1线性回归

1.1简单线性回归

在简单线性回归中，输入x只有一个特征，通过调整a和b的参数值，来拟合从x到y的线性关系。下图为进行拟合所需要优化的目标，也即是MES（Mean Squared Error)，只不过省略了平均的部分（除以m）。

对于简单线性回归，只有两个参数a和b，通过对MSE优化目标求导后取极值点极值（最小二乘法），即可求得最优a和b如下，所以在训练简单线性回归模型时，也只需要根据数据求解这两个参数值即可。

下面使用波士顿房价数据集中，索引为5（第6个特征）的特征RM (average number of rooms per dwelling：每个住宅的平均房间数)来进行简单线性回归。其中使用的评价指标为：

以及R方指标。这里R方指标一般情况下可以将对模型的评价压缩到[0,1]之间（也可能是负数），值越接近1效果越好。在式子分子分母中的两项都可以认作在算均方误差，只不过分子使用我们训练的线性回归模型的输出，分母则使用真实标签数据的平均值来做均方误差，表示对每个样本，都是一个固定值瞎猜结果，其值等价于实际标签值的方差。通过模型的均方误差和瞎猜产生的误差做对比，就可以将评价指标限定在[0,1]（如果模型比瞎猜还差就会成为负数）。

# 以sklearn的形式对simple linear regression 算法进行封装
import numpy as np
import sklearn.datasets as datasets
from sklearn.model_selection import train_test_split
import matplotlib.pyplot as plt
from sklearn.metrics import mean_squared_error,mean_absolute_error
np.random.seed(123)

class SimpleLinearRegression():
    def __init__(self):
        """
        initialize model parameters
        """
        self.a_=None
        self.b_=None

    def fit(self,x_train,y_train):
        """
        training model parameters

        Parameters
        ----------
            x_train:train x ,shape:data [N,]
            y_train:train y ,shape:data [N,]
        """
        assert (x_train.ndim==1 and y_train.ndim==1),\
            """Simple Linear Regression model can only solve single feature training data"""
        assert len(x_train)==len(y_train),\
            """the size of x_train must be equal to y_train"""

        x_mean=np.mean(x_train)
        y_mean=np.mean(y_train)
        self.a_=np.vdot((x_train-x_mean),(y_train-y_mean))/np.vdot((x_train-x_mean),(x_train-x_mean))
        self.b_=y_mean-self.a_*x_mean

    def predict(self,input_x):
        """
        make predictions based on a batch of data

        Parameters
        ----------
            input_x:shape->[N,]
        """
        assert input_x.ndim==1 ,\
            """Simple Linear Regression model can only solve single feature data"""

        return np.array([self.pred_(x) for x in input_x])

    def pred_(self,x):
        """
        give a prediction based on single input x
        """
        return self.a_*x+self.b_

    def __repr__(self):
        return "SimpleLinearRegressionModel"

if __name__ == '__main__':
    boston_data = datasets.load_boston()
    print(boston_data['DESCR'])
    x = boston_data['data'][:, 5]  # total x data (506,)
    y = boston_data['target']  # total y data (506,)
    # keep data with target value less than 50.
    x = x[y < 50]  # total x data (490,)
    y = y[y < 50]  # total x data (490,)

    plt.scatter(x, y)
    plt.show()

    # train size:(343,) test size:(147,)
    x_train, x_test, y_train, y_test = train_test_split(x, y, test_size=0.3)

    regs = SimpleLinearRegression()
    regs.fit(x_train, y_train)
    y_hat = regs.predict(x_test)

    rmse = np.sqrt(np.sum((y_hat - y_test) ** 2) / len(x_test))
    mse = mean_squared_error(y_test, y_hat)
    mae = mean_absolute_error(y_test, y_hat)
    # notice
    R_squared_Error = 1 - mse / np.var(y_test)

    print('mean squared error:%.2f' % (mse))
    print('root mean squared error:%.2f' % (rmse))
    print('mean absolute error:%.2f' % (mae))
    print('R squared Error:%.2f' % (R_squared_Error))

    a = regs.a_
    b = regs.b_
    x_plot = np.linspace(4, 8, 50)
    y_plot = x_plot * a + b
    plt.scatter(x, y)
    plt.plot(x_plot, y_plot, color='red')
    plt.show()

输出结果：

mean squared error:26.74
root mean squared error:5.17
mean absolute error:3.85
R squared Error:0.50

