回归分析

最小二乘法和极大似然估计 :思路不同推出同一个公式

最小二次法是均方误差最小化进行模型求解，极大似然估计则是利用了联合分布及似然函数 得到公式
$ E_{(w,b)}= $

利用最优化的思路当$ E_{(w,b)}= $最小时w,b的值
求解方法包括梯度下降法，根据推到公式直接解
损失函数：逐点计算平方损失误差，然后求平均数
导数： 西瓜书上 公式3.30，
这几个方法都没有用牛顿法，海森矩阵计算比较复杂，梯度下降法基本上就是乘了一个学习率，如果用海森矩阵不需要用学习率，相当于一个渐变步长的学习率，下降速度更快。
数据载入

import numpy as np
from sklearn import datasets
from sklearn.model_selection import train_test_split
import matplotlib.pyplot as plt
X,y = datasets.load_iris(return_X_y=True)
X= X[:,0]
X.shape
X_train,X_test,y_train,y_test = train_test_split(X,y,test_size=0.3)

一元回归

直接解

def formula_calculate(X,y):
    m=len(X)
    X_mean=np.mean(X)
    y_mean=np.mean(y)
    w=np.sum(np.multiply(y,X-X_mean)/(np.sum(np.multiply(X,X))-np.sum(X)*np.sum(X)/m))
    b=y_mean-w*X_mean
    return w,b
formula_calculate(X_train,y_train)

plt.scatter(X_train,y_train)
# plt.plot(X_test,0.76*X_test-3.47)
plt.plot(X_test,0.37*X_test-1.15)

梯度下降法

损失函数是系数的函数，另外还要传入数据的x，y
参考链接

def compute_cost(w, b, points):
    total_cost = 0
    M = len(points)
 
    # 逐点计算平方损失误差，然后求平均数
    for i in range(M):
        x = points[i, 0]
        y = points[i, 1]
        total_cost += ( y - w * x - b ) ** 2
 
    return total_cost/M

def grad_desc(points, initial_w, initial_b, alpha, num_iter,cost_limi):
    w = initial_w
    b = initial_b
    # 定义一个list保存所有的损失函数值，用来显示下降的过程
    cost_list = []
 
    for i in range(num_iter):
        cost = compute_cost(w, b, points)
        cost_list.append( cost )
        w, b = step_grad_desc( w, b, alpha, points )
        if cost<=cost_limi:
            break
 
    return [w, b, cost_list]
 
def step_grad_desc( current_w, current_b, alpha, points ):
    sum_grad_w = 0
    sum_grad_b = 0
    M = len(points)
 
    # 对每个点，代入公式求和
    for i in range(M):
        x = points[i, 0]
        y = points[i, 1]
        # w 和b 的导数
        sum_grad_w += ( current_w * x + current_b - y ) * x
        sum_grad_b += current_w * x + current_b - y
 
    # 用公式求当前梯度
    grad_w = 2/M * sum_grad_w
    grad_b = 2/M * sum_grad_b
 
    # 梯度下降，更新当前的w和b
    updated_w = current_w - alpha * grad_w
    updated_b = current_b - alpha * grad_b
 
    return updated_w, updated_b


initial_w=0
initial_b=0
alpha=0.01
num_iter=10000
points = np.array([X_train,y_train]).T
cost_limi=0.1
li_x=grad_desc(points, initial_w, initial_b, alpha, num_iter,cost_limi)

多元回归

多元直接解

X,y = datasets.load_iris(return_X_y=True)
def Multi_calculate(X,y):
    num_feature=X.shape[1]
    beta = np.ones((num_feature+1))
    X = np.c_[X,np.ones(len(X))]
    beta=np.dot(np.dot(np.linalg.pinv(np.dot(X.T,X)),X.T),y)
    return beta
Multi_calculate(X_train,y_train)

