classoptStruct: def__init__(self,dataMatIn, classLabels, C, toler, kTup):# Initialize the structure with the parameters self.X = dataMatIn self.labelMat = classLabels self.C = C self.tol = toler self.m = np.shape(dataMatIn)[0] self.alphas = np.mat(np.zeros((self.m,1))) self.b = 0 # 误差缓存 self.eCache = np.mat(np.zeros((self.m,2))) #first column is valid flag self.K = np.mat(np.zeros((self.m,self.m))) for i in range(self.m): self.K[:,i] = kernelTrans(self.X, self.X[i,:], kTup) defcalcEk(oS, k): """ 计算E值并返回 """ fXk = float(np.multiply(oS.alphas,oS.labelMat).T*oS.K[:,k] + oS.b) Ek = fXk - float(oS.labelMat[k]) return Ek defselectJ(i, oS, Ei):#this is the second choice -heurstic, and calcs Ej """ 用于选择第二个alpha或者说内循环的alpha值 """ # 内循环中的启发式方法 maxK = -1; maxDeltaE = 0; Ej = 0 # eCache的第一列给出的是eCache是否有效的标志位,而第二列给出的是实际的E值 oS.eCache[i] = [1,Ei] #set valid #choose the alpha that gives the maximum delta E validEcacheList = np.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 E if k == i: continue#don't calc for i, waste of time Ek = calcEk(oS, k) deltaE = abs(Ei - Ek) # (以下两行)选择具有最大步长的j if (deltaE > maxDeltaE): maxK = k; maxDeltaE = deltaE; Ej = Ek return maxK, Ej else: #in this case (first time around) we don't have any valid eCache values j = selectJrand(i, oS.m) Ej = calcEk(oS, j) return j, Ej
defupdateEk(oS, k):#after any alpha has changed update the new value in the cache Ek = calcEk(oS, k) oS.eCache[k] = [1,Ek]
classoptStruct: def__init__(self,dataMatIn, classLabels, C, toler, kTup):# Initialize the structure with the parameters self.X = dataMatIn self.labelMat = classLabels self.C = C self.tol = toler self.m = np.shape(dataMatIn)[0] self.alphas = np.mat(np.zeros((self.m,1))) self.b = 0 # 误差缓存 self.eCache = np.mat(np.zeros((self.m,2))) #first column is valid flag self.K = np.mat(np.zeros((self.m,self.m))) for i in range(self.m): self.K[:,i] = kernelTrans(self.X, self.X[i,:], kTup) defcalcEk(oS, k): """ 计算E值并返回 """ fXk = float(np.multiply(oS.alphas,oS.labelMat).T*oS.K[:,k] + oS.b) Ek = fXk - float(oS.labelMat[k]) return Ek defselectJ(i, oS, Ei):#this is the second choice -heurstic, and calcs Ej """ 用于选择第二个alpha或者说内循环的alpha值 """ # 内循环中的启发式方法 maxK = -1; maxDeltaE = 0; Ej = 0 # eCache的第一列给出的是eCache是否有效的标志位,而第二列给出的是实际的E值 oS.eCache[i] = [1,Ei] #set valid #choose the alpha that gives the maximum delta E validEcacheList = np.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 E if k == i: continue#don't calc for i, waste of time Ek = calcEk(oS, k) deltaE = abs(Ei - Ek) # (以下两行)选择具有最大步长的j if (deltaE > maxDeltaE): maxK = k; maxDeltaE = deltaE; Ej = Ek return maxK, Ej else: #in this case (first time around) we don't have any valid eCache values j = selectJrand(i, oS.m) Ej = calcEk(oS, j) return j, Ej
defupdateEk(oS, k):#after any alpha has changed update the new value in the cache Ek = calcEk(oS, k) oS.eCache[k] = [1,Ek] print('完成')