Question

与sklearn相比，具有梯度下降的线性回归在相同数据集上给出的结果不同。

想知道为什么会这样。是局部极小值的问题

数据集如下

Sklearn计算截距为-266.53439537，系数为6.13758146

而梯度下降给出的截距为-1.49087014，系数为2.3239637

import numpy as np
import pandas as pd

from sklearn.linear_model import LinearRegression 

import matplotlib.pyplot as plt

def cost (m,b , data_size):
    x = IN
    y = OUT
    totalError = 0
    for i in range (data_size):
    x = IN[i]
    y = OUT[i]
    totalError += ((m*x + b) - y) ** 2
    return totalError/ float(data_size)


def compute_gradient(X , Y, theta_1 ,theta_0 , N, learning_rate):

    gradient_theta_0 = 0
    gradient_theta_1 = 0

    #print (X.shape, Y.shape, N)

    Y_pred = theta_1*X + theta_0

    gradient_theta_1 = ((-2/N) * sum(X * (Y - Y_pred)))
    gradient_theta_0 = ((-2/N) * sum(Y - Y_pred))


    #print (gradient_theta_0 , gradient_theta_1, gradient_theta_0 * 
    learning_rate, gradient_theta_1 * learning_rate)    
    new_theta_0 = theta_0 - (gradient_theta_0 * learning_rate)
    new_theta_1 = theta_1 - (gradient_theta_1 * learning_rate)

    return (new_theta_1,new_theta_0)

IN = np.array([63 , 64, 66, 69, 69, 71, 71, 72, 73, 75])
OUT = np.array([127,121,142,157,162,156,169,165,181,208])

X = IN[:,np.newaxis]
Y = OUT[:,np.newaxis]

iterations       = 10000
initial_theta_0  = 0 
initial_theta_1  = 0
learning_rate    = 0.00001  
theta_0          = initial_theta_0
theta_1          = initial_theta_1

fig,ax = plt.subplots(figsize=(12,8))
cost_history = []

for i in range (iterations):
    #print ("iteration {} m {} b {}".format(i, theta_1, theta_0))
    [theta_1, theta_0] = compute_gradient(X , Y , theta_1 ,theta_0, 
data_size, learning_rate) 
    totalError = cost (theta_1,theta_0, data_size)
    #print (totalError)
    cost_history.append (totalError)

ax.plot(range(iterations),cost_history,'b.')    

print ("iteration {} m {} b {}".format(i, theta_1, theta_0))

reg_line = [(theta_1 * x) + theta_0 for x in IN]

lm = LinearRegression()
lm.fit(X, Y)

print ("SKLEARN coeff {}".format(lm.coef_))
print ("SKLEARN intercept {}".format(lm.intercept_))

#reg_line = [(lm.coef_[0] * x) + lm.intercept_ for x in IN]

ax3.plot (IN, reg_line , color='red')  
plt.show()

print ("SKLEARN coeff {}".format(lm.coef_))
print ("SKLEARN intercept {}".format(lm.intercept_)) 

RESULTS
iteration 99999 m [2.3239637] b [-1.49087014]
SKLEARN coeff [[6.13758146]]
SKLEARN intercept [-266.53439537]

Answer 1

您的初始条件不好（0,0），并跌至接近该点的局部最小值。更加直观的初始条件基于ht和wt的最大值和最小值，即

initial_theta_0 = np.min(Y)+np.min(X)*(np.min(Y)-np.max(Y))/(np.max(X)-np.min(X)) #-335.75
initial_theta_1 = (np.max(Y)-np.min(Y))/(np.max(X)-np.min(X)) # 7.25    

#initial_theta_0 = 121+63*(121-208)/(75-63) # -335.75
#initial_theta_1 = (208-121)/(75-63) # 7.25

带有梯度的线性回归未给出正确的结果

1 个答案: