Matplotlib.artist.Artist.get_url() in Python

# Implementation of matplotlib function
from matplotlib.artist import Artist 
import matplotlib.pyplot as plt 
   

f, ax = plt.subplots() 
ax.scatter([1, 2, 3], [4, 5, 6]) 
ax.set_url('http://www.google.com') 
f.canvas.print_figure('scatter.svg') 
  
ax.text(1.5, 5.5, "URL : "
        + str( Artist.get_url(ax)),  
        fontweight ="bold") 

f.suptitle("""matplotlib.artist.Artist.get_url()
function Example""", fontweight ="bold")

plt.show()

# Implementation of matplotlib function
from matplotlib.artist import Artist 
import matplotlib.pyplot as plt
import numpy as np 

   
f, ax = plt.subplots() 
delta = 0.025
  
x = y = np.arange(-3.0, 3.0, delta) 
X, Y = np.meshgrid(x, y) 
  
Z1 = np.exp(-X**2 - Y**2) 
Z2 = np.exp(-(X - 1)**2 - (Y - 1)**2) 
Z = (Z1 - Z2) * 2
   
ax.imshow(Z, interpolation ='bilinear',  
          cmap = "PuBuGn_r", 
          origin ='lower', 
          extent =[-3, 3, -3, 3]) 
   
ax.set_url('https://www.geeksforgeeks.org/') 
f.savefig('image.svg') 
  
ax.set_title("URL : "+str(Artist.get_url(ax)), fontweight ="bold")

f.suptitle("""matplotlib.artist.Artist.get_url()
function Example""", fontweight ="bold")


plt.show()