[FIXED] Slider with for loop in Matplotlib

Issue

I want to find the two roots of a quadratic equation ax^2 + bx + c =0 and plot them against coefficient c while keeping a as a changeable parameter. To change a and see what happens to the plots with varying the parameter, I would like to create a Silder for a from Python's Matplotlib module.

I have the following, however, it doesn't seem to be working.

# Solve the quadratic equation ax**2 + bx + c = 0

from matplotlib.widgets import Slider  # import the Slider widget
import numpy as np
import matplotlib.pyplot as plt
import cmath


a_min = -2   
a_max = 2 
a_init = -2  


fig = plt.figure(figsize=(8,3))

# Slider layout
slider_ax = plt.axes([0.1, 0.05, 0.8, 0.05])

b = 10

def sol1(a,b,c):
      d = (b**2) - (4*a*c) # Discriminant
      return (-b-np.sqrt(d))/(2*a)
def sol2(a,b,c): 
      d = (b**2) - (4*a*c) # Discriminant
      return (-b+np.sqrt(d))/(2*a)

for c in np.linspace(-2, 2, 11):

  print('c=',c,' sol1=',sol1(a_init,b,c),' sol2=',sol2(a_init,b,c))


  # Plot with initial parameter value 
  #plt.axes(func_ax) 
  plt.xlabel('$c$') 
  plt.title('Roots of  $ax^2 + bx + c = 0$')
  plot1, = plt.plot(c, sol1(a_init,b,c), 'r')
  plot2, = plt.plot(c, sol2(a_init,b,c), 'b')

# Create a slider 
a_slider = Slider(slider_ax,      # the axes object containing the slider
                  '$a$',          # the name of the slider parameter
                  a_min,          # minimal value of the parameter
                  a_max,          # maximal value of the parameter
                  valinit=a_init  # initial value of the parameter
                 )

# Update function
def update(a):
    plot1.set_ydata(sol1(a,b,c)) 
    plot2.set_ydata(sol2(a,b,c)) 
    fig.canvas.draw_idle() # redraw the plot

# Execute when parameter gets updated 
a_slider.on_changed(update)

plt.show()

Any help?

Solution

I made a couple of modifications to your code to make the slider work:

• First, plot1 and plot2 are defined in a loop in your code. This means that they are deleted and created again at each iteration. Instead, it makes more sense to compute all the variables you want to plot and then plot them outside the loop. This is performed in the update_sols function.
• Second, I changed the update function of your slider. The variable updated by the slider should be accessed by using a_slider.val (see an example here).
• Finally, to make sure that you can see everything that you want to plot, you need to update the y limits of your plot each time you move your slider.

Overall, the updated code looks like that:

from matplotlib.widgets import Slider  
import numpy as np
import matplotlib.pyplot as plt
import cmath

#Initial values
a_min = -2   
a_max = 2 
a_init = -2  
b = 10
c_list=np.linspace(-2, 2, 11)

fig, ax = plt.subplots()
plt.subplots_adjust(left=0.25, bottom=0.25)

# Slider layout
slider_ax = plt.axes([0.25, 0.1, 0.65, 0.03])

def sol1(a,b,c):
      d = (b**2) - (4*a*c) # Discriminant
      return (-b-np.sqrt(d))/(2*a)

def sol2(a,b,c): 
      d = (b**2) - (4*a*c) # Discriminant
      return (-b+np.sqrt(d))/(2*a)

#Function to update solutions after modifying c
def update_sols(a):
  res1=[]
  res2=[]
  for c in c_list:
    res1.append(sol1(a,b,c))
    res2.append(sol2(a,b,c))
  return res1,res2

#Initialising plot with solutions for a_init
sols1,sols2=update_sols(a_init)
plot1,=ax.plot(c_list, sols1, 'r')
plot2,=ax.plot(c_list, sols2, 'b')
ax.set_xlabel('$c$') 
ax.set_title('Roots of  $ax^2 + bx + c = 0$')
 
# Create a slider 
a_slider = Slider(ax=slider_ax,label='$a$',valmin=a_min,valmax=a_max,valinit=a_init)

# Update function
def update(val):
    #updating y data
    sols1,sols2=update_sols(a_slider.val)
    plot1.set_ydata(sols1) 
    plot2.set_ydata(sols2) 
   
    #updating y limits
    sols1_abs=np.abs(sols1)
    sols2_abs=np.abs(sols2)
    max_ylim=np.amax(np.concatenate((sols1_abs,sols2_abs)))
    ax.set_ylim([-1.1*max_ylim,1.1*max_ylim])
    fig.canvas.draw_idle() # redraw the plot

# Execute when parameter gets updated 
a_slider.on_changed(update)

plt.show()

And the output looks like that:

Answered By - jylls