流程：给定麦子可以产出一定量的面包，然后做一元回归，进行预测。
程序：在迭代次数中不断求解成本函数（cost function）对b和m的偏导，然后根据偏导更新b和m，来使得b和m 达到合适的位置，尽量使的cost function足够小。
如图
J就是cost function，θ可代表m或b
代码
wheat_and_bread = [[0.5, 5], [0.6, 5.5], [0.8, 6], [1.1, 6.8], [1.4, 7]]#麦子产生出面包量的数据
#y=m*x+b
#梯度下降 给定b 和 m 训练数据 学习率
def step_gradient(b_current, m_current, points, learningRate):
b_gradient = 0
m_gradient = 0
N = float(len(points))
for i in range(0, len(points)):
x = points[i][0]
y = points[i][1]
#loss function (y - ((m_current * x) + b_current))^2
#分别对 b 和 m 求偏导，然后求平均 所有训练数据对b，m产生梯度的平均
b_gradient += -(2 / N) * (y - ((m_current * x) + b_current))
m_gradient += -(2 / N) * x * (y - ((m_current * x) + b_current))
#梯度下降 沿着梯度找最合适的 b和m
new_b = b_current - (learningRate * b_gradient)
new_m = m_current - (learningRate * m_gradient)
return [new_b, new_m]
#数据 初始 b，m 学习率 迭代次数
def gradient_descent_runner(points, starting_b, starting_m, learning_rate, num_iterations):
b = starting_b
m = starting_m
for i in range(num_iterations):
b, m = step_gradient(b, m, points, learning_rate)
return [b, m]
b,m=gradient_descent_runner(wheat_and_bread, 1, 1, 0.01, 100)
画出损失函数和拟合函数
拟合函数
cost function和迭代次数的关系
完整版
这里用了plotly包，作图流程类似R语言中的ggplot，便于人的理解。
wheat_and_bread = [[0.5, 5], [0.6, 5.5], [0.8, 6], [1.1, 6.8], [1.4, 7]]#麦子产生出面包量的数据
#y=m*x+b
#梯度下降 给定b 和 m 训练数据 学习率
import plotly
import plotly.plotly as py
import plotly.graph_objs as go
plotly.tools.set_credentials_file(username='username', api_key='api_key')
def step_gradient(b_current, m_current, points, learningRate):
b_gradient = 0
m_gradient = 0
N = float(len(points))
for i in range(0, len(points)):
x = points[i][0]
y = points[i][1]
#loss function (y - ((m_current * x) + b_current))^2
#分别对 b 和 m 求偏导，然后求平均 所有训练数据对b，m产生梯度的平均
b_gradient += -(2 / N) * (y - ((m_current * x) + b_current))
m_gradient += -(2 / N) * x * (y - ((m_current * x) + b_current))
#梯度下降 沿着梯度找最合适的 b和m
new_b = b_current - (learningRate * b_gradient)
new_m = m_current - (learningRate * m_gradient)
return [new_b, new_m]
#数据 初始 b，m 学习率 迭代次数
def gradient_descent_runner(points, starting_b, starting_m, learning_rate, num_iterations):
b = starting_b
m = starting_m
cost_list=[]
for i in range(num_iterations):
b, m = step_gradient(b, m, points, learning_rate)
cost_function = 0.0
for x,y in points:
cost_function+=(y-m*x-b)*(y-m*x-b)/len(points)
cost_list.append(cost_function)
#cost_function 不断下降的散点图
trace = go.Scatter(
x = list(range(num_iterations)),
y = cost_list,
mode = 'markers'
)
data=[trace]
py.iplot(data, filename='basic-scatter')
return [b, m]
b,m=gradient_descent_runner(wheat_and_bread, 1, 1, 0.01, 100)
#画预测的图
x=[]
y=[]
for i in range(0, len(wheat_and_bread)):
x.append(wheat_and_bread[i][0])
y.append(wheat_and_bread[i][1])
trace0=go.Scatter(
x = x,
y = y,
mode = 'markers',
name = 'markers'
)
x_predict=[]
y_predict=[]
for i in range(0, len(wheat_and_bread)):
x_predict.append(wheat_and_bread[i][0])
y_predict.append((wheat_and_bread[i][0])*m+b)
trace1=go.Scatter(
x=x_predict,
y=y_predict,
mode='lines+markers',
name='lines+markers'
)
data = [trace0, trace1]
py.iplot(data, filename='combine')