梯度下降法

参考链接

def compute_cost(beta,X,Y):
    '''计算当前损失函数值 ,平方损失误差的平均
    input: X: 
            beta : w 和 b
            Y: 实际值 label
    output: err/m(float): 错误率
    '''
    X = np.c_[X,np.ones(len(X))]
    total_cost= np.mean(np.power( Y - np.dot(X,beta),2 ))
    return total_cost

def grad_desc(X,Y, alpha, num_iter,cost_limi=0):
    '''梯度下降
    input: X，Y
            alpha:学习率
            num_iter: 最大迭代次数
            cost_limi:最小损失值
    output: list，list[1]：更新后的beta，损失函数的list
    '''
    num_feature=X.shape[1]
    beta = np.ones((num_feature+1))
    X = np.array(X)
    Y = np.array(Y)
    # 定义一个list保存所有的损失函数值，用来显示下降的过程
    cost_list = []
 
    for i in range(num_iter):
        cost = compute_cost(beta ,X,Y )
        if i==0:
            cost_between=cost_limi+1
        else:
            cost_between=np.abs(cost_list[-1]-cost)
        cost_list.append(cost)
        beta = step_grad_desc( beta, alpha, X,Y )
        if cost_between<=cost_limi:
            break
    return [beta, cost_list]
def step_grad_desc( current_beta, alpha, X,Y ):
    '''梯度更新
    input: X,Y 
            current_beta :现在的 w 和 b
            alpha:学习率
    output: updated_beta： 更新后的 w 和 b'''
    # 求导数，南瓜书14页
    M = len(X)
    X = np.c_[X,np.ones(len(X))]
    #     /M 是因为用M个样本更新的参数，要得到的是每个样本的参数，所以需要除以样本数
    # 可以不除以M但是相对学习率需要下降几个数量级否则可能不收敛
    grad_beta= 2*np.dot(X.T,np.dot(X,current_beta)-Y)/M
    #     print(grad_beta)
    # 梯度下降，更新当前的w和b
    updated_beta = current_beta - alpha * grad_beta
    return updated_beta
initial_w=0
initial_b=0
alpha=0.01
num_iter=40000
X=X_train
Y=y_train
cost_limi=0
li_x=grad_desc(X,Y, alpha, num_iter,cost_limi)

sklearn

# from sklearn.model_selection import train_test_split
# X_train, X_test, y_train, y_test = train_test_split(X, y, test_size = 0.2, random_state = 0)
from sklearn.linear_model import LinearRegression
regressor = LinearRegression()  # 创建回归器
regressor.fit(X_train, y_train)  # 用训练集去拟合回归器
# Predicting the Test set results
y_pred = regressor.predict(X_test) #获得预测结果

迭代次数增加之后结果基本没差别

公式直接解	梯度下降	sklearn
array([-0.10627533, -0.0397204 , 0.22894234, 0.61123074, 0.16149541])	array([-0.10646638, -0.03981776, 0.22896909, 0.61130829, 0.16272824])	[array([-0.10627533, -0.0397204 , 0.22894234, 0.61123074]), 0.16149541375178667]

梯度下降明显不如直接解公式和sklearn 的运行速度快

逻辑回归

逻辑回归西瓜树上有详细的公式可以参考这里不再列出

二分类

梯度下降

# 非线性可以用梯度下降，二分类问题
# https://blog.csdn.net/Weary_PJ/article/details/107380387
# 公式可以参考 ：https://blog.csdn.net/Weary_PJ/article/details/107589709
def sigmoid(x):
    ''' Sigmoid函数
    input: x(mat): feature * w
    output: sigmoid(x)(mat): Sigmoid值
    '''
    return 1/(1+np.exp(-x))

def gradient_descent_batch(current_w, alpha,h, X,Y):
    '''梯度更新和损失值
    input: X,Y 
            current_w :现在的 w 和 b
            alpha:学习率
    output: updated_beta： 更新后的 w 和 b
    err: 错误率'''
    h = sigmoid(np.dot(X,current_w))
     # 损失函数
    cost= -np.mean(label_data*np.log(h) +\
                 (1 - label_data) * np.log(1 - h))
    error = Y - h
    w = current_w + alpha *np.dot(X.T,error)  # 权值修正 batch gradient descent

