Perceptron Algorithm plotting with matplotlib

Question

In an ML course, I m taking, I have 100 entries of data, and I'm using it in a Perceptron Algorithm. What I want is to show a plot like this one.

As you can see above we have the data represented by point in red and blue and the different calculated lines that minimize the error. This is the output that I want.. Here is my Data and my code.

data.csv

0.78051,-0.063669,1
0.28774,0.29139,1
0.40714,0.17878,1
0.2923,0.4217,1
0.50922,0.35256,1
0.27785,0.10802,1
0.27527,0.33223,1
0.43999,0.31245,1
0.33557,0.42984,1
0.23448,0.24986,1
0.0084492,0.13658,1
0.12419,0.33595,1
0.25644,0.42624,1
0.4591,0.40426,1
0.44547,0.45117,1
0.42218,0.20118,1
0.49563,0.21445,1
0.30848,0.24306,1
0.39707,0.44438,1
0.32945,0.39217,1
0.40739,0.40271,1
0.3106,0.50702,1
0.49638,0.45384,1
0.10073,0.32053,1
0.69907,0.37307,1
0.29767,0.69648,1
0.15099,0.57341,1
0.16427,0.27759,1
0.33259,0.055964,1
0.53741,0.28637,1
0.19503,0.36879,1
0.40278,0.035148,1
0.21296,0.55169,1
0.48447,0.56991,1
0.25476,0.34596,1
0.21726,0.28641,1
0.67078,0.46538,1
0.3815,0.4622,1
0.53838,0.32774,1
0.4849,0.26071,1
0.37095,0.38809,1
0.54527,0.63911,1
0.32149,0.12007,1
0.42216,0.61666,1
0.10194,0.060408,1
0.15254,0.2168,1
0.45558,0.43769,1
0.28488,0.52142,1
0.27633,0.21264,1
0.39748,0.31902,1
0.5533,1,0
0.44274,0.59205,0
0.85176,0.6612,0
0.60436,0.86605,0
0.68243,0.48301,0
1,0.76815,0
0.72989,0.8107,0
0.67377,0.77975,0
0.78761,0.58177,0
0.71442,0.7668,0
0.49379,0.54226,0
0.78974,0.74233,0
0.67905,0.60921,0
0.6642,0.72519,0
0.79396,0.56789,0
0.70758,0.76022,0
0.59421,0.61857,0
0.49364,0.56224,0
0.77707,0.35025,0
0.79785,0.76921,0
0.70876,0.96764,0
0.69176,0.60865,0
0.66408,0.92075,0
0.65973,0.66666,0
0.64574,0.56845,0
0.89639,0.7085,0
0.85476,0.63167,0
0.62091,0.80424,0
0.79057,0.56108,0
0.58935,0.71582,0
0.56846,0.7406,0
0.65912,0.71548,0
0.70938,0.74041,0
0.59154,0.62927,0
0.45829,0.4641,0
0.79982,0.74847,0
0.60974,0.54757,0
0.68127,0.86985,0
0.76694,0.64736,0
0.69048,0.83058,0
0.68122,0.96541,0
0.73229,0.64245,0
0.76145,0.60138,0
0.58985,0.86955,0
0.73145,0.74516,0
0.77029,0.7014,0
0.73156,0.71782,0
0.44556,0.57991,0
0.85275,0.85987,0
0.51912,0.62359,0

And now this is my code. The first part

import numpy as np
import pandas as pd
# Setting the random seed, feel free to change it and see different solutions.
np.random.seed(42)
import matplotlib.pyplot as plt


def stepFunction(t):
    return 1 if t >= 0 else 0


def prediction(X, W, b):
    return stepFunction((np.matmul(X, W) + b)[0])

# TODO: Fill in the code below to implement the perceptron trick.
# INPUTS
# data X, the labels y,
# the weights W (as an array), and the bias b,
# The function  weights and bias W, b, according to the perceptron algorithm,
# and return W and b.

