1 x - 1'

What else do you mean by 'actual equation'?

polyval will evaluate f at a specific set of x. Thus to recreate y, use polyval(f,x):

In [107]: np.allclose(np.polyval(f,x),y)
Out[107]: True