吳恩達 machine learning 編程作業 python實現 ex1

# -*- coding: utf-8 -*-
"""
Created on Mon Jun 29 16:01:13 2020

@author: cheetah023
"""
import numpy as np
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D

#函數定義
#輸出對角矩陣
def warmUpExercise(n):
    A = np.eye(n)
    return A
#損失函數
def computeCost(X, y, theta):
    J = np.sum((X.dot(theta) - y) ** 2) / (2 * len(y))
    return J
#梯度下降函數
def gradientDescent(X, y, theta, alpha, num_iters):
    m = X.shape[0]
    J_history = np.zeros((num_iters,1))
    for i in range(1,num_iters):
        h = np.dot(X,theta)
        theta = theta - (alpha / m) * np.dot(X.T,h-y)
        J_history[i] = computeCost(X, y, theta) 
    return theta,J_history

#Part 1: Basic Function
print('Part1.\n')
A = warmUpExercise(5)
print(A)

#Part 2: Plotting
print('Part2.\n')

data = np.loadtxt('ex1data1.txt',delimiter=',')
X = data[:,0]
y = data[:,1:2]#使用data[:,1]的話，y.shape打印出來是(97,)
print('y:')
print(y.shape)
plt.scatter(X, y,marker='x',c='r',)
plt.ylabel('Profit in $10,000s')
plt.xlabel('Population of City in 10,000s')

#Part 3: Cost and Gradient descent
print('Part3.\n')
m = len(y)
ones = np.ones(m)
theta = np.zeros((2,1))
X = np.column_stack((ones,X))#在X的首列插入全是1的列
print('X:')
print(X.shape)
iterations = 1500
alpha = 0.01
J = computeCost(X, y, theta)
print('With theta = [0 ; 0],Cost computed = %f' %(J))
print('Expected cost value (approx) 32.07\n')

J = computeCost(X, y, np.array([[-1],[2]]))
print('With theta = [-1 ; 2],Cost computed = %f' %(J))
print('Expected cost value (approx) 54.24\n')

(theta,J_history) = gradientDescent(X, y, theta, alpha, iterations)

print('Theta found by gradient descent:')
print(theta)
print('Expected theta values (approx):\n-3.6303\n  1.1664')

#plot data
plt.plot(X[:,1],np.dot(X,theta),c='b')

#predict
predict1 = np.dot([1,3.5],theta)
print('For population = 35,000, we predict a profit of %f' %(predict1 * 10000))
predict2 = np.dot([1,7],theta)
print('For population = 70,000, we predict a profit of %f' %(predict2 * 10000))

# Part 4: Visualizing J(theta_0, theta_1)
theta0_vals = np.linspace(-10,10,100)
theta1_vals = np.linspace(-1,4,100)
J_val = np.zeros((len(theta0_vals),len(theta1_vals)))

for i in range(1,len(theta0_vals)):
    for j in range(1,len(theta1_vals)):
        t = np.row_stack((theta0_vals[i],theta1_vals[j]))
        J_val[i, j] = computeCost(X, y, t)

J_val = J_val.T

fig = plt.figure()
ax = Axes3D(fig)
ax.plot_surface(theta1_vals, theta0_vals, J_val,
                rstride=1, cstride=1, cmap='rainbow')
plt.ylabel('theta_0')
plt.xlabel('theta_1')

plt.figure()

plt.contourf(theta0_vals, theta1_vals, J_val,
             20, alpha=0.6, cmap=plt.cm.hot)
plt.plot(theta[0], theta[1], c='r', marker='x', markerSize=10, LineWidth=2)

運行結果：

Part1.

[[1. 0. 0. 0. 0.]
 [0. 1. 0. 0. 0.]
 [0. 0. 1. 0. 0.]
 [0. 0. 0. 1. 0.]
 [0. 0. 0. 0. 1.]]
Part2.

y:
(97, 1)
Part3.

X:
(97, 2)
With theta = [0 ; 0],Cost computed = 32.072734
Expected cost value (approx) 32.07

With theta = [-1 ; 2],Cost computed = 54.242455
Expected cost value (approx) 54.24

Theta found by gradient descent:
[[-3.62981201]
 [ 1.16631419]]
Expected theta values (approx):
-3.6303
  1.1664
For population = 35,000, we predict a profit of 4522.876458
For population = 70,000, we predict a profit of 45343.872966

參考資料：

https://blog.csdn.net/lccflccf/category_8379707.html

https://blog.csdn.net/Cowry5/article/details/83302646

https://blog.csdn.net/weixin_44027820/category_9754493.html

總結：遇到bug或是計算結果相差太大，注意檢查矩陣的維度，多打印出來看看