本文共 4442 字，大约阅读时间需要 14 分钟。

课程地址：

最小二乘法的推导博客点击

代码实现（参考Bobo实现，如果要看BoBo老师源代码，请点击）：

# -*- 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 npclass 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 npfrom timeit import timeit as timeitimport matplotlib.pyplot as pltfrom simplelinerregression import SimpleLineRegession1, SimpleLineRegession2x = 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 reg1def test_reg2():    reg2 = SimpleLineRegession2()    reg2.fit(x, y)    reg2.predict(x)    print reg2def 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)0.0413969199446SimpleLineRegession2(a=0.8018889242367646, b=0.39478340695594794)SimpleLineRegession2(a=0.8018889242367646, b=0.39478340695594794)SimpleLineRegession2(a=0.8018889242367646, b=0.39478340695594794)0.0128730256884