
#1




Chapter 1 Problem 1.5
Hello All,
I am simply generating data points in first and third quadrant with below code: import random import numpy as np import matplotlib.pyplot as plt plt.rcParams['axes.unicode_minus'] = False #generate a data set of 100. #for simplicity, 50 in the first quadrant, another 50 in the third quadrant train_X1 = [] train_Y1 = [] train_X2 = [] train_Y2 = [] for i in range(50): train_X1.append(random.uniform(0,1)) train_Y1.append(random.uniform(0,1)) train_X2.append(random.uniform(1,0)) train_Y2.append(random.uniform(1,0)) #label the data train_data1 = [np.array([1,train_X1[i],train_Y1[i],1]) for i in range(50)] train_data2 = [np.array([1,train_X2[i],train_Y2[i],1]) for i in range(50)] train_data = train_data1 + train_data2 I have my perceptron algorithm as below: #Problem 1.5 class Perceptron(object): def __init__(self, data): self.W = np.zeros(len(data[0:3])) self.update = 0 self.learning_rate = 0.01 def predict(self, x): activation = np.dot(self.W.T,x) return np.sign(activation) def fit(self, data): count = 0 X = np.array(data)[:,0:3] d = np.array(data)[:, 3:4] while self.update < 1000: #self.update = 0 for i in range(len(data)): predicted_value_y = self.predict(X[i]) expected_value = d[i] if expected_value * predicted_value_y <=1: self.W = self.W + self.learning_rate*(expected_value  predicted_value_y) * X[i] #print(self.W) self.update += 1 print("Number of iterations for converging:",count) For me it seems correct according to mention of update rule in problem 1.5 in book. But some how even if my learning rate changes my target function 'g' does not change and classifies some data points incorrectly. Is this expected? 
#2




Re: Chapter 1 Problem 1.5
You should perhaps considering picking random points within each iteration instead of following the original (data) order of points. Hope this helps.
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#3




Re: Chapter 1 Problem 1.5
Thanks I got it, Thank you so much! It's great to have you guys around. There will be a lot of questions from me on this book as I am going through it and have no background.

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problem 1.5, variation of adaline 
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