人工智能—Python实现线性回归

1、概述

（1）人工智能学习          

（2）机器学习 

（3）有监督学习 

（4）线性回归 

2、线性回归 

（1）实现步骤

• 根据随机初始化的 w x b 和 y 来计算 loss
• 根据当前的 w x b 和 y 的值来计算梯度
• 更新梯度，循环将新的 w′ 和 b′ 复赋给 w 和 b ，最终得到一个最优的 w′ 和 b′ 作为方程最终的

（2）数学表达式        

3、代码实现（python）

（1）机器学习库（sklearn.linear_model）

代码：

from sklearn import linear_model
from sklearn.linear_model import LinearRegression
import matplotlib.pyplot as plt#用于作图
from pylab import *
mpl.rcParams['font.sans-serif'] = ['SimHei']
mpl.编程客栈rcParams['axes.unicode_minus'] = False
import numpy as np#用于创建向量


reg=linear_model.LinearRegression(fit_intercept=True,normalize=False)
x=[[32.50235],[53.4268],[61.53036],[47.47564],[59.81321],[55.14219],[52.14219],[39.29957],
[48.10504],[52.55001],[45.41873],[54.35163],[44.16405],[58.16847],[56.72721]]
y=[31.70701,68.7776,62.56238,71.54663,87.23093,78.21152,79.64197,59.17149,75.33124,71.30088,55.16568,82.47885,62.00892
,75.39287,81.43619]
reg.fit(x,y)
k=reg.coef_#获取斜率w1,w2,w3,...,wn
b=reg.intercept_#获取截距w0
x0=np.arange(30,60,0.2)
y0=k*x0+b
print("k={0},b={1}".format(k,b))
plt.scatter(x,y)
plt.plot(x0,y0,label='Line编程客栈arRegression')
plt.xlabel('X')
plt.ylabel('Y')
plt.legend()
plt.show()

结果：

k=[1.36695374],b=0.13079331831460195

（2）Python详细实现（方法1）

代码：

#方法1
import numpy as np
import matplotlib.pyplot as plt
from pylab import *
mpl.rcParams['font.sans-serif'] = ['SimHei']
mpl.rcParams['axes.unicode_minus'] = False
#数据生成
data = []
for i in range(100):
  x = np.random.uniform(3., 12.)
  # mean=0, std=1
  eps = np.random.normal(0., 1)
  y = 1.677 * x + 0.039 http://www.cppcns.com+ eps
  data.append([x, y])

data = np.array(data)

#统计误差
# y = wx + b
def compute_error_for_line_given_points(b, w, points):
  totalError = 0
  for i in range(0, len(points)):
    x = points[i, 0]
    y = points[i, 1]
    # computer mean-squared-error
    totalError += (y - (w * x + b)) ** 2
  # average loss for each point
  return totalError / float(len(points))


#计算梯度
def step_gradient(b_current, w_current, points, learningRate):
  b_gradient = 0
  w_gradient = 0
  N = float(len(points))
  for i in range(0, len(points)):
    x = points[i, 0]
    y = points[i, 1]
    # grad_b = 2(wx+b-y)
    b_gradient += (2/N) * ((w_current * x + b_current) - y)
    # grad_w = 2(wx+b-y)*x
    w_gradient += (2/N) * x * ((w_current * x + b_current) - y)
  # update w'
  new_b = b_current - (learningRate * b_gradient)
  new_w = w_current - (learningRate * w_gradient)
  return [new_b, new_w]

#迭代更新
def gradient_descent_runner(points, starting_b, starting_w, learning_rate, num_iterations):
  b = starting_b
  w = starting_w
  # update for several times
  for i in range(num_iterations):
    b, w = step_gradient(b, w, np.array(points), learning_rate)
  return [b, w]


def main():

  learning_rate = 0.0001
  initial_b = 0 # initial y-intercept guess
  initial_w = 0 # initial slope guess
  num_iterations = 1000
  print("迭代前 b = {0}, w = {1}, error = {2}"
     .format(initial_b, initial_w,
         compute_error_for_line_given_points(initial_b, initial_w, data))
     )
  print("Running...")
  [b, w] = gradient_descent_runner(data, initial_b, initial_w, learning_rate, num_iterations)
  print("第 {0} 次迭代结果 b = {1}, w = {2}, error = {3}".
     format(num_iterations, bhttp://www.cppcns.com, w,
        compute_error_for_line_given_points(b, w, data))
     )
  plt.plot(data[:,0],data[:,1], color='b', marker='+', linestyle='--',label='true')
  plt.plot(data[:,0],w*data[:,0]+b,color='r',label='predict')
  plt.xlabel('X')
  plt.ylabel('Y')
  plt.legend()
  plt.show()


if __name__ == '__main__':
  main()

 结果：

迭代前 ：b = 0, w = 0, error = 186.61000821356697

Running...