def perceptronStep(X, y, W, b, learn_rate=0.01):
    for i in range(len(X)):
        y_hat = prediction(X[i], W, b)
        if y[i] - y_hat == 1:
            W[0] += X[i][0] * learn_rate
            W[1] += X[i][1] * learn_rate
            b += learn_rate
        elif y[i] - y_hat == -1:
            W[0] -= X[i][0] * learn_rate
            W[1] -= X[i][1] * learn_rate
            b -= learn_rate
    return W, b

# This function runs the perceptron algorithm repeatedly on the dataset,
# and returns a few of the boundary lines obtained in the iterations,
# for plotting purposes.
# Feel free to play with the learning rate and the num_epochs,
# and see your results plotted below.
def trainPerceptronAlgorithm(X, y, learn_rate=0.01, num_epochs=25):
    x_min, x_max = min(X.T[0]), max(X.T[0])
    y_min, y_max = min(X.T[1]), max(X.T[1])
    W = np.array(np.random.rand(2, 1))
    b = np.random.rand(1)[0] + x_max
    # These are the solution lines that get plotted below.
    boundary_lines = []
    for i in range(num_epochs):
        # In each epoch, we apply the perceptron step.
        W, b = perceptronStep(X, y, W, b, learn_rate)
        # Here I have a doubt . Why if y = W0*x1 + W1*x2 + b
        # So we can get  x2 =y/W1 -(W0*x1)/W1 -b/W1 + y/W1)
        # If we remove y/W1 we just get intercept and slope
        # But why we are not using the last term y/W1
        boundary_lines.append((-W[0] / W[1], -b / W[1]))
    return boundary_lines

# Get data and plot the points
data = pd.read_csv('data.csv', header = None)
X = data.iloc[:, :2].values
y = data.iloc[:, -1].values

x1 = X[:, 0]
x2 = X[:, 1]
color = ['red' if value == 1 else 'blue' for value in y]
plt.scatter(x1, x2, marker='o', color=color)
plt.xlabel('X1 input feature')
plt.ylabel('X2 input feature')
plt.title('Perceptron regression for X1, X2')
plt.show()

When you run this code you correctly get

So now I want to plot the line in the same plot the lines that represent the best function for each iteration.For that, I commented the last line above plt.show() and did

# So now lets plot the lines that represent the best function for each iteration
boundary_lines = trainPerceptronAlgorithm(X, y)
x_lin = np.linspace(0, 1, 100)
for line in boundary_lines:
    Θo, Θ1  = line
    Θ1 = Θ1[0]
    Θo = Θo[0]
    # TODO: The equation of the error function is
    # y = W0*x1 + W1*x2 + b
    # So we can get  x2 =y/W1 -(W0*x1)/W1 -b/W1 + y/W1)
    # If we remove y/W1 we just get intercept and slope
    # boundary_lines.append((-W[0] / W[1], -b / W[1])
    # plt.axes([-0.5, -0.5, 1.5, 1.5])
    plt.plot(x_lin, (Θ1 * x_lin / Θo))
    plt.draw()
    plt.pause(5)
    input("Press enter to continue")
    plt.close()

But that does not get me the expected result. Why doesn't this get the expected result?

Answer 1

The mistake is in plt.plot(x_lin, (Θ1 * x_lin / Θo)) where instead of Θ1 * x_lin / Θo you should have Θo * x_lin + Θ1.

fig, ax = plt.subplots(1, 1, figsize=(8,5))
ax.set_xlim(0, 1)
ax.set_ylim(0, 1)
ax.scatter(x1, x2, marker='o', color=color)
for i, line in enumerate(boundary_lines):
    Θo, Θ1  = line
    if i == len(boundary_lines) - 1:
        c, ls, lw = 'k', '-', 2
    else:
        c, ls, lw = 'g', '--', 1.5
    ax.plot(x_lin, Θo * x_lin + Θ1, c=c, ls=ls, lw=lw)
plt.show()

Result: