Python3入门机器学习 - 线性回归与knn算法处理bost

简单线性回归

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最小二乘法实现原理

使用最小二乘法计算a、b的值，实现线性回归的拟合
# _*_ encoding:utf-8 _*_
import numpy as np

class SimpleLinearRegression1:   //该类使用for循环方法计算a、b值，效率较低
    def __init__(self):
        self.a_ = None
        self.b_ = None

    def fit(self,X_train,y_train):
        X_mean = np.mean(X_train)
        y_mean = np.mean(y_train)
        num = 0.0
        d = 0.0
        for (x,y) in zip(X_train,y_train):
            num += (x-X_mean)*(y-y_mean)
            d += (x-X_mean)**2
        self.a_ = num/d
        self.b_ = y_mean - self.a_*X_mean

    def predict(self,X_test):
        return np.array([self._predict(x) for x in X_test ])

    def _predict(self,x):
        return self.a_*x+self.b_

    def __repr__(self):
        return "SimpleLinearRegression1()"



class SimpleLinearRegression2:          // 该类使用向量乘积方法计算a、b值，效率较高 
    def __init__(self):
        self.a_ = None
        self.b_ = None

    def fit(self, X_train, y_train):
        X_mean = np.mean(X_train)
        y_mean = np.mean(y_train)
        num = (X_train-X_mean).dot(y_train-y_mean)
        d = (X_train-X_mean).dot(X_train-X_mean)
        self.a_ = num / d
        self.b_ = y_mean - self.a_ * X_mean

    def predict(self, X_test):
        return np.array([self._predict(x) for x in X_test])

    def _predict(self, x):
        return self.a_ * x + self.b_

    def __repr__(self):
        return "SimpleLinearRegression2()"

测试

import numpy as np
from matplotlib import pyplot

x = np.random.random(size=100)
y = 3.0*x+4.0+np.random.normal(size=100)

%run MyScripts/SimpleLinearRegression.py
reg1 = SimpleLinearRegression1()
reg2 = SimpleLinearRegression2()

%timeit reg1.fit(x,y)
%timeit reg2.fit(x,y)

y1 = reg1.predict(x)
y2 = reg2.predict(x)

pyplot.scatter(x,y)
pyplot.plot(x,y1,color="r",alpha=0.5)
pyplot.plot(x,y2,color='g')

简单线性回归处理boston数据集

仅以boston数据集的第六个特征作为x轴

衡量指标

MSE

mse = np.sum((y_predict-y_test)**2)/len(y_test)

RMSE

rmse = sqrt(mse)

MAE

mae = np.sum(np.absolute(y_predict-y_test))/len(y_test)

R Square

1-mean_squared_error(y_test,y_predict)/np.var(y_test)

多元线性回归模型

---

# _*_ encoding:utf-8 _*_
import numpy as np
from sklearn.metrics import r2_score

class LinearRegression:
    def __init__(self):
        self.coef_ = None
        self.interception_ = None
        self._theta = None

    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.interception_ = self._theta[0]
        self.coef_ = self._theta[1:]
        return self

    def predict(self,X_predict):
        X_b = np.hstack([np.ones((len(X_predict),1)),X_predict])
        return X_b.dot(self._theta)

    def score(self,X_test,y_test):
        return r2_score(y_test,self.predict(X_test))

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

KNN算法处理回归问题

knn_reg = KNeighborsRegressor()
params=[
    {
        'weights':['uniform'],
        'n_neighbors':[i for i in range(1,11)]
    },
    {
        'weights':['distance'],
        'n_neighbors':[i for i in range(1,11)],
        'p':[i for i in range(1,6)]
    }
]
grid_search = GridSearchCV(knn_reg,params,n_jobs=-1,verbose=1)
grid_search.fit(X_train,y_train)

• grid_search.best_params_ {'n_neighbors': 5, 'p': 1, 'weights': 'distance'}
• grid_search.best_score_ 0.634093080186858
• grid_search.best_estimator_.score(X_test,y_test) 0.7044357727037996