机器学习第四篇:SVM,SMO算法的实现

这两天摸索了SVM,刚开始的时候接触SMO的时候就很懵,但是我有机器学习三大法宝护体,最终还是搞懂了一些。

前面的细节部分就不予阐述了,直接从SMO算法部分开始讲起:

下面讨论具体方法:

\underset{\alpha }{min } \frac{1}{2}\sum\sum\alpha _i\alpha _jy_iy_jK(x_i,x_j) - \sum\alpha _i                                       (1)

s.t. \sum\alpha _iy_i =0                                                                                (2)

        0\leq \alpha _i\leqslant C , i = 1,2,...,N                                                     (3)

\omega =\sum\alpha _iy_ix_i                                                                                    (4)

\alpha_1y_1+\alpha_2y_2 = -\sum_{i=3}^{N}y_i\alpha_i = \xi                                                        (5)

g(x) = \sum\alpha_iy_iK(x_i,x) +b                                                           (6)

利用上面这些等式来计算:

首先假设我们的初始可行解为\alpha_1 ^{old},\alpha_2^{old},最优解为\alpha_1 ^{new},\alpha_2^{new},未经处理过的\alpha_2的最优解为\alpha_2^{new,unc},根据(5),

\alpha_1y_1+\alpha_2y_2 = \xi,y只有两种情况1和-1,然后在两种情况下我们来确定\alpha_2^{new,unc}的范围:L\leq \alpha _2^{new,unc}\leqslant H

然后分情况讨论L,H的取值:

此时y_1\neq y_2,则:

\fn_jvn L=max(0,\alpha_2^{old}-\alpha_1^{old})H=min(C,C+\alpha_2^{old}-\alpha_1^{old})

y_1= y_2时:

L=max(0,\alpha_2^{old}+\alpha_1^{old}-C)H=min(C,\alpha_2^{old}+\alpha_1^{old})

具体的详细的过程就不予阐述,可以参考李航的《统计学习方法》:

\alpha_2^{new,unc} = \alpha_2^{old} + \frac{y_2(E_1 - E_2)}{\eta }

其中\eta = K_{11} +K_{22}-2K_{12} = \left \| \Phi (x_1) - \Phi(x_2) \right \|^2,所以\eta \geq 0

E_i = g(x_i) - y_i = (\sum \alpha_iy_iK(x_j,x_i)+b)-y_i

得到的\alpha_2^{new,unc},对其进行处理可以得到:

\alpha_2^{new}\left\{\begin{matrix} H, \alpha_2^{old,unc}>H\\ \alpha_2^{new,unc},L\leq \alpha_2^{old,unc}\leq H\\ L,\alpha_2^{old,unc}<L \end{matrix}\right.

得到\alpha_2^{new}后,利用下面公式得到\alpha_1^{new}

\alpha_1^{new} = \alpha_1^{old} +y_1y_2(\alpha_2^{old}-\alpha_2^{new})

现在我们可以进行对\alpha的更新了,那我们如何选择它呢:

第一个变量的选择:在代码当中用外层循环实现,外层循环中选取违反KKT条件最严重的样本点

第二个变量的选择:内层循环,假设已经找到了第一个变量,本意选择是希望能使第二个变量有足够大的变化,但在代码当中一般就选取\left | E_1 -E_2 \right |最大的

更新了\alpha之后,我们也需要更新我们的b。

b_1^{new} = -E_1-y_1K_{11}(\alpha_1^{new} - \alpha_1^{old})-y_2K_{21}(\alpha_2^{new} - \alpha_2^{old})+b^{old}

b_2^{new} = -E_2-y_1K_{12}(\alpha_1^{new} - \alpha_1^{old})-y_2K_{22}(\alpha_2^{new} - \alpha_2^{old})+b^{old}

那么新的b应该选取谁比较合适呢?

如果\alpha_1^{new}在界内,那么就选b_1^{new},如何\alpha_2^{new}在界内,就选b_2^{new},如果都在界内,那么b_1^{new}=b_2^{new},如果都在界上,一般选取b^{new} = \frac{b_1^{new}+b_2^{new}}{2}

在完成两个变量的优化之后,还必须更新我们的E

E_i^{new}= (\sum_{i=3}^N \alpha_iy_iK(x_j,x_i)+b^{new})-y_i

在编程之前再介绍一下SVM的损失函数:合页损失函数。

L(y(w\cdot x+b))=[1-y(w\cdot x+b)]_+

称为合页损失函数,下标+表示取正值的函数.

[z]_+\left\{\begin{matrix} z,z>0\\ 0,z\leq 0 \end{matrix}\right.

也就是说,当样本正确分类时,y(w\cdot x+b)大于等于1的,损失为0,否则损失,1-y(w\cdot x+b)

接下来就接受SMO算法的源代码:

先接受精简版的SMO:

输入我们的数据:

def loadDataSet(fileName):dataMat = []; labelMat = []#x存在datamat当中,y就是我们的标签,存在labelmat中fr = open(fileName)for line in fr.readlines():lineArr = line.strip().split('\t')dataMat.append([float(lineArr[0]), float(lineArr[1])])labelMat.append(float(lineArr[2]))#数据一共有三列,最后一列为yreturn dataMat,labelMat

精简版就随机选择我们的alpha:

def selectJrand(i,m):j=i #we want to select any J not equal to iwhile (j==i):j = int(random.uniform(0,m))return j

提前定义一个\alpha_2^{new}的范围,以便后面使用:

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

然后就是我们的大餐了:

def smoSimple(dataMatIn, classLabels, C, toler, maxIter):dataMatrix = mat(dataMatIn); labelMat = mat(classLabels).transpose()b = 0; m,n = shape(dataMatrix)#返回他的行列alphas = mat(zeros((m,1)))iter = 0 while (iter < maxIter):alphaPairsChanged = 0for i in range(m):fXi = float(multiply(alphas,labelMat).T*(dataMatrix*dataMatrix[i,:].T)) + b     #这一步就是在计算我们的g(x),其中.T是np中的转置,可能并没有完全按照公式来,但只要能求得最后的结果你想怎么转就怎么转,注意矩阵下标,multiply是对位相乘,不是矩阵相乘运算。Ei = fXi - float(labelMat[i])if ((labelMat[i]*Ei < -toler) and (alphas[i] < C)) or ((labelMat[i]*Ei > toler) and (alphas[i] > 0)):#就是检查误差E是否足够大,我们选择alpha的时候就是希望有很大的变化j = selectJrand(i,m)#随机挑选第二个alphafXj = float(multiply(alphas,labelMat).T*(dataMatrix*dataMatrix[j,:].T)) + bEj = fXj - float(labelMat[j])#计算出第二个EalphaIold = alphas[i].copy(); alphaJold = alphas[j].copy();if (labelMat[i] != labelMat[j]):#按照前面的公式,确定L,HL = 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"); continueeta = 2.0 * dataMatrix[i,:]*dataMatrix[j,:].T - dataMatrix[i,:]*dataMatrix[i,:].T - dataMatrix[j,:]*dataMatrix[j,:].T#这里是在计算eta,没有完全按照公式,把公式前面的正号变成负号就可以了if eta >= 0: print ("eta>=0"); continue#按照公式计算,eta>=0,但是没有按照公式计算就是eta<=0,所以出现eta>=0肯定错误alphas[j] -= labelMat[j]*(Ei - Ej)/eta#alpha2的更新公式alphas[j] = clipAlpha(alphas[j],H,L)if (abs(alphas[j] - alphaJold) < 0.00001): print ("j not moving enough"); continue#变化足够小,就不用变化了,迅速增加迭代次数,以退出程序alphas[i] += labelMat[j]*labelMat[i]*(alphaJold - alphas[j])
#update i by the same amount as jb1 = b - Ei- labelMat[i]*(alphas[i]-alphaIold)*dataMatrix[i,:]*dataMatrix[i,:].T - labelMat[j]*(alphas[j]-alphaJold)*dataMatrix[i,:]*dataMatrix[j,:].Tb2 = b - Ej- labelMat[i]*(alphas[i]-alphaIold)*dataMatrix[i,:]*dataMatrix[j,:].T - labelMat[j]*(alphas[j]-alphaJold)*dataMatrix[j,:]*dataMatrix[j,:].Tif (0 < alphas[i]) and (C > alphas[i]): b = b1elif (0 < alphas[j]) and (C > alphas[j]): b = b2else: b = (b1 + b2)/2.0#b的选择我在前面也有详细介绍alphaPairsChanged += 1print ("iter: %d i:%d, pairs changed %d" % (iter,i,alphaPairsChanged))if (alphaPairsChanged == 0): iter += 1else: iter = 0print ("iteration number: %d" % iter)return b,alphas

