机器学习 线性模型里面的线性回归

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线性模型 Python 实现

import numpy as np
from sklearn import linear_model

class LinearRegression:
    def __init__(self):
        self.w = None  # 要训练的参数
        self.n_features = None   # 特征的个数

    def fit(self,X,y):    # 计算权重 W 
        """
        w=(X^TX)^{-1}X^Ty
        """
        assert isinstance(X,np.ndarray) and isinstance(y,np.ndarray)
        assert X.ndim==2 and y.ndim==1    # assert 断言，就是说后面的条件成立时执行下面代码，不满足时返回错误
        assert y.shape[0]==X.shape[0]
        n_samples = X.shape[0]            #样本数量
        self.n_features = X.shape[1]      #特征个数，X.shape是个元组，元组的第二位代表列，也就是特征个数
        extra = np.ones((n_samples,))
        X = np.c_[X,extra]                #是在列方向扩展连接两个矩阵，就是把两矩阵左右相加，要求行数相等，相当于 hstack https://blog.csdn.net/qq_43657442/article/details/108030183
        if self.n_features < n_samples:
            self.w = np.linalg.inv(X.T.dot(X)).dot(X.T).dot(y)  # 其实就是X的广义逆乘Y，得到权重，详解见https://blog.csdn.net/qq_43657442/article/details/108032355
        else:
            raise ValueError('dont have enough samples')

    def predict(self, X):
        n_samples=X.shape[0]
        extra = np.ones((n_samples,))  #产生一个二维数组，用这样的方式代表这个数组可行向量可列向量
        X = np.c_[X, extra]
        if self.w is None:
            raise RuntimeError('cant predict before fit')
        y_=X.dot(self.w)
        return y_

if __name__ == '__main__':
    X = np.array([[1.0,0.5,0.5],[1.0,1.0,0.3],[-0.1,1.2,0.5],[1.5,2.4,3.2],[1.3,0.2,1.4]])
    y = np.array([1,0.5,1.5,2,-0.3])
    lr = LinearRegression()
    lr.fit(X,y)
    X_test = np.array([[1.3,1,3.2],[-1.2,1.2,0.8]])
    y_pre = lr.predict(X_test)
    print(y_pre)

    sklearn_lr = linear_model.LinearRegression()
    sklearn_lr.fit(X,y)
    sklearn_y_pre = sklearn_lr.predict(X_test)
    print(sklearn_y_pre)

    ridge_reg = linear_model.Ridge(alpha=0.05, solver='lsqr')  # 岭回归，具有L2正则化的线性最小二乘法回归模型，损失函数是线性最小二乘函数，而正则化由l2-范数给出, alpha 是正则化的罚系数
    ridge_reg.fit(X, y)
    ridge_y_pre=ridge_reg.predict(X_test)
    print(ridge_y_pre)

机器学习 线性模型里面的线性回归