Bobo老师机器学习笔记第五课-简单线性回归
课程地址：https://coding.imooc.com/class/169.html
最小二乘法的推导博客点击此处
代码实现（参考Bobo实现，如果要看BoBo老师源代码，请点击此处）：
# -*- encoding: utf-8 -*-
"""
实现简单的线性回归,
自己实现SimpleLineRegession1过程中的2个错误：
1、deno += (x - x_mean) ** 2 写成 deno = (x - x_mean) ** 2 这里要注意： deno是所有计算结果的累计值
2、 方程方式self.a_ * x + self.b_ 写成 self.a_ * x - self.b_。 计算b的公式b=y_mean - a * x_mean, 但是整个方程是 y = ax+b
"""
import numpy as np
class SimpleLineRegession1(object):
"""
不使用向量化实现简单的线性回归
"""
def __init__(self):
"""
在过程中计算出来的变量统一命令,后缀加上_
"""
self.a_ = None # 表示线性的斜率
self.b_ = None # 表示线
def fit(self, X_train, y_train):
"""
训练模型
:param X_train:
:return:
"""
assert X_train.ndim == 1 and y_train.ndim == 1, 'X和Y必须为1维'
assert len(X_train) == len(y_train), 'X和Y的训练个数不相同'
x_mean = np.mean(X_train)
y_mean = np.mean(y_train)
num = 0.0 # 分子 Numerator and denominator
deno = 0.0
for x, y in zip(X_train, y_train):
num += (x - x_mean) * (y - y_mean)
deno += (x - x_mean) ** 2
self.a_ = num / deno
self.b_ = y_mean - self.a_ * x_mean
def _predict(self, x):
"""
预测单个X的结果 线性方程y = a*x + b
:param x:
:return:
"""
return self.a_ * x + self.b_
def predict(self, X_test):
"""
预测X，X是一维的数据
:param X_test:
:return:
"""
assert X_test.ndim == 1, 'X_test必须是一维数组'
assert self.a_ is not None and self.b_ is not None , '在predict之前请先fit'
y_pridect = [self._predict(x) for x in X_test]
return np.array(y_pridect)
def __repr__(self):
return ('SimpleLineRegession1(a=%s, b=%s)' %(self.a_, self.b_))
class SimpleLineRegession2(object):
"""
不使用向量化实现简单的线性回归
"""
def __init__(self):
"""
在过程中计算出来的变量统一命令,后缀加上_
"""
self.a_ = None # 表示线性的斜率
self.b_ = None # 表示线
def fit(self, X_train, y_train):
"""
训练模型
:param X_train:
:return:
"""
assert X_train.ndim == 1 and y_train.ndim == 1, 'X和Y必须为1维'
assert len(X_train) == len(y_train), 'X和Y的训练个数不相同'
x_mean = np.mean(X_train)
y_mean = np.mean(y_train)
self.a_ = (X_train - x_mean).dot(y_train - y_mean) / (X_train - x_mean).dot(X_train - x_mean)
self.b_ = y_mean - self.a_ * x_mean
def _predict(self, x):
"""
预测单个X的结果 线性方程y = a*x + b
:param x:
:return:
"""
return self.a_ * x + self.b_
def predict(self, X_test):
"""
预测X，X是一维的数据
:param X_test:
:return:
"""
assert X_test.ndim == 1, 'X_test必须是一维数组'
assert self.a_ is not None and self.b_ is not None , '在predict之前请先fit'
y_pridect = [self._predict(x) for x in X_test]
return np.array(y_pridect)
def __repr__(self):
return 'SimpleLineRegession2(a=%s, b=%s)' %(self.a_, self.b_)测试代码：
import numpy as np
from timeit import timeit as timeit
import matplotlib.pyplot as plt
from simplelinerregression import SimpleLineRegession1, SimpleLineRegession2
x = np.random.randint(1.0, 6, 10000) + np.random.normal(size=10000)
y = 0.8 * x + 0.4 + np.random.normal(size=len(x))
def test_reg1():
reg1 = SimpleLineRegession1()
reg1.fit(x, y)
reg1.predict(x)
print reg1
def test_reg2():
reg2 = SimpleLineRegession2()
reg2.fit(x, y)
reg2.predict(x)
print reg2
def draw_graph():
x = np.array([1., 2., 3., 4., 5.])
y = np.array([1., 3., 2., 3.0, 5.0])
plt.scatter(x, y)
plt.scatter(x, y, color='green')
plt.axis([0, 6, 0, 6])
reg1 = SimpleLineRegession1()
reg1.fit(x, y)
y_predict = reg1.predict(x)
line_mark = 'y=%sx+%s' % (np.round(reg1.a_, 2), np.round(reg1.b_, 2))
plt.plot(x, y_predict, color='red', label=line_mark)
plt.legend()
plt.show()
if __name__ == '__main__':
print timeit('test_reg1()', "from __main__ import test_reg1", number=3)
print timeit('test_reg2()', "from __main__ import test_reg2", number=3)
draw_graph()运行结果：
运行结果，明显SimpleLineRegession2效率要比SimpleLineRegession1高很多 SimpleLineRegession1(a=0.8018889242367586, b=0.39478340695596614) SimpleLineRegession1(a=0.8018889242367586, b=0.39478340695596614) SimpleLineRegession1(a=0.8018889242367586, b=0.39478340695596614) 0.0413969199446 SimpleLineRegession2(a=0.8018889242367646, b=0.39478340695594794) SimpleLineRegession2(a=0.8018889242367646, b=0.39478340695594794) SimpleLineRegession2(a=0.8018889242367646, b=0.39478340695594794) 0.0128730256884