使用正规化方程实现多元线性回归

笔记

接下来在代码中实现多元线性回归：

import numpy as np
from sklearn.metrics import r2_score
"""
多元线性回归的实现
"""

class LinearRegression:
    def __init__(self):
        self.coef_=None
        self.interception=None
        self.theta=None
#训练模型
    def fit_normal(self,X_train,y_train):
        X_b=np.hstack([np.ones((len(X_train),1)),X_train])
        self.theta=np.linalg.inv(X_b.T.dot(X_b)).dot(X_b.T).dot(y_train))
        self.coef_=self.theta[0]
        self.interception=self.theta[1:]
        return self
#预测
    def predict(self,X_predict):

        X_b=np.hstack([np.ones((len(X_predict),1)),X_predict])
        return X_b.dot(self.theta)
#计算精确度
    def score(self,X_test,y_test):
        y_predict=self.predict(X_test)
        return r2_score(y_test,y_predict)

测试过程：

import numpy as np
from sklearn import datasets

boston=datasets.load_boston()

X=boston.data
y=boston.target

X=X[y<50]
y=y[y<50]

from Play.model_selection import train_test_split

X_train,X_test,y_train,y_test=train_test_split(X,y,seed=666)

X_train.shape

(392, 13)

y_train.shape

(392,)

X_test.shape

(98, 13)

from Play.LinearDay2 import LinearRegression

reg=LinearRegression()
reg.fit_normal(X_train,y_train)

<Play.LinearDay2.LinearRegression at 0x10a6d5518>

reg.predict(X_test)

array([18.08047724, 25.52374702, 12.93068154, 32.89616169, 24.17956679,
        2.67010028, 26.64700396, 32.23851244, 13.96168643, 24.04280799,
       14.93247906, 10.58513734, 30.28710828, 16.2782111 , 23.67817017,
       25.64047759, 18.67821777, 24.02076592, 28.77437534, 26.93946254,
       12.81354434, 27.22770353, 26.0804716 , 23.41900039, 20.79727917,
       31.96786535, 14.90862058, 20.954883  , 12.92314457, 29.80207127,
       35.28684545,  5.03624207, 13.10143242, 35.54317123, 15.98890703,
       21.53597166, 12.47621364, 29.12864349, 27.36022467, 24.05031901,
       14.35220626, 23.61433371, 10.90347719, 22.38154099, 18.62937294,
       16.37126778, 24.43078261, 33.06293684, 19.19809767, 27.04404675,
       18.05674457, 14.85136715, 25.08935314, 16.0884098 , 21.74619772,
       16.3194766 , 24.25591698, 11.72935395, 27.92260116, 31.05867941,
       20.17444189, 24.9964365 , 25.99734127, 12.14007801, 16.58246637,
       27.30690012, 22.26787948, 21.72492458, 31.50402544, 14.03991351,
       16.42672344, 24.77534313, 25.18133042, 18.65228238, 17.34412768,
       27.90749795, 23.71553798, 14.62906156, 11.22231617, 31.4243847 ,
       33.66552044, 17.66375929, 18.69989012, 17.79423031, 25.15668084,
       23.66633124, 24.55578753, 26.09123112, 25.49718056, 20.28898727,
       24.87506605, 33.48356492, 36.08610386, 23.07558528, 18.79472835,
       31.04138456, 35.78577626, 20.84603282])

系数：

reg.coef_

34.16143549622471

截距：

reg.interception

array([-1.18919477e-01,  3.63991462e-02, -3.56494193e-02,  5.66737830e-02,
       -1.16195486e+01,  3.42022185e+00, -2.31470282e-02, -1.19509560e+00,
        2.59339091e-01, -1.40112724e-02, -8.36521175e-01,  7.92283639e-03,
       -3.81966137e-01])

测试精确度：

reg.score(X_test,y_test)

0.8129802602658466

bobo代码：

这里写代码片import numpy as np
from .metrics import r2_score


class LinearRegression:

    def __init__(self):
        """初始化Linear Regression模型"""
        self.coef_ = None
        self.intercept_ = None
        self._theta = None

        #训练
    def fit_normal(self, X_train, y_train):
        """根据训练数据集X_train, y_train训练Linear Regression模型"""
        assert X_train.shape[0] == y_train.shape[0], \
            "the size of X_train must be equal to the size of y_train"

        X_b = np.hstack([np.ones((len(X_train), 1)), X_train])#X_b要多加一列数为1，通过hstack添加（全为1（传入行数为X_train的长度，列只有一列），X_train），两者组合在一起
        self._theta = np.linalg.inv(X_b.T.dot(X_b)).dot(X_b.T).dot(y_train)#根据公式求解

        self.intercept_ = self._theta[0]#截距是第0个元素
        self.coef_ = self._theta[1:]#从第1个元素直至末尾

        return self

    #预测
    def predict(self, X_predict):
        """给定待预测数据集X_predict，返回表示X_predict的结果向量"""
        assert self.intercept_ is not None and self.coef_ is not None, \
            "must fit before predict!"
        assert X_predict.shape[1] == len(self.coef_), \
            "the feature number of X_predict must be equal to X_train"

        X_b = np.hstack([np.ones((len(X_predict), 1)), X_predict])
        return X_b.dot(self._theta)#X_b*theta

    def score(self, X_test, y_test):
        """根据测试数据集 X_test 和 y_test 确定当前模型的准确度"""

        y_predict = self.predict(X_test)
        return r2_score(y_test, y_predict)

    def __repr__(self):
        return "LinearRegression()"