    return w,cost
    
    
def optimal_model_batch(feature_data, label_data,cost_limi, epoch, alpha):
    '''利用梯度下降法训练LR模型
    input: feature_data(mat)样例数据
            label_data(mat)标签数据
            epoch(int)训练的轮数
            alpha(float)学习率
    output: w(mat):权重
    '''
    X = np.c_[feature_data,np.ones(len(feature_data))]
    y = label_data
    n = np.shape(feature_data)[1]  # 特征个数
    w = np.ones((n+1))  # 初始化权重
    i = 0
    li_cost=[]
    while i <= epoch:
        i += 1
         # sigma函数
        w,cost = gradient_descent_batch(w, alpha,h, X,y) #梯度更新 
        if i==1:
            cost_between=cost_limi+1
        else:
            cost_between= np.abs(li_cost[-1]-cost)
        li_cost.append(cost)
        if i % 100 == 0:
            print("\t-----iter = " + str(i) + \
                  ", train error rate = " + str(errorl))
        elif  cost_between<=cost_limi:
            print("\t-----iter = " + str(i) + \
                  ", train error rate = " + str(errorl))
            break
    return w,li_cost
y1 = y_train[(y_train==1)|(y_train==0)]
X1= X_train[(y_train==1)|(y_train==0),:]
alpha=0.01
num_iter=40000
X=X1
Y=y1
cost_limi=0.0001
li_x=lr_train_bgd(X,Y,cost_limi, num_iter, alpha)
plt.plot(li_x[1][:150])

sklearn

# sklearn 
from sklearn.linear_model import LogisticRegression
clf = LogisticRegression(penalty='none',solver="sag",tol=cost_limi)
clf.fit(X,Y)
[clf.coef_,clf.intercept_]

结果相差比较大，主要是逻辑回归方法太多了找不到针对的方法

多元多分类

损失函数

J ( W ) = − 1 m [ ∑ i = 1 m ∑ j = 1 k I { y ( i ) = j } log ⁡ e W j T X ( i ) ∑ l = 1 k e W l T X ( i ) ] J(W) = -\frac{1}{m} \left [\sum_{i=1}^{m} \sum_{j=1}^{k} I \{ y^{(i)}=j \} \log \frac{e^{W_j^TX^{(i)}}}{\sum _{l=1}^{k}e^{W_l^TX^{(i)}}} \right ] J(W)=−m1​[i=1∑m​j=1∑k​I{y(i)=j}log∑l=1k​eWlT​X(i)eWjT​X(i)​]