第 1000 次迭代结果：b = 0.20558501549252192, w = 1.6589067569038516, error = 0.9963685680112963

（3）Python详细实现（方法2）

代码：

#方法2

import numpy as np
import pandas as pd
import matplotlib.pyplot as plt
import matplotlib as mpl
mpl.rcParams["font.sans-serif"]=["SimHei"]
mpl.rcParams["axes.unicode_minus"]=False


# y = wx + b
#Import data
file=pd.read_csv("data.csv")

def compute_error_for_line_given(b, w):
  totalError = np.sum((file['y']-(w*file['x']+b))**2)
  return np.mean(totalError)

def step_gradient(b_current, w_current, learningRate):
  b_gradient = 0
  w_gradient = 0
  N = float(len(file['x']))
  for i in range (0,len(file['x'])):
    # grad_b = 2(wx+b-y)
    b_gradient += (2 / N) * ((w_current * file['x'] + b_current) - file['y'])
    # grad_w = 2(wx+b-y)*x
    w_gradient += (2 / N) * file['x'] * ((w_current * file['x'] + b_current) - file['x'])
  # update w'
  new_b = b_current - (learningRate * b_gradient)
  new_w = w_current - (learningRate * w_gradient)
  return [new_b, new_w]


def gradient_descent_runner( starting_b, starting_w, learning_rate, num_iterations):
  b = starting_b
  w = starting_w
  # update for several times
  for i in range(num_iterations):
    b, w = step_gradient(b, w, learning_rate)
  return [b, w]


def main():
  learning_rate = 0.0001
  initial_b = 0 # initial y-intercept guess
  initial_w = 0 # initial slope guess
  num_iterations = 100
  print("Starting gradient descent at b = {0}, w = {1}, error = {2}"
     .format(initial_b, initial_w,
         compute_error_for_line_given(initial_b, initial_w))
     )
  print("Running...")
  [b, w] = gradient_descent_runner(initial_b, initial_w, learning_rate, num_iterations)
  print("After {0} iterations b = {1}, w = {2}, error = {3}".
     format(num_iterations, b, w,
        compute_error_for_line_given(b, w))
     )
  plt.plot(file['x'],file['y'],'ro',label='线性回归')
  plt.xlabel('X')
  plt.ylabel('Y')
  plt.legend()
  plt.show()




if __name__ == '__main__':
  main()

结果：

Starting gradient descent at b = 0, w = 0, error = 75104.71822821398
Running...
编程客栈After 100 iterations b = 0   0.014845
1   0.325621
2   0.036883
3   0.502265
4   0.564917
5   0.479366
6   0.568968
7   0.422619
8   0.565073
9   0.393907
10  0.216854
11  0.580750
12  0.379350
13  0.361574
14  0.511651
dtype: float64, w = 0   0.999520
1   0.994006
2   0.999405
3   0.989645
4   0.990683
5   0.991444
6   0.989282
7   0.989573
8   0.988498
9   0.992633
10  0.995329
11  0.989490
12  0.991617
13  0.993872
14  0.991116
dtype: float64, error = 6451.5510231710905

数据： 

（4）Python详细实现（方法3） 

#方法3

import numpy as np

points = np.genfromtxt("data.csv", delimiter=",")
#从数据读入到返回需要两个迭代循环，第一个迭代将文件中每一行转化为一个字符串序列，
#第二个循环迭代对每个字符串序列指定合适的数据类型:
# y = wx + b
def compute_error_for_line_given_points(b, w, points):
  totalError = 0
  for i in range(0, len(points)):
    x = points[i, 0]
    y = points[i, 1]
    # computer mean-squared-error
    totalError += (y - (w * x + b)) ** 2
  # average loss for each point
  return totalError / float(len(points))