再附上完整的优化的SMO代码,其实变化不大,主要的变化在于E的更新与alpha的选择上面,上面懂了下面的也不会在话下:

class optStruct:def __init__(self,dataMatIn, classLabels, C, toler, kTup):  # Initialize the structure with the parameters self.X = dataMatInself.labelMat = classLabelsself.C = Cself.tol = tolerself.m = shape(dataMatIn)[0]self.alphas = mat(zeros((self.m,1)))self.b = 0self.eCache = mat(zeros((self.m,2))) #first column is valid flagself.K = mat(zeros((self.m,self.m)))for i in range(self.m):self.K[:,i] = kernelTrans(self.X, self.X[i,:], kTup)def calcEk(oS, k):fXk = float(multiply(oS.alphas,oS.labelMat).T*oS.K[:,k] + oS.b)Ek = fXk - float(oS.labelMat[k])return Ekdef selectJ(i, oS, Ei):         #this is the second choice -heurstic, and calcs EjmaxK = -1; maxDeltaE = 0; Ej = 0oS.eCache[i] = [1,Ei]  #set valid #choose the alpha that gives the maximum delta EvalidEcacheList = nonzero(oS.eCache[:,0].A)[0]if (len(validEcacheList)) > 1:for k in validEcacheList:   #loop through valid Ecache values and find the one that maximizes delta Eif k == i: continue #don't calc for i, waste of timeEk = calcEk(oS, k)deltaE = abs(Ei - Ek)if (deltaE > maxDeltaE):maxK = k; maxDeltaE = deltaE; Ej = Ekreturn maxK, Ejelse:   #in this case (first time around) we don't have any valid eCache valuesj = selectJrand(i, oS.m)Ej = calcEk(oS, j)return j, Ejdef updateEk(oS, k):#after any alpha has changed update the new value in the cacheEk = calcEk(oS, k)oS.eCache[k] = [1,Ek]def innerL(i, oS):Ei = calcEk(oS, i)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)):j,Ej = selectJ(i, oS, Ei) #this has been changed from selectJrandalphaIold = oS.alphas[i].copy(); alphaJold = oS.alphas[j].copy();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 0eta = 2.0 * oS.K[i,j] - oS.K[i,i] - oS.K[j,j] #changed for kernelif eta >= 0: print "eta>=0"; return 0oS.alphas[j] -= oS.labelMat[j]*(Ei - Ej)/etaoS.alphas[j] = clipAlpha(oS.alphas[j],H,L)updateEk(oS, j) #added this for the Ecacheif (abs(oS.alphas[j] - alphaJold) < 0.00001): print "j not moving enough"; return 0oS.alphas[i] += oS.labelMat[j]*oS.labelMat[i]*(alphaJold - oS.alphas[j])#update i by the same amount as jupdateEk(oS, i) #added this for the Ecache                    #the update is in the oppostie directionb1 = oS.b - Ei- oS.labelMat[i]*(oS.alphas[i]-alphaIold)*oS.K[i,i] - oS.labelMat[j]*(oS.alphas[j]-alphaJold)*oS.K[i,j]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]if (0 < oS.alphas[i]) and (oS.C > oS.alphas[i]): oS.b = b1elif (0 < oS.alphas[j]) and (oS.C > oS.alphas[j]): oS.b = b2else: oS.b = (b1 + b2)/2.0return 1else: return 0def smoP(dataMatIn, classLabels, C, toler, maxIter,kTup=('lin', 0)):    #full Platt SMOoS = optStruct(mat(dataMatIn),mat(classLabels).transpose(),C,toler, kTup)iter = 0entireSet = True; alphaPairsChanged = 0while (iter < maxIter) and ((alphaPairsChanged > 0) or (entireSet)):alphaPairsChanged = 0if entireSet:   #go over allfor i in range(oS.m):        alphaPairsChanged += innerL(i,oS)print "fullSet, iter: %d i:%d, pairs changed %d" % (iter,i,alphaPairsChanged)iter += 1else:#go over non-bound (railed) alphasnonBoundIs = nonzero((oS.alphas.A > 0) * (oS.alphas.A < C))[0]for i in nonBoundIs:alphaPairsChanged += innerL(i,oS)print "non-bound, iter: %d i:%d, pairs changed %d" % (iter,i,alphaPairsChanged)iter += 1if entireSet: entireSet = False #toggle entire set loopelif (alphaPairsChanged == 0): entireSet = True  print "iteration number: %d" % iterreturn oS.b,oS.alphas

好了,SMO的话基本就这样了,有问题的小伙伴欢迎给我留言


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