数据的可视化：

1.2 多元线性回归的正规方程解

多元线性回归中，单个x的样本拥有了多个特征，也就是上图中带下标的x。
其结构可以用向量乘法表示出来：
为了便于计算，一般会将x增加一个为1的特征，方便与截距bias计算。

而多元线性回归的优化目标与简单线性回归一致。

通过矩阵求导计算，可以得到方程解，但求解的时间复杂度很高。

下面使用正规方程解的形式，来对波士顿房价的所有特征做多元线性回归。

from math import sqrt
import numpy as np
#from PlayML.metrics import r2_score
from sklearn.model_selection import train_test_split
import sklearn.datasets as datasets
#from PlayML.metrics import  root_mean_squared_error
np.random.seed(123)


def mean_squared_error(y_true, y_predict):
    """计算y_true和y_predict之间的MSE"""
    assert len(y_true) == len(y_predict), \
        "the size of y_true must be equal to the size of y_predict"

    return np.sum((y_true - y_predict)**2) / len(y_true)
def r2_score(y_true, y_predict):
    """计算y_true和y_predict之间的R Square"""
    return 1 - mean_squared_error(y_true, y_predict)/np.var(y_true)

def root_mean_squared_error(y_true, y_predict):
    """计算y_true和y_predict之间的RMSE"""

    return sqrt(mean_squared_error(y_true, y_predict))

class LinearRegression():
    def __init__(self):
        self.coef_=None # coeffient
        self.intercept_=None # interception
        self.theta_=None

    def fit_normal(self, x_train, y_train):
        """
        use normal equation solution for multiple linear regresion as model parameters

        Parameters
        ----------
        theta=(X^T * X)^-1 * X^T * y
        """
        assert x_train.shape[0] == y_train.shape[0],\
            """size of the x_train must be equal to 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) # (featere,1)
        self.coef_=self.theta_[1:]
        self.intercept_=self.theta_[0]

    def predict(self,x_pred):
        """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_pred.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_pred),1)),x_pred])
        return X_b.dot(self.theta_)

    def score(self,x_test,y_test):
        """
        Calculate evaluating indicator socre

        Parameters
        ---------
            x_test:x test data
            y_test:true label y for x test data
        """
        y_pred=self.predict(x_test)
        return r2_score(y_test,y_pred)

    def __repr__(self):
        return "LinearRegression"


if __name__ == '__main__':
    # use boston house price dataset for test
    boston_data = datasets.load_boston()
    x = boston_data['data']  # total x data (506,)
    y = boston_data['target']  # total y data (506,)
    # keep data with target value less than 50.
    x = x[y < 50]  # total x data (490,)
    y = y[y < 50]  # total x data (490,)
    # train size:(343,) test size:(147,)
    x_train, x_test, y_train, y_test = train_test_split(x, y, test_size=0.3,random_state=123)

    regs = LinearRegression()
    regs.fit_normal(x_train, y_train)

    # calc error
    score=regs.score(x_test,y_test)
    rmse=root_mean_squared_error(y_test,regs.predict(x_test))

    print('R squared error:%.2f' % (score))
    print('Root mean squared error:%.2f' % (rmse))

输出结果：

R squared error:0.79
Root mean squared error:3.36

1.3 使用梯度下降求解多元线性回归

使用梯度下降的方法，在数据比较多的情况下会比方程解快很多。代码也不难，只是关于参数梯度的向量化推导有一点复杂，还有就是训练前要先进行数据标准化。
损失函数与之前相同，还是使用MSE损失。

为了让梯度为列向量，再做依次转置，得到最终的结果。

import numpy as np
from math import sqrt
import sklearn.datasets as datasets
from sklearn.model_selection import train_test_split
from sklearn.preprocessing import StandardScaler

def accuracy_score(y_true, y_predict):
    """计算y_true和y_predict之间的准确率"""
    assert len(y_true) == len(y_predict), \
        "the size of y_true must be equal to the size of y_predict"

    return np.sum(y_true == y_predict) / len(y_true)


def mean_squared_error(y_true, y_predict):
    """计算y_true和y_predict之间的MSE"""
    assert len(y_true) == len(y_predict), \
        "the size of y_true must be equal to the size of y_predict"

    return np.sum((y_true - y_predict)**2) / len(y_true)


def root_mean_squared_error(y_true, y_predict):
    """计算y_true和y_predict之间的RMSE"""

    return sqrt(mean_squared_error(y_true, y_predict))


def mean_absolute_error(y_true, y_predict):
    """计算y_true和y_predict之间的MAE"""
    assert len(y_true) == len(y_predict), \
        "the size of y_true must be equal to the size of y_predict"

    return np.sum(np.absolute(y_true - y_predict)) / len(y_true)


def r2_score(y_true, y_predict):
    """计算y_true和y_predict之间的R Square"""

    return 1 - mean_squared_error(y_true, y_predict)/np.var(y_true)


class LinearRegression:

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

    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])
        self._theta = np.linalg.inv(X_b.T.dot(X_b)).dot(X_b.T).dot(y_train)

        self.intercept_ = self._theta[0]
        self.coef_ = self._theta[1:]

        return self

    def fit_gd(self, X_train, y_train, eta=0.01, n_iters=1e4):
        """根据训练数据集X_train, y_train, 使用梯度下降法训练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"

        def J(theta, X_b, y):
            try:
                return np.sum((y - X_b.dot(theta)) ** 2) / len(y)
            except:
                return float('inf')
            
        def dJ(theta, X_b, y):
            return X_b.T.dot(X_b.dot(theta) - y) * 2. / len(y)

        def gradient_descent(X_b, y, initial_theta, eta, n_iters=1e4, epsilon=1e-8):

            theta = initial_theta
            cur_iter = 0

            while cur_iter < n_iters:
                gradient = dJ(theta, X_b, y)
                last_theta = theta
                theta = theta - eta * gradient
                if (abs(J(theta, X_b, y) - J(last_theta, X_b, y)) < epsilon):
                    break

                cur_iter += 1

            return theta

        X_b = np.hstack([np.ones((len(X_train), 1)), X_train])
        initial_theta = np.zeros(X_b.shape[1])
        self._theta = gradient_descent(X_b, y_train, initial_theta, eta, n_iters)

        self.intercept_ = self._theta[0]
        self.coef_ = self._theta[1:]

        return self

    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)

    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()"

if __name__ == '__main__':
    # use boston house price dataset
    boston_data = datasets.load_boston()
    x = boston_data['data']  # total x size (506,)
    y = boston_data['target']  # total y size (506,)
    # keep data with target value less than 50.
    x = x[y < 50]  # total x size (490,)
    y = y[y < 50]  # total x size (490,)

    # notice! need data normalize
    scaler=StandardScaler().fit(x)
    x=scaler.transform(x)

    # train size:(343,) test size:(147,)
    x_train, x_test, y_train, y_test = train_test_split(x, y, test_size=0.3, random_state=123)

    regs = LinearRegression()
    regs.fit_gd(x_train, y_train)

    # calc error
    score = regs.score(x_test, y_test)
    rmse = root_mean_squared_error(y_test, regs.predict(x_test))

    print('R squared error:%.2f' % (score))
    print('Root mean squared error:%.2f' % (rmse))

    print('coeffient:', regs.coef_.shape)
    print('interception:', regs.intercept_.shape)

1.4 sklearn中的线性回归模型

import sklearn.datasets as datasets
from sklearn.linear_model import LinearRegression
import numpy as np
from sklearn.model_selection import train_test_split
from math import sqrt
np.random.seed(123)

def mean_squared_error(y_true, y_predict):
    """计算y_true和y_predict之间的MSE"""
    assert len(y_true) == len(y_predict), \
        "the size of y_true must be equal to the size of y_predict"

    return np.sum((y_true - y_predict)**2) / len(y_true)

def root_mean_squared_error(y_true, y_predict):
    """计算y_true和y_predict之间的RMSE"""
    return sqrt(mean_squared_error(y_true, y_predict))

if __name__ == '__main__':
    # use boston house price dataset
    boston_data = datasets.load_boston()
    x = boston_data['data']  # total x size (506,)
    y = boston_data['target']  # total y size (506,)
    # keep data with target value less than 50.
    x = x[y < 50]  # total x size (490,)
    y = y[y < 50]  # total x size (490,)
    # train size:(343,) test size:(147,)
    x_train, x_test, y_train, y_test = train_test_split(x, y, test_size=0.3, random_state=123)

    regs = LinearRegression()
    regs.fit(x_train, y_train)

    # calc error
    score = regs.score(x_test, y_test)
    rmse = root_mean_squared_error(y_test, regs.predict(x_test))

    print('R squared error:%.2f' % (score))
    print('Root mean squared error:%.2f' % (rmse))

    print('coeffient:',regs.coef_.shape)
    print('interception:',regs.intercept_.shape)

R squared error:0.79
Root mean squared error:3.36
coeffient: (13,)
interception: ()