机器学习-SVM

时间 2019-11-20

标签机器学习 svm 繁體版

原文原文链接

\[此篇文章介绍关于SVM中的一些不懂的地方的公式推导，以及代码实现和一些SVM问题，经过作题检验掌握的效果。\]python

1、代码实现

\[调用sklearn包，进行SVM分类\]算法

#!/usr/bin/python
# -*- coding utf-8 -*-


import numpy as np
import matplotlib.pyplot as plt
import pandas as pd
import matplotlib as mpl
from sklearn import svm
from sklearn.model_selection import train_test_split
from sklearn.metrics import accuracy_score


def load_data():
    path = 'E:\数据挖掘\Machine learning\[小象学院]机器学习课件\8.Regression代码\8.Regression\iris.data'  
    # 读取文件路径
    data = pd.read_csv(path, header = None)
    # 从data 读取数据, x为前4列的全部数据, y为第5列数据
    x, y = data[range(4)], data[4]
    # 返回字符类别的位置索引, 因y数组包含三类, 对应返回下标值
    y = pd.Categorical(y).codes
    # 取x的前两列数据, 通常SVM只作二特征分类, 多特征的转化为多个二特征分类再bagging?
    x = x[[0, 1]]
    # x = x[[0 ,2]]
    return x, y



def classifier(x,y):
    # 鸢尾花包含四个特征属性, 包含三类标签, 山鸢尾(0), 变色鸢尾(1), 维吉尼亚鸢尾(2)
    iris_feature = u'花萼长度', u'花萼宽度', u'花瓣长度', u'花瓣宽度'
    # 按 0.6 的比例，test_data 占40%， train_data 占60%, random_state随机数的种子, 1为产生相同随机数, 产生不一样随机数
    x_train, x_test, y_train, y_test = train_test_split(x, y, random_state=1, train_size=0.6)
    # 使用SVM进行分类训练, 包含关键字, C, gamma, kernel
    # kernel='linear'时，为线性核，C越大分类效果越好, kernel= 'rbf' 时（default）, 为高斯核
    # gamma值越小，分类界面越连续；gamma值越大，分类界面越“散”，分类效果越好
    # decision_function_shape = 'ovr' 时，为one vs rest, 即一个类别与其余类别进行划分，decision_function_shape = 'ovo'
    # 为one vs one,即将类别两两之间进行划分，用二分类的方法模拟多分类的结果
    clf = svm.SVC(C=0.8, kernel='rbf', gamma=20, decision_function_shape='ovr')
    clf.fit(x_train, y_train.ravel())
    # score函数返回返回该次预测的系数R2, 在（0, 1)之间、accuracy_score指的是分类准确率，即分类正确占全部分类的百分比
    # recall_score 召回率 = 提取出的正确信息条数 / 样本中的信息条数
    print(clf.score(x_train, y_train))
    print('训练集准确率：', accuracy_score(y_train, clf.predict(x_train)))
    print(clf.score(x_test, y_test))
    print('测试集准确率：', accuracy_score(y_test, clf.predict(x_test)))

    # decision_function()的功能: 计算样本点到分割超平面的函数距离, 每一列的值表明距离各种别的距离
    print('decision_function:\n', clf.decision_function(x_train))
    print('\npredict:\n', clf.predict(x_train))

    # 画图
    x1_min, x2_min = x.min()  # 第0列的范围
    x1_max, x2_max = x.max()  # 第1列的范围
    x1, x2 = np.mgrid[x1_min:x1_max:500j, x2_min:x2_max:500j]  # 生成网格采样点
    grid_test = np.stack((x1.flat, x2.flat), axis=1)  # 测试点
    # print 'grid_test = \n', grid_test
    # Z = clf.decision_function(grid_test)    # 样本到决策面的距离
    # print Z
    grid_hat = clf.predict(grid_test)  # 预测分类值
    grid_hat = grid_hat.reshape(x1.shape)  # 使之与输入的形状相同
    mpl.rcParams['font.sans-serif'] = [u'SimHei']
    mpl.rcParams['axes.unicode_minus'] = False

    cm_light = mpl.colors.ListedColormap(['#A0FFA0', '#FFA0A0', '#A0A0FF'])
    cm_dark = mpl.colors.ListedColormap(['g', 'r', 'b'])
    plt.figure(facecolor='w')
    plt.pcolormesh(x1, x2, grid_hat, cmap=cm_light)
    plt.scatter(x[0], x[1], c=y, edgecolors='k', s=50, cmap=cm_dark)  # 样本
    plt.scatter(x_test[0], x_test[1], s=120, facecolors='none', zorder=10)  # 圈中测试集样本
    plt.xlabel(iris_feature[0], fontsize=13)
    plt.ylabel(iris_feature[1], fontsize=13)
    plt.xlim(x1_min, x1_max)
    plt.ylim(x2_min, x2_max)
    plt.title(u'鸢尾花SVM二特征分类', fontsize=16)
    plt.grid(b=True, ls=':')
    plt.tight_layout(pad=1.5)
    plt.show()


if __name__ == "__main__":
    x, y = load_data()
    classifier(x, y)

\[SMO算法\]数组

# -*- coding: utf-8 -*-
import numpy as np
import matplotlib.pyplot as plt


def loadDataSet(fileName):
    # 数据矩阵
    dataMat = []
    # 标签向量
    labelMat = []
    # 打开文件
    fr = open(fileName)
    # 逐行读取
    for line in fr.readlines():
        # 去掉每一行首尾的空白符，例如'\n','\r','\t',' '
        # 将每一行内容根据'\t'符进行切片
        lineArr = line.strip().split('\t')
        # 添加数据(100个元素排成一行)
        dataMat.append([float(lineArr[0]), float(lineArr[1])])
        # 添加标签(100个元素排成一行)
        labelMat.append(float(lineArr[2]))
    return dataMat, labelMat

def selectJrand(i, m):
    # i为第一个alpha的下标，m是全部alpha的数目
    j = i
    while (j == i):
        # uniform()方法将随机生成一个实数，它在[x, y)范围内
        j = int(np.random.uniform(0, m))
    return j


def clipAlpha(aj, H, L):
    if aj > H:
        aj = H
    if L > aj:
        aj = L
    return aj

def smoSimple(dataMatIn, classLabels, C, toler, maxIter):
    # 转换为numpy的mat矩阵存储(100,2)
    dataMatrix = np.mat(dataMatIn)
    # 转换为numpy的mat矩阵存储并转置(100,1)
    labelMat = np.mat(classLabels).transpose()
    # 初始化b参数，统计dataMatrix的维度,m:行；n:列
    b = 0
    # 统计dataMatrix的维度,m:100行；n:2列
    m, n = np.shape(dataMatrix)
    # 初始化alpha参数，设为0
    alphas = np.mat(np.zeros((m, 1)))
    # 初始化迭代次数
    iter_num = 0
    # 最多迭代maxIter次
    while (iter_num < maxIter):
        alphaPairsChanged = 0
        for i in range(m):
            # 步骤1：计算偏差Ei
            # multiply(a,b)就是个乘法，若是a,b是两个数组，那么对应元素相乘
            # .T为转置
            fxi = float(np.multiply(alphas, labelMat).T * (dataMatrix * dataMatrix[i, :].T)) + b
            # 偏差项计算公式
            Ei = fxi - float(labelMat[i])
            # 优化alpha，设定必定的容错率
            if ((labelMat[i] * Ei < -toler) and (alphas[i] < C)) or ((labelMat[i] * Ei > toler) and (alphas[i] > 0)):
                # 随机选择另外一个alpha_i成对比优化的alpha_j
                j = selectJrand(i, m)
                # 步骤1，计算偏差Ej
                fxj = float(np.multiply(alphas, labelMat).T * (dataMatrix * dataMatrix[j, :].T)) + b
                # 偏差项计算公式
                Ej = fxj - float(labelMat[j])
                # 保存更新前的alpha值，使用深拷贝(彻底拷贝)A深层拷贝为B，A和B是两个独立的个体
                alphaIold = alphas[i].copy()
                alphaJold = alphas[j].copy()
                # 步骤2：计算上下界H和L
                if (labelMat[i] != labelMat[j]):
                    L = max(0, alphas[j] - alphas[i])
                    H = min(C, C + alphas[j] - alphas[i])
                else:
                    L = max(0, alphas[j] + alphas[i] - C)
                    H = min(C, alphas[j] + alphas[i])
                if (L == H):
                    print("L == H")
                    continue
                # 步骤3：计算eta
                eta = 2.0 * dataMatrix[i, :] * dataMatrix[j, :].T - dataMatrix[i, :] * dataMatrix[i, :].T - dataMatrix[
                                                                                                            j,
                                                                                                            :] * dataMatrix[
                                                                                                                 j, :].T
                if eta >= 0:
                    print("eta>=0")
                    continue
                # 步骤4：更新alpha_j
                alphas[j] -= labelMat[j] * (Ei - Ej) / eta
                # 步骤5：修剪alpha_j
                alphas[j] = clipAlpha(alphas[j], H, L)
                if (abs(alphas[j] - alphaJold) < 0.00001):
                    print("alpha_j变化过小")
                    continue
                # 步骤6：更新alpha_i
                alphas[i] += labelMat[j] * labelMat[i] * (alphaJold - alphas[j])
                # 步骤7：更新b_1和b_2
                b1 = b - Ei - labelMat[i] * (alphas[i] - alphaIold) * dataMatrix[i, :] * dataMatrix[i, :].T - labelMat[
                    j] * (alphas[j] - alphaJold) * dataMatrix[j, :] * dataMatrix[i, :].T
                b2 = b - Ej - labelMat[i] * (alphas[i] - alphaIold) * dataMatrix[i, :] * dataMatrix[j, :].T - labelMat[
                    j] * (alphas[j] - alphaJold) * dataMatrix[j, :] * dataMatrix[j, :].T
                # 步骤8：根据b_1和b_2更新b
                if (0 < alphas[i] < C):
                    b = b1
                elif (0 < alphas[j] < C):
                    b = b2
                else:
                    b = (b1 + b2) / 2.0
                # 统计优化次数
                alphaPairsChanged += 1
                # 打印统计信息
                print("第%d次迭代 样本：%d， alpha优化次数：%d" % (iter_num, i, alphaPairsChanged))
        # 更新迭代次数
        if (alphaPairsChanged == 0):
            iter_num += 1
        else:
            iter_num = 0
        print("迭代次数：%d" % iter_num)
    return b, alphas


