西瓜书习题3.3

3.3 编程实现对率回归，并给出西瓜数据集 3.0α 上的结果.
 

数据集

密度	甜度	好or 坏
0.676	0.460	1
0.774	0.376	1
0.634	0.264	1
0.608	0.318	1
0.556	0.215	1
0.403	0.237	1
0.481	0.149	1
0.437	0.211	1
0.666	0.091	0
0.243	0.267	0
0.245	0.057	0
0.343	0.099	0
0.639	0.161	0
0.657	0.198	0
0.360	0.370	0
0.593	0.042	0
0.719	0.103	0

import string
import math
def sigmoid(x):
    return 1 / (1 + math.e**(-x))

def classify(x):
    if x <= 0.5:
        return 0
    else:
        return 1

def object_function(w1, w2, x1, x2, b):
    return sigmoid(w1 * x1 + w2 * x2 + b)

file = open("D:\Desktop\watermelon.txt", "r")
training_set = list()
num = 0

#读入数据
for line in file.readlines():
    density, sugar, label = line.split()
    #print(density, sugar, label)
    training_set.append((density, sugar, label))
    #print(training_set)
    num += 1

#开始训练
w1, w2 = 0.0, 0.0
b = 1.0
learning_rate = 0.5
for epoch in range(100000):
    correct = 0.0
    deltaW1, deltaW2, deltaB = 0.0, 0.0, 0.0
    for i in range(num):
        #计算正确率
        x1, x2 = float(training_set[i][0]), float(training_set[i][1])
        y = int(training_set[i][2])
        #print("x1: %f, x2: %f" % (x1, x2))
        predict_y = object_function(w1, w2, x1, x2, b)
        #print("w1 * x1 + w2 * x2 + b = %f" % (w1 * x1 + w2 * x2 + b))
        #print("y = %f, predict_y = %f" % (y, predict_y))
        correct += (classify(predict_y) == y)
        #更新参数
        temp = learning_rate * (predict_y - y)
        deltaW1 += temp * x1
        deltaW2 += temp * x2
        deltaB += temp
    #print("epoch %d has accuracy: %f" % (epoch, float(correct / num)))
    w1 -= deltaW1
    w2 -= deltaW2
    b -= deltaB

print("epoch %d has accuracy: %f%%" % (epoch, float(correct / num) * 100))
print("w1 = %f, w2 = %f, b = %f" % (w1, w2, b))

最终的结果为

epoch 99999 has accuracy: 70.588235%
w1 = 3.156182, w2 = 12.541840, b = -4.431327