导数 求的是 θ j \theta _j θj​的偏导

∂ J ( θ ) ∂ θ j = − 1 m [ ∑ i = 1 m ▽ θ j { ∑ j = 1 k I ( y ( i ) = j ) log ⁡ e θ j T X ( i ) ∑ l = 1 k e θ l T X ( i ) } ] \frac{\partial J(\theta )}{\partial \theta _j} = -\frac{1}{m}\left [ \sum_{i=1}^{m}\triangledown _{\theta_j}\left \{ \sum_{j=1}^{k}I(y^{(i)}=j) \log\frac{e^{\theta_j^TX^{(i)}}}{\sum _{l=1}^{k}e^{\theta_l^TX^{(i)}}} \right \} \right ] ∂θj​∂J(θ)​=−m1​[i=1∑m​▽θj​​{j=1∑k​I(y(i)=j)log∑l=1k​eθlT​X(i)eθjT​X(i)​}]
如果 y ( i ) = j y^{(i)}=j y(i)=j
∂ J ( θ ) ∂ θ j = − 1 m ∑ i = 1 m [ ( 1 − e θ j T X ( i ) ∑ l = 1 k e θ l T X ( i ) ) X ( i ) ] \frac{\partial J(\theta )}{\partial \theta _j} = -\frac{1}{m}\sum_{i=1}^{m}\left [\left ( 1-\frac{e^{\theta_j^TX^{(i)}}}{\sum _{l=1}^{k}e^{\theta_l^TX^{(i)}}} \right )X^{(i)} \right ] ∂θj​∂J(θ)​=−m1​i=1∑m​[(1−∑l=1k​eθlT​X(i)eθjT​X(i)​)X(i)]
如果 y ( i ) ≠ j y^{(i)}\neq j y(i)​=j
∂ J ( θ ) ∂ θ j = − 1 m ∑ i = 1 m [ ( − e θ j T X ( i ) ∑ l = 1 k e θ l T X ( i ) ) X ( i ) ] \frac{\partial J(\theta )}{\partial \theta _j} = -\frac{1}{m}\sum_{i=1}^{m}\left [\left ( -\frac{e^{\theta_j^TX^{(i)}}}{\sum _{l=1}^{k}e^{\theta_l^TX^{(i)}}} \right )X^{(i)} \right ] ∂θj​∂J(θ)​=−m1​i=1∑m​[(−∑l=1k​eθlT​X(i)eθjT​X(i)​)X(i)]
参考链接1，链接2

def softmax(a,k):
    c = np.max(a,axis=1)
    exp_a = np.exp(a - (np.ones([k,1])*c).T)  # 防止溢出的对策
    sum_exp_a = np.sum(exp_a,axis=1)
    y = exp_a / (np.ones([k,1])*sum_exp_a).T
    return y
def gradient_descent_batch(current_w, alpha,k, X,Y):
    '''梯度更新和损失值
    input: X,Y 
            current_w :现在的 w 和 b
            alpha:学习率
    output: updated_beta： 更新后的 w 和 b
    err: 错误率'''
    h = softmax(np.dot(X,current_w),k)
     # 损失函数
    cost= -np.mean(np.eye(k)[Y]*np.log(h))
    error = np.eye(k)[Y] - h
    w = current_w + (alpha/len(X))*np.dot(X.T,error)  # 权值修正 batch gradient descent
    return w,cost
    
    
def optimal_model_batch(feature_data, label_data,cost_limi, epoch, alpha):
    '''利用梯度下降法训练LR模型
    input: feature_data(mat)样例数据
            label_data(mat)标签数据
            epoch(int)训练的轮数
            alpha(float)学习率
    output: w(mat):权重
    '''
    k=np.unique(label_data).shape[0]
    X = np.c_[feature_data,np.ones(len(feature_data))]
    y = label_data
    n = np.shape(feature_data)[1]  # 特征个数
    w = np.ones([n+1,k])  # 初始化权重
    i = 0
    li_cost=[]

    while i <= epoch:
        i += 1
         # sigma函数
        w,cost = gradient_descent_batch(w, alpha,k, X,y) #梯度更新 
        if i==1:
            cost_between=cost_limi+1
        else:
            cost_between= np.abs(li_cost[-1]-cost)

        li_cost.append(cost)
        if i % 100 == 0:
            print("\t-----iter = " + str(i) + \
                  ", train error rate = " + str(cost))
        elif  cost_between<=cost_limi:
            print("\t-----iter = " + str(i) + \
                  ", train error rate = " + str(cost))
            break
    return w,li_cost
X,y = datasets.load_iris(return_X_y=True)
alpha=0.1
num_iter=40000
X=X_train
Y=y_train
cost_limi=0.000001
li_x=optimal_model_batch(X,Y,cost_limi, num_iter, alpha)

sklearn

# sklearn 
from sklearn.linear_model import LogisticRegression
clf = LogisticRegression(penalty='none',solver="sag",multi_class="multinomial",tol=cost_limi)
clf.fit(X,Y)
[clf.coef_,clf.intercept_]

LDA 线性判别

其实线性判别和PCA有很大的相似之处，尤其都是针对某一个矩阵求解方程，但是在线性判别分析中特征向量有明确的意义是针对d维数据的d’投影，尤其是使类与类之间差别最大的结果，所以是一种监督降维的方法
对于多分类问题LDA：
Ｗ是 S w − 1 S b S_{w}^{-1}S_{b} Sw−1​Sb​的d’个最大非零广义特征值多对应的特征向量组成的矩阵 d’<=N-1
引用链接 基本没有改动

import numpy as np
from sklearn.datasets import load_iris
import matplotlib.pyplot as plt
from sklearn.discriminant_analysis import LinearDiscriminantAnalysis


def LDA_dimensionality(X, y, k):
    '''
    X为数据集，y为label，k为目标维数
    '''
    label_ = list(set(y))
    X_classify = {}
    # 每一类对应的样本
    for label in label_:
        Xi = np.array([X[i] for i in range(len(X)) if y[i] == label])
        X_classify[label] = Xi
    
    mju = np.mean(X, axis=0)
    mju_classify = {}
    # 每一类的mju
    for label in label_:
        mjui = np.mean(X_classify[label], axis=0)
        mju_classify[label] = mjui

    # 计算类内散度矩阵
    # Sw = np.dot((X - mju).T, X - mju)
    Sw = np.zeros((len(mju), len(mju)))
    for i in label_:
        Sw += np.dot((X_classify[i] - mju_classify[i]).T,
                     X_classify[i] - mju_classify[i])

    # 计算类间散度矩阵
    # Sb=St-Sw
    Sb = np.zeros((len(mju), len(mju)))
    for i in label_:
        Sb += len(X_classify[i]) * np.dot((mju_classify[i] - mju).reshape(
            (len(mju), 1)), (mju_classify[i] - mju).reshape((1, len(mju))))

    eig_vals, eig_vecs = np.linalg.eig(
        np.linalg.inv(Sw).dot(Sb))  # 计算Sw-1*Sb的特征值和特征矩阵

    # 按从小到大排序，输出排序指示值
    sorted_indices = np.argsort(eig_vals)
    # 反转
    sorted_indices = sorted_indices[::-1]
    # 提取前k个特征向量
    topk_eig_vecs = eig_vecs[:, sorted_indices[0:k:1]]
    """https://blog.csdn.net/qq_41973536/article/details/82690242
    s[i:j:k]，i起始位置,j终止位置，k表示步长，默认为1
    s[::-1]是从最后一个元素到第一个元素复制一遍（反向）
    """
    """也可使用如下方法"""
    # # 按从小到大排序，输出排序指示值
    # sorted_indices = np.argsort(eig_vals)
    # # 提取前k个特征向量
    # topk_eig_vecs = eig_vecs[:, sorted_indices[:-k - 1:-1]]

    return topk_eig_vecs


if '__main__' == __name__:

    iris = load_iris()
    X = iris.data
    y = iris.target

    W = LDA_dimensionality(X, y, 2)
    X_new = np.dot((X), W)  # 估计值

    plt.figure(1)
    plt.scatter(X_new[:, 0], X_new[:, 1], marker='o', c=y)
    plt.title("LDA reducing dimension - our method")
    plt.xlabel("x1")
    plt.ylabel("x2")

    # 与sklearn中的LDA函数对比
    lda = LinearDiscriminantAnalysis(n_components=2)
    lda.fit(X, y)
    X_new = lda.transform(X)
    X_new = - X_new  # 为了对比方便，取个相反数，并不影响分类结果
    print(X_new)
    plt.figure(2)
    plt.scatter(X_new[:, 0], X_new[:, 1], marker='o', c=y)
    plt.title("LDA reducing dimension - sklearn method")
    plt.xlabel("x1")
    plt.ylabel("x2")

    plt.show()