def get_w(dataMat, labelMat, alphas):
    alphas, dataMat, labelMat = np.array(alphas), np.array(dataMat), np.array(labelMat)
    # 咱们不知道labelMat的shape属性是多少，
    # 可是想让labelMat变成只有一列，行数不知道多少，
    # 经过labelMat.reshape(1, -1)，Numpy自动计算出有100行，
    # 新的数组shape属性为(100, 1)
    # np.tile(labelMat.reshape(1, -1).T, (1, 2))将labelMat扩展为两列(将第1列复制获得第2列)
    # dot()函数是矩阵乘，而*则表示逐个元素相乘
    # w = sum(alpha_i * yi * xi)
    w = np.dot((np.tile(labelMat.reshape(1, -1).T, (1, 2)) * dataMat).T, alphas)
    return w.tolist()


def showClassifer(dataMat, w, b):
    # 正样本
    data_plus = []
    # 负样本
    data_minus = []
    for i in range(len(dataMat)):
        if labelMat[i] > 0:
            data_plus.append(dataMat[i])
        else:
            data_minus.append(dataMat[i])
    # 转换为numpy矩阵
    data_plus_np = np.array(data_plus)
    # 转换为numpy矩阵
    data_minus_np = np.array(data_minus)
    # 正样本散点图（scatter）
    # transpose转置
    plt.scatter(np.transpose(data_plus_np)[0], np.transpose(data_plus_np)[1], s=30, alpha=0.7)
    # 负样本散点图（scatter）
    plt.scatter(np.transpose(data_minus_np)[0], np.transpose(data_minus_np)[1], s=30, alpha=0.7)
    # 绘制直线
    x1 = max(dataMat)[0]
    x2 = min(dataMat)[0]
    a1, a2 = w
    b = float(b)
    a1 = float(a1[0])
    a2 = float(a2[0])
    y1, y2 = (-b - a1 * x1) / a2, (-b - a1 * x2) / a2
    plt.plot([x1, x2], [y1, y2])
    # 找出支持向量点
    # enumerate在字典上是枚举、列举的意思
    for i, alpha in enumerate(alphas):
        # 支持向量机的点
        if (abs(alpha) > 0):
            x, y = dataMat[i]
            plt.scatter([x], [y], s=150, c='none', alpha=0.7, linewidth=1.5, edgecolors='red')
    plt.show()


if __name__ == '__main__':
    dataMat, labelMat = loadDataSet('E:\\数据挖掘\\Machine learning\\代码\\SVM_Project1\\testSet.txt')
    b, alphas = smoSimple(dataMat, labelMat, 0.6, 0.001, 40)
    w = get_w(dataMat, labelMat, alphas)
    showClassifer(dataMat, w, b)

\[核函数测试\]缓存

# -*- coding: utf-8 -*-

import matplotlib.pyplot as plt
import numpy as np
import random

class optStruct:
    def __init__(self, dataMatIn, classLabels, C, toler, kTup):
        # 数据矩阵
        self.X = dataMatIn
        # 数据标签
        self.labelMat = classLabels
        # 松弛变量
        self.C = C
        # 容错率
        self.tol = toler
        # 矩阵的行数
        self.m = np.shape(dataMatIn)[0]
        # 根据矩阵行数初始化alphas矩阵，一个m行1列的全零列向量
        self.alphas = np.mat(np.zeros((self.m, 1)))
        # 初始化b参数为0
        self.b = 0
        # 根据矩阵行数初始化偏差缓存矩阵，第一列为是否有效标志位，第二列为实际的偏差E的值
        self.eCache = np.mat(np.zeros((self.m, 2)))
        # 初始化核K
        self.K = np.mat(np.zeros((self.m, self.m)))
        # 计算全部数据的核K
        for i in range(self.m):
            self.K[:, i] = kernelTrans(self.X, self.X[i, :], kTup)

def kernelTrans(X, A, kTup):
    # 读取X的行列数
    m, n = np.shape(X)
    # K初始化为m行1列的零向量
    K = np.mat(np.zeros((m, 1)))
    # 线性核函数只进行内积
    if kTup[0] == 'lin':
        K = X * A.T
    # 高斯核函数，根据高斯核函数公式计算
    elif kTup[0] == 'rbf':
        for j in range(m):
            deltaRow = X[j, :] - A
            K[j] = deltaRow * deltaRow.T
        K = np.exp(K / (-1 * kTup[1] ** 2))
    else:
        raise NameError('核函数没法识别')
    return K


def loadDataSet(fileName):
    # 数据矩阵
    dataMat = []
    # 标签向量
    labelMat = []
    # 打开文件
    fr = open(fileName)
    # 逐行读取
    for line in fr.readlines():
        # 去掉每一行首尾的空白符，例如'\n','\r','\t',' '
        # 将每一行内容根据'\t'符进行切片
        lineArr = line.strip().split('\t')
        # 添加数据(100个元素排成一行)
        dataMat.append([float(lineArr[0]), float(lineArr[1])])
        # 添加标签(100个元素排成一行)
        labelMat.append(float(lineArr[2]))
    return dataMat, labelMat

def calcEk(oS, k):
    # multiply(a,b)就是个乘法，若是a,b是两个数组，那么对应元素相乘
    # .T为转置
    fXk = float(np.multiply(oS.alphas, oS.labelMat).T * oS.K[:, k] + oS.b)
    # 计算偏差项
    Ek = fXk - float(oS.labelMat[k])
    # 返回偏差项
    return Ek


def selectJrand(i, m):
    j = i
    while (j == i):
        # uniform()方法将随机生成一个实数，它在[x, y)范围内
        j = int(random.uniform(0, m))
    return j


def selectJ(i, oS, Ei):
    # 初始化
    maxK = -1
    maxDeltaE = 0
    Ej = 0
    # 根据Ei更新偏差缓存
    oS.eCache[i] = [1, Ei]
    # 对一个矩阵.A转换为Array类型
    # 返回偏差不为0的数据的索引值
    validEcacheList = np.nonzero(oS.eCache[:, 0].A)[0]
    # 有不为0的偏差
    if (len(validEcacheList) > 1):
        # 遍历，找到最大的Ek
        for k in validEcacheList:
            # 不计算k==i节省时间
            if k == i:
                continue
            # 计算Ek
            Ek = calcEk(oS, k)
            # 计算|Ei - Ek|
            deltaE = abs(Ei - Ek)
            # 找到maxDeltaE
            if (deltaE > maxDeltaE):
                maxK = k
                maxDeltaE = deltaE
                Ej = Ek
        # 返回maxK，Ej
        return maxK, Ej
    # 没有不为0的偏差
    else:
        # 随机选择alpha_j的索引值
        j = selectJrand(i, oS.m)
        # 计算Ej
        Ej = calcEk(oS, j)
    # 返回j，Ej
    return j, Ej


def updateEk(oS, k):
    # 计算Ek
    Ek = calcEk(oS, k)
    # 更新偏差缓存
    oS.eCache[k] = [1, Ek]


def clipAlpha(aj, H, L):
    if aj > H:
        aj = H
    if L > aj:
        aj = L
    return aj

def innerL(i, oS):
    # 步骤1：计算偏差Ei
    Ei = calcEk(oS, i)
    # 优化alpha,设定必定的容错率
    if ((oS.labelMat[i] * Ei < -oS.tol) and (oS.alphas[i] < oS.C)) or (
            (oS.labelMat[i] * Ei > oS.tol) and (oS.alphas[i] > 0)):
        # 使用内循环启发方式2选择alpha_j,并计算Ej
        j, Ej = selectJ(i, oS, Ei)
        # 保存更新前的alpha值，使用深层拷贝
        alphaIold = oS.alphas[i].copy()
        alphaJold = oS.alphas[j].copy()
        # 步骤2：计算上界H和下界L
        if (oS.labelMat[i] != oS.labelMat[j]):
            L = max(0, oS.alphas[j] - oS.alphas[i])
            H = min(oS.C, oS.C + oS.alphas[j] - oS.alphas[i])
        else:
            L = max(0, oS.alphas[j] + oS.alphas[i] - oS.C)
            H = min(oS.C, oS.alphas[j] + oS.alphas[i])
        if L == H:
            print("L == H")
            return 0
        # 步骤3：计算eta
        eta = 2.0 * oS.K[i, j] - oS.K[i, i] - oS.K[j, j]
        if eta >= 0:
            print("eta >= 0")
            return 0
        # 步骤4：更新alpha_j
        oS.alphas[j] -= oS.labelMat[j] * (Ei - Ej) / eta
        # 步骤5：修剪alpha_j
        oS.alphas[j] = clipAlpha(oS.alphas[j], H, L)
        # 更新Ej至偏差缓存
        updateEk(oS, j)
        if (abs(oS.alphas[j] - alphaJold) < 0.00001):
            print("alpha_j变化过小")
            return 0
        # 步骤6：更新alpha_i
        oS.alphas[i] += oS.labelMat[i] * oS.labelMat[j] * (alphaJold - oS.alphas[j])
        # 更新Ei至偏差缓存
        updateEk(oS, i)
        # 步骤7：更新b_1和b_2:
        b1 = oS.b - Ei - oS.labelMat[i] * (oS.alphas[i] - alphaIold) * oS.K[i, i] - oS.labelMat[j] * (
                    oS.alphas[j] - alphaJold) * oS.K[j, i]
        b2 = oS.b - Ej - oS.labelMat[i] * (oS.alphas[i] - alphaIold) * oS.K[i, j] - oS.labelMat[j] * (
                    oS.alphas[j] - alphaJold) * oS.K[j, j]
        # 步骤8：根据b_1和b_2更新b
        if (0 < oS.alphas[i] < oS.C):
            oS.b = b1
        elif (0 < oS.alphas[j] < oS.C):
            oS.b = b2
        else:
            oS.b = (b1 + b2) / 2.0
        return 1
    else:
        return 0


def smoP(dataMatIn, classLabels, C, toler, maxIter, kTup=('lin', 0)):
    # 初始化数据结构
    oS = optStruct(np.mat(dataMatIn), np.mat(classLabels).transpose(), C, toler, kTup)
    # 初始化当前迭代次数
    iter = 0
    entrieSet = True
    alphaPairsChanged = 0
    # 遍历整个数据集alpha都没有更新或者超过最大迭代次数，则退出循环
    while (iter < maxIter) and ((alphaPairsChanged > 0) or (entrieSet)):
        alphaPairsChanged = 0
        # 遍历整个数据集
        if entrieSet:
            for i in range(oS.m):
                # 使用优化的SMO算法
                alphaPairsChanged += innerL(i, oS)
                print("全样本遍历:第%d次迭代 样本:%d, alpha优化次数:%d" % (iter, i, alphaPairsChanged))
            iter += 1
        # 遍历非边界值
        else:
            # 遍历不在边界0和C的alpha
            nonBoundIs = np.nonzero((oS.alphas.A > 0) * (oS.alphas.A < C))[0]
            for i in nonBoundIs:
                alphaPairsChanged += innerL(i, oS)
                print("非边界遍历:第%d次迭代 样本:%d, alpha优化次数:%d" % (iter, i, alphaPairsChanged))
            iter += 1
        # 遍历一次后改成非边界遍历
        if entrieSet:
            entrieSet = False
        # 若是alpha没有更新，计算全样本遍历
        elif (alphaPairsChanged == 0):
            entrieSet = True
        print("迭代次数:%d" % iter)
    # 返回SMO算法计算的b和alphas
    return oS.b, oS.alphas

def testRbf(k1=1.3):
    # 加载训练集
    dataArr, labelArr = loadDataSet('E:\\数据挖掘\\Machine learning\\代码\\SVM_Project3\\testSetRBF.txt')
    # 根据训练集计算b, alphas
    b, alphas = smoP(dataArr, labelArr, 200, 0.0001, 100, ('rbf', k1))
    datMat = np.mat(dataArr)
    labelMat = np.mat(labelArr).transpose()
    # 得到支持向量
    svInd = np.nonzero(alphas.A > 0)[0]
    sVs = datMat[svInd]
    labelSV = labelMat[svInd]
    print("支持向量个数:%d" % np.shape(sVs)[0])
    m, n = np.shape(datMat)
    errorCount = 0
    for i in range(m):
        # 计算各个点的核
        kernelEval = kernelTrans(sVs, datMat[i, :], ('rbf', k1))
        # 根据支持向量的点计算超平面，返回预测结果
        predict = kernelEval.T * np.multiply(labelSV, alphas[svInd]) + b
        # 返回数组中各元素的正负号，用1和-1表示，并统计错误个数
        if np.sign(predict) != np.sign(labelArr[i]):
            errorCount += 1
    # 打印错误率
    print('训练集错误率:%.2f%%' % ((float(errorCount) / m) * 100))
    # 加载测试集
    dataArr, labelArr = loadDataSet('E:\\数据挖掘\\Machine learning\\代码\\SVM_Project3\\testSetRBF2.txt')
    errorCount = 0
    datMat = np.mat(dataArr)
    labelMat = np.mat(labelArr).transpose()
    m, n = np.shape(datMat)
    for i in range(m):
        # 计算各个点的核
        kernelEval = kernelTrans(sVs, datMat[i, :], ('rbf', k1))
        # 根据支持向量的点计算超平面，返回预测结果
        predict = kernelEval.T * np.multiply(labelSV, alphas[svInd]) + b
        # 返回数组中各元素的正负号，用1和-1表示，并统计错误个数
        if np.sign(predict) != np.sign(labelArr[i]):
            errorCount += 1
    # 打印错误率
    print('训练集错误率:%.2f%%' % ((float(errorCount) / m) * 100))


def showDataSet(dataMat, labelMat):
    # 正样本
    data_plus = []
    # 负样本
    data_minus = []
    for i in range(len(dataMat)):
        if labelMat[i] > 0:
            data_plus.append(dataMat[i])
        else:
            data_minus.append(dataMat[i])
    # 转换为numpy矩阵
    data_plus_np = np.array(data_plus)
    # 转换为numpy矩阵
    data_minus_np = np.array(data_minus)
    # 正样本散点图（scatter）
    # transpose转置
    plt.scatter(np.transpose(data_plus_np)[0], np.transpose(data_plus_np)[1])
    # 负样本散点图（scatter）
    plt.scatter(np.transpose(data_minus_np)[0], np.transpose(data_minus_np)[1])
    # 显示
    plt.show()


if __name__ == '__main__':
    testRbf()

2、公式推导

\(在样本空间中任意点x到超平面(w,b)的距离可写为：\)数据结构

\[ r = \frac{|w^Tx + b|}{||w||} \]app

\[推导以下：\\ 取x_0为任意点x在超平面y= w^Tx + b的投影\\ wx_0 +b = 0 \Longrightarrow |w\vec {xx_0}| = |w\vec r|= ||w||r \\ 另外一方面:|w\vec{xx_0}| = |w(x_0 -x)|=|-b-wx|=|b+wx|\\ \therefore r = \frac{|w^Tx + b|}{||w||}\]
\[ \hat r=yf(x)=y(w^Tx + b)\\ \tilde r = ry = y\frac{|w^Tx + b|}{||w||}=\frac {\hat r}{||w||}\\ \\定义\hat r为函数间隔，\tilde r为几何间隔 \]dom

\[ L(\boldsymbol{w}, b, \boldsymbol{\alpha})=\frac{1}{2}\|\boldsymbol{w}\|^{2}+\sum_{i=1}^{m} \alpha_{i}\left(1-y_{i}\left(\boldsymbol{w}^{\top} \boldsymbol{x}_{i}+b\right)\right) \]机器学习

\[原问题为极小极大问题\min_{\boldsymbol{w,b}}\quad \max_{\boldsymbol{\alpha}}\quad L(w,b,\alpha)\\ 转化为极大极小问题\max_{\boldsymbol{\alpha}}\quad \min_{\boldsymbol{w,b}}\quad L(w,b,\alpha)\]函数

\[推导以下：\\ 目标函数：min\frac{1}{2}||w||^2\\ 约束条件：y_i(w^Tx_i + b) \geq 1\\ \therefore 对每一个在y_i(w^Tx_i+b)-1的i乘以\alpha_i\\ \therefore L(\boldsymbol{w}, b, \boldsymbol{\alpha})=\frac{1}{2}\|\boldsymbol{w}\|^{2}+\sum_{i=1}^{m} \alpha_{i}\left(1-y_{i}\left(\boldsymbol{w}^{\top} \boldsymbol{x}_{i}+b\right)\right)\]学习

\[在其余的机器学习上述公式是L(\boldsymbol{w}, b, \boldsymbol{\alpha})=\frac{1}{2}\|\boldsymbol{w}\|^{2}-\sum_{i=1}^{m} \alpha_{i}\left(y_{i}\left(\boldsymbol{w}^{\top} \boldsymbol{x}_{i}+b\right)-1\right)，二者等价\]
\[ \begin{aligned} w &= \sum_{i=1}^m\alpha_iy_i\boldsymbol{x}_i \\ 0 &=\sum_{i=1}^m\alpha_iy_i \end{aligned} \]

\[推导以下：\\ \begin{aligned} L(\boldsymbol{w},b,\boldsymbol{\alpha}) &= \frac{1}{2}||\boldsymbol{w}||^2+\sum_{i=1}^m\alpha_i(1-y_i(\boldsymbol{w}^T\boldsymbol{x}_i+b)) \\ & = \frac{1}{2}||\boldsymbol{w}||^2+\sum_{i=1}^m(\alpha_i-\alpha_iy_i \boldsymbol{w}^T\boldsymbol{x}_i-\alpha_iy_ib)\\ & =\frac{1}{2}\boldsymbol{w}^T\boldsymbol{w}+\sum_{i=1}^m\alpha_i -\sum_{i=1}^m\alpha_iy_i\boldsymbol{w}^T\boldsymbol{x}_i-\sum_{i=1}^m\alpha_iy_ib \end{aligned}\]

\[\frac {\partial L}{\partial \boldsymbol{w}}=\frac{1}{2}\times2\times\boldsymbol{w} + 0 - \sum_{i=1}^{m}\alpha_iy_i \boldsymbol{x}_i-0= 0 \Longrightarrow \boldsymbol{w}=\sum_{i=1}^{m}\alpha_iy_i \boldsymbol{x}_i\]

\[\frac {\partial L}{\partial b}=0+0-0-\sum_{i=1}^{m}\alpha_iy_i=0 \Longrightarrow \sum_{i=1}^{m}\alpha_iy_i=0\]
\[ \begin{aligned} \max_{\boldsymbol{\alpha}} & \sum_{i=1}^m\alpha_i - \frac{1}{2}\sum_{i = 1}^m\sum_{j=1}^m\alpha_i \alpha_j y_iy_j\boldsymbol{x}_i^T\boldsymbol{x}_j \\ s.t. & \sum_{i=1}^m \alpha_i y_i =0 \\ & \alpha_i \geq 0 \quad i=1,2,\dots ,m \end{aligned} \]
\(推导以下：\\计算拉格朗日函数，即将求得的两个公式代入\)

\[\begin{aligned} \min_{\boldsymbol{w},b} L(\boldsymbol{w},b,\boldsymbol{\alpha}) &=\frac{1}{2}\boldsymbol{w}^T\boldsymbol{w}+\sum_{i=1}^m\alpha_i -\sum_{i=1}^m\alpha_iy_i\boldsymbol{w}^T\boldsymbol{x}_i-\sum_{i=1}^m\alpha_iy_ib \\ &=\frac {1}{2}\boldsymbol{w}^T\sum _{i=1}^m\alpha_iy_i\boldsymbol{x}_i-\boldsymbol{w}^T\sum _{i=1}^m\alpha_iy_i\boldsymbol{x}_i+\sum _{i=1}^m\alpha_ i -b\sum _{i=1}^m\alpha_iy_i \\ & = -\frac {1}{2}\boldsymbol{w}^T\sum _{i=1}^m\alpha_iy_i\boldsymbol{x}_i+\sum _{i=1}^m\alpha_i -b\sum _{i=1}^m\alpha_iy_i \end{aligned}\]

\[\begin{aligned} \min_{\boldsymbol{w},b} L(\boldsymbol{w},b,\boldsymbol{\alpha}) &= -\frac {1}{2}\boldsymbol{w}^T\sum _{i=1}^m\alpha_iy_i\boldsymbol{x}_i+\sum _{i=1}^m\alpha_i \\ &=-\frac {1}{2}(\sum_{i=1}^{m}\alpha_iy_i\boldsymbol{x}_i)^T(\sum _{i=1}^m\alpha_iy_i\boldsymbol{x}_i)+\sum _{i=1}^m\alpha_i \\ &=-\frac {1}{2}\sum_{i=1}^{m}\alpha_iy_i\boldsymbol{x}_i^T\sum _{i=1}^m\alpha_iy_i\boldsymbol{x}_i+\sum _{i=1}^m\alpha_i \\ &=\sum _{i=1}^m\alpha_i-\frac {1}{2}\sum_{i=1 }^{m}\sum_{j=1}^{m}\alpha_i\alpha_jy_iy_j\boldsymbol{x}_i^T\boldsymbol{x}_j \end{aligned}\]
\[ \begin{aligned} & \min_{\boldsymbol{\alpha}}\frac{1}{2}\sum_{i = 1}^m\sum_{j=1}^m\alpha_i \alpha_j y_iy_j\boldsymbol{x}_i^T\boldsymbol{x}_j- \sum_{i=1}^m\alpha_i\\ & s.t. \sum_{i=1}^m \alpha_i y_i =0 \\ & \alpha_i \geq 0 \quad i=1,2,\dots ,m \end{aligned} \]
\[在原\max_{\boldsymbol{\alpha}}\quad \min_{\boldsymbol{w,b}}\quad L(w,b,\alpha)加负号，一样转化为约束最优化问题，为了求解最优解\alpha^*\]
\[ 计算获得\\w^* = \sum_{i =1}^m{\alpha_i}^*y_ix_i\\ b^* = y_i -\sum_{i=1}^m{\alpha_i}^*y_i（x_ix_j）\\ 分离获得超平面：\\ w^*x+ b^* =0\\ 分类决策函数：\\ f(x) =sign(w^*x+b^*) \]

\[引入松弛因子\xi_i的目标函数以下:\\\]
\[ \begin{aligned} & \min_{\boldsymbol{w,b,\xi}}\frac{1}{2}||w||^2 +C\sum_{i = 1}^m\xi_i\\ s.t. & y_i(w.x_i+b)\geq1-\xi_i, i=1,2,\dots ,m \\ & \xi_i \geq 0 \quad i=1,2,\dots ,m \end{aligned} \]

\[同理如上式，构造拉格朗日函数L，再对w,b,\xi分别求偏导，再代入L\]
\[ \begin{aligned} L(\boldsymbol{w},b,\boldsymbol{\alpha},\boldsymbol{\xi},\boldsymbol{\mu}) &= \frac{1}{2}||\boldsymbol{w}||^2+C\sum_{i=1}^m \xi_i+\sum_{i=1}^m \alpha_i(1-\xi_i-y_i(\boldsymbol{w}^T\boldsymbol{x}_i+b))-\sum_{i=1}^m\mu_i \xi_i \end{aligned} \]
\[对w,b,\xi求偏导\]
\[ \begin{aligned} w &= \sum_{i=1}^m\alpha_iy_i\boldsymbol{x}_i \\ 0 &=\sum_{i=1}^m\alpha_iy_i\\ C & = a_i+\mu_i \end{aligned} \]

\[代入L\]
\[ \begin{aligned} \min_{\boldsymbol{w},b,\boldsymbol{\xi}}L(\boldsymbol{w},b,\boldsymbol{\alpha},\boldsymbol{\xi},\boldsymbol{\mu}) &= \frac{1}{2}||\boldsymbol{w}||^2+C\sum_{i=1}^m \xi_i+\sum_{i=1}^m \alpha_i(1-\xi_i-y_i(\boldsymbol{w}^T\boldsymbol{x}_i+b))-\sum_{i=1}^m\mu_i \xi_i \\ &=\frac{1}{2}||\boldsymbol{w}||^2+\sum_{i=1}^m\alpha_i(1-y_i(\boldsymbol{w}^T\boldsymbol{x}_i+b))+C\sum_{i=1}^m \xi_i-\sum_{i=1}^m \alpha_i \xi_i-\sum_{i=1}^m\mu_i \xi_i \\ & = -\frac {1}{2}\sum_{i=1}^{m}\alpha_iy_i\boldsymbol{x}_i^T\sum _{i=1}^m\alpha_iy_i\boldsymbol{x}_i+\sum _{i=1}^m\alpha_i +\sum_{i=1}^m C\xi_i-\sum_{i=1}^m \alpha_i \xi_i-\sum_{i=1}^m\mu_i \xi_i \\ & = -\frac {1}{2}\sum_{i=1}^{m}\alpha_iy_i\boldsymbol{x}_i^T\sum _{i=1}^m\alpha_iy_i\boldsymbol{x}_i+\sum _{i=1}^m\alpha_i +\sum_{i=1}^m (C-\alpha_i-\mu_i)\xi_i \\ &=\sum _{i=1}^m\alpha_i-\frac {1}{2}\sum_{i=1 }^{m}\sum_{j=1}^{m}\alpha_i\alpha_jy_iy_j\boldsymbol{x}_i^T\boldsymbol{x}_j \end{aligned} \]
\[再求\alpha极大\max\]
\[ \begin{aligned} &\max_{\alpha}\sum _{i=1}^m\alpha_i-\frac {1}{2}\sum_{i=1 }^{m}\sum_{j=1}^{m}\alpha_i\alpha_jy_iy_j\boldsymbol{x}_i^T\boldsymbol{x}_j\\ 转化为\\ &\min_{\alpha}\frac {1}{2}\sum_{i=1 }^{m}\sum_{j=1}^{m}\alpha_i\alpha_jy_iy_j\boldsymbol{x}_i^T\boldsymbol{x}_j-\sum _{i=1}^m\alpha_i\\ &s.t.\sum_{i=1}^m \alpha_i y_i=0 \\ &0 \leq\alpha_i \leq C \quad i=1,2,\dots ,m \end{aligned} \]
\[求最优解\alpha^*\]
\[ 计算获得\\w^* = \sum_{i =1}^m{\alpha_i}^*y_ix_i\\ b^* = (\max_{i: y_i =1} w^*.x_i + \min_{i: y_i =-1} w^x* +x_i)/2\\ 分离获得超平面：\\ w^*x+ b^* =0\\ 分类决策函数：\\ f(x) =sign(w^*x+b^*) \]

\[ \left\{\begin{array}{l} {\alpha_{i}\left(f\left(\boldsymbol{x}_{i}\right)-y_{i}-\epsilon-\xi_{i}\right)=0} \\ {\hat{\alpha}_{i}\left(y_{i}-f\left(\boldsymbol{x}_{i}\right)-\epsilon-\hat{\xi}_{i}\right)=0} \\ {\alpha_{i} \hat{\alpha}_{i}=0, \xi_{i} \hat{\xi}_{i}=0} \\ {\left(C-\alpha_{i}\right) \xi_{i}=0,\left(C-\hat{\alpha}_{i}\right) \hat{\xi}_{i}=0} \end{array}\right. \]
\[推导以下：\\\]
\[ \left\{\begin{array}{l}2{f\left(\boldsymbol{x}_{i}\right)-y_{i}-\epsilon-\xi_{i} \leq 0 } \\ 3{y_{i}-f\left(\boldsymbol{x}_{i}\right)-\epsilon-\hat{\xi}_{i} \leq 0 } \\ 4{-\xi_{i} \leq 0} \\5{-\hat{\xi}_{i} \leq 0}6\end{array}\right. \]

\[ \left\{\begin{array}{l} {\alpha_i\left(f\left(\boldsymbol{x}_{i}\right)-y_{i}-\epsilon-\xi_{i} \right) = 0 } \\ {\hat{\alpha}_i\left(y_{i}-f\left(\boldsymbol{x}_{i}\right)-\epsilon-\hat{\xi}_{i} \right) = 0 } \\ {-\mu_i\xi_{i} = 0 \Rightarrow \mu_i\xi_{i} = 0 } \\ {-\hat{\mu}_i \hat{\xi}_{i} = 0 \Rightarrow \hat{\mu}_i \hat{\xi}_{i} = 0 } \end{array}\right. \]

\[\because\begin{aligned} \mu_i=C-\alpha_i \\ \hat{\mu}_i=C-\hat{\alpha}_i \end{aligned}\]
\[ \left\{\begin{array}{l} {\alpha_i\left(f\left(\boldsymbol{x}_{i}\right)-y_{i}-\epsilon-\xi_{i} \right) = 0 } \\ {\hat{\alpha}_i\left(y_{i}-f\left(\boldsymbol{x}_{i}\right)-\epsilon-\hat{\xi}_{i} \right) = 0 } \\ {(C-\alpha_i)\xi_{i} = 0 } \\ {(C-\hat{\alpha}_i) \hat{\xi}_{i} = 0 } \end{array}\right. \]
\[前面硬间隔与软间隔均处理线性问题，而对非线性问题须要将低维空间映射到高维空间，引入核函数\]

\[多项式核函数\]
\[ k(\vec x,\vec y)= (\vec x,\vec y +c)^2\\ =(\vec x, \vec y)^2+2c\vec x\vec y+c^2\\ =\sum_{i =1}^n \sum_{j=1}^m(x_ix_j)(y_iy_j)+\sum_{i=1}^m(\sqrt {2c}x_i \sqrt{2cy_i})+c^2 \]
\(高斯核函数\)
\[ k(\vec x_1,\vec x_2) = e^-\frac{x_1^2+x_2^2}{2\sigma^2}(1+\frac {x_1 x_2}{\sigma^2}+\frac{x_1^2+x_2^2}{2\sigma^2 \sigma^2}+...+\frac{x_1^n+x_2^n}{n!\sigma^n\sigma^n}) \]

3、练习题目

3.1 给定三个数据点，正例点\(x_1 = (3, 3)^T\), \(x_2 = (4, 3)^T\),负例点\(x_3 = (1, 1)^T\)，求线性可分SVM

3.2 SVM能否用于多分类

3.3 SVM和Logistic回归的比较

3.4 核函数是什么？高斯核映射到无穷维是怎么回事？

3.5 如何理解SVM的损失函数？

3.6 使用高斯核函数，请描述SVM的参数C和 \(\sigma\) 对分类器的影响

3.7 比较感知机的对偶性形式与线性可分支持向量机的对偶形式

3-8 证实内积的正整数幂函数：

\[ K(x,z) = (x,z)^p\\ 是正定核函数，此处p为正整数，x,z为R \]

3.9 线性支持向量机还可定义为如下形式：

\[ \begin{aligned} \min_{\boldsymbol{w,b,\xi}}\quad \frac{1}{2}||w||^2+C\sum_{i=1}^N{\xi_i}^2\\ s.t.\quad y_i(\boldsymbol w.{x_i}+b)\geq 1-\xi_i, i= 1,2,...,N\\ {\xi}_i \geq 0, i=1, 2,...,N \end{aligned}\\ 求其对偶形式 \]

3.10 给定数据点，正例点\(x_1 = (1, 2)^T\), \(x_2 = (2, 3)^T\), \(x_3 = (1, 1)^T,\)负例点\(x_4 = (1, 1)^T\),\(x_5 = (1, 1)^T\),求最大间隔分离超平面和分类决策函数，并在图上画出分离超平面，间隔边界及支持向量

3.11 分析SVM对噪声敏感的缘由

3.12 使用核技巧推广对数概率回归，产生核对率回归

3.13 给出式（6.52）的KKT条件

3.13 讨论线性判别分析与线性核支持向量机在何种条件等价

4、参看文献

[1] 《机器学习》邹博

[2] 《SVM的三重境界》 July

[3] 《pumpkin-book》 Datawhale

[4] 《机器学习》周志华

[5] 《机器学习实战》Peter

[6] 《统计学习方法》李航，清华大学出版社，2012

[7] 《机器学习算法精讲》秦曾昌，2018