Python求解回归问题

y=wx+b

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import numpy as np
import matplotlib.pyplot as plt
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# 计算给定(w,b)的平均误差
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))
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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]
# 求导数 除N取平均值
# 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 b' w'
# 梯度指向极大值方向 因此反方向更新梯度
new_b = b_current - (learningRate * b_gradient)
new_w = w_current - (learningRate * w_gradient)
temploss = compute_error_for_line_given_points(new_b, new_w, points)
loss.append(temploss)
return [new_b, new_w]
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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]

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def run():
points = np.genfromtxt("data.csv", delimiter=",")
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))
)
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loss = []
run()
Starting gradient descent at b = 0, w = 0, error = 5565.107834483211
Running...
After 1000 iterations b = 0.08893651993741346, w = 1.4777440851894448, error = 112.61481011613473
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x = [i for i in range(0, len(loss))]
plt.plot(x,loss)
[<matplotlib.lines.Line2D at 0x7f82b2c8bd90>]

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