線性迴歸

• 我們可以通過測量損耗來衡量線路的適合程度。
• 線性迴歸的目標是最小化損失。
• 為了找到最佳擬合線，我們嘗試找到最小化損失的b值（截距）和m值（斜率）。
• 收斂是指引數在每次迭代時停止變化時的引數
• 學習率是指每次迭代時引數的變化程度。
• 我們可以使用Scikit-learn的LinearRegression()模型對一組點進行線性迴歸。

Scikit-Learn庫

line_fitter = LinearRegression()  建立模型
line_fitter.fit(temperature, sales)  傳入引數
sales_predict = line_fitter.predict(temperature) 預測模型

import codecademylib3_seaborn
from sklearn.linear_model import LinearRegression
import matplotlib.pyplot as plt
import numpy as np

temperature = np.array(range(60, 100, 2))
temperature = temperature.reshape(-1, 1)
sales = [65, 58, 46, 45, 44, 42, 40, 40, 36, 38, 38, 28, 30, 22, 27, 25, 25, 20, 15, 5]

line_fitter = LinearRegression()
line_fitter.fit(temperature, sales)
sales_predict = line_fitter.predict(temperature)

plt.plot(temperature, sales, 'o')
plt.plot(temperature,sales_predict)

plt.show()
 

原理 

預測直線 直線上會有loss

import codecademylib3_seaborn
import matplotlib.pyplot as plt
months = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12]
revenue = [52, 74, 79, 95, 115, 110, 129, 126, 147, 146, 156, 184]

#slope:
m = 12
#intercept:
b = 35

plt.plot(months, revenue, "o")

y = [m*month + b for month in months]

plt.plot(months,y)

plt.show()
 

LOSS

計算loss時 要使用平方距離 如下圖 A的loss是 9(3^2) B的loss是1(1^2)

總loss=10 如果發現一條線路使loss小於10 那麼這條線路會成為更好的線路

for i in range(len(y)):
  total_loss+=(y_predicted[i]-y[i])**2

減少loss

grandient descent 梯度下降

找到一點斜率向下的方向說明可以減少損失,所以應該漸變向下

公式

• N is the number of points we have in our dataset
• m is the current gradient guess  斜率
• b is the current intercept guess  截距

找到當截距=b時的梯度的函式

def get_gradient_at_b(x,y,m,b):
    diff=0
    for i in range(len(x)):
      diff+=(y[i]-(m*x[i]+b))
    b_gradient=diff*(-2)/len(x)
    return b_gradient

公式

• N is the number of points we have in our dataset
• m is the current gradient guess  斜率
• b is the current intercept guess  截距

找到當斜率=m時的梯度的函式 

def get_gradient_at_m(x, y, m, b):
    diff = 0
    N = len(x)
    for i in len(x):
      diff += x[i]*(y[i]-(m*x[i]+b))
    m_gradient = -2/N * diff
    return m_gradient

得到合適的梯度

def get_gradient_at_b(x, y, b, m):
  N = len(x)
  diff = 0
  for i in range(N):
    x_val = x[i]
    y_val = y[i]
    diff += (y_val - ((m * x_val) + b))
  b_gradient = -(2/N) * diff  
  return b_gradient

def get_gradient_at_m(x, y, b, m):
  N = len(x)
  diff = 0
  for i in range(N):
      x_val = x[i]
      y_val = y[i]
      diff += x_val * (y_val - ((m * x_val) + b))
  m_gradient = -(2/N) * diff  
  return m_gradient

#Your step_gradient function here
def step_gradient(x, y, b_current, m_current):
    b_gradient = get_gradient_at_b(x, y, b_current, m_current)
    m_gradient = get_gradient_at_m(x, y, b_current, m_current)
    b = b_current - (0.01 * b_gradient)
    m = m_current - (0.01 * m_gradient)
    return [b, m]

months = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12]
revenue = [52, 74, 79, 95, 115, 110, 129, 126, 147, 146, 156, 184]

# current intercept guess:
b = 0
# current slope guess:
m = 0

b, m = step_gradient(months, revenue, b, m)
print(b, m)