def step_gradient(b_current, w_current, points, learningRate):
  b_gradient = 0
  w_gradient = 0
  N = float(len(points))
  for i in range(0, len(points)):
    x = points[i, 0]
    y = points[i, 1]
    # grad_b = 2(wx+b-y)
    b_gradient += (2 / N) * ((w_current * x + b_current) - y)
    # grad_w = 2(wx+b-y)*x
    w_gradient += (2 / N) * x * ((w_current * x + b_current) - y)
  # update w'
  new_b = b_current - (learningRate * b_gradient)
  new_w = w_current - (learningRate * w_gradient)
  return [new_b, new_w]


def gradient_descent_runner(points, starting_b, starting_w, learning_rate, num_iterations):
  b = starting_b
  w = starting_w
  # update for several times
  for i in range(num_iterations):
    b, w = step_gradient(b, w, np.array(points), learning_rate)
  return [b, w]


def main():
  learning_rate = 0.0001
  initial_b = 0 # initial y-intercept guess
  initial_w = 0 # initial slope guess
  num_iterations = 1000
  print("Starting gradient descent at b = {0}, w = {1}, error = {2}"
     .format(initial_b, initial_w,
         compute_error_for_line_given_points(initial_b, initial_w, points))
     )
  print("Running...")
  [b, w] = gradient_descent_runner(points, initial_b, initial_w, learning_rate, num_iterations)
  print("After {0} iterations b = {1}, w = {2}, error = {3}".
     format(num_iterations, b, w,
        compute_error_for_line_given_points(b, w, points))
     )


if __name__ == '__main__':
  main()

4、案例——房屋与价格、尺寸

（1）代码 

#1.导入包
import matplotlib.pyplot as plt
import numpy as np
import pandas as pd
from sklearn import linear_model

#2.加载训练数据，建立回归方程
# 取数据集(1)
datasets_X = []   #存放房屋面积
datasets_Y = []   #存放交易价格
fr = open('房价与房屋尺寸.csv','r')  #读取文件，r: 以只读方式打开文件，w: 打开一个文件只用于写入。
lines = fr.readlines()       #一次读取整个文件。
for line in lines:         #逐行进行操作，循环遍历所有数据
  items = line.strip().split(',')  #去除数据文件中的逗号,strip()用于移除字符串头尾指定的字符（默认为空格或换行符）或字符序列。
                   #split（‘ '）: 通过指定分隔符对字符串进行切片，如果参数 num 有指定值，则分隔 num+1 个子字符串。
  datasets_X.append(int(items[0]))  #将读取的数据转换为int型，并分别写入
  datasets_Y.append(int(items[1]))

length = len(datasets_X)       #求得datasets_X的长度，即为数据的总数
datasets_X = np.array(datasets_X).reshape([length,1])  #将datasets_X转化为数组，并变为1维，以符合线性回归拟合函数输入参数要求
datasets_Y = np.array(datasets_Y)          #将datasets_Y转化为数组

#取数据集(2)
'''fr = pd.read_csv('房价与房屋尺寸.csv',encoding='utf-8')
datasets_X=fr['房屋面积']
datasets_Y=fr['交易价格']'''

minX = min(datasets_X)
maxX = max(datasets_X)
X = np.arange(minX,maxX).reshape([-1,1])    #以数据datasets_X的最大值和最小值为范围，建立等差数列，方便后续画图。
                        #reshape([-1,1]），转换成１列，reshape([２,－1]）：转换成两行
linear = linear_model.LinearRegression()   #调用线性回归模块，建立回归方程，拟合数据
linear.fit(datasets_X, datasets_Y)

#3.斜率及截距
print('Coefficients:', linear.coef_)   #查看回归方程系数(k)
print('intercept:', linear.intercept_)  ##查看回归方程截距(b)
print('y={0}x+{1}'.format(linear.coef_,linear.intercept_)) #拟合线

# 4.图像中显示
plt.scatter(datasets_X, datasets_Y, color = 'red')
plt.plot(X, linear.predict(X), color = 'blue')
plt.xlabel('Area')
plt.ylabel('Price')
plt.show()

（2）结果

Coefficients: [0.14198749]
intercept: 53.43633899175563
y=[0.14198749]x+53.43633899175563

（3）数据 

第一列是房屋面积，第二列是交易价格：

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