svm+python实现

#利用svm求解逻辑回归，然后画图（详细版）
import pandas as pd
import numpy as np
import matplotlib.pyplot as plt

#读取数据
traindata=pd.read_csv("E:\\caffe\\study\\3_data.csv")
traindata=traindata.iloc[0:100,:]

#准备函数
def ui(No,traindata,alpha,b):#求ui
    a=0
    for i in range(len(traindata)):
        a=a+alpha[i]*traindata.iloc[i,2]*(traindata.iloc[i,0]*traindata.iloc[No,0]+traindata.iloc[i,1]*traindata.iloc[No,0])
    a=a+b
    return a
def Ei(No,traindata,alpha,b):#求Ei=ui-yi
    a=ui(No,traindata,alpha,b)-traindata.iloc[No,2]
    return a
def alpha2(i,traindata,alpha,b):#找第二个alpha2在alpha向量中的位置，通过max|Ei-Ej|
    ei=Ei(i,traindata,alpha,b)
    a=0
    c=0
    for j in range(len(traindata)):
        ej=Ei(j,traindata,alpha,b)
        bi=abs(ei-ej)
        if bi>a:
            a=bi
            c=j
    return c
def eta(traindata,i,j):#求分母eta
    a=traindata.iloc[i,0]**2+traindata.iloc[i,1]**2+traindata.iloc[j,0]**2+traindata.iloc[j,1]**2-2*(traindata.iloc[i,0]*traindata.iloc[j,0]+traindata.iloc[i,1]*traindata.iloc[j,1])
    return a
def alpha2new(traindata, i,j,alpha,b):#求alpha2new，这里直接做约束
    a=alpha[j]+traindata.iloc[j,2]*(Ei(i,traindata,alpha,b)-Ei(j,traindata,alpha,b))/eta(traindata,i,j)
    if traindata.iloc[i,2]==traindata.iloc[j,2]:
        L=0
        H=alpha[i]+alpha[j]
        if a>H:
            return H
        elif a<L:
            return L
        else:
            return a
    else:
        L=np.array([0,alpha[j]-alpha[i]]).max()
        if a<L:
            return L
        else:
            return a
def alpha1new(traindata, i,j,alpha,b):#把alpha2new带进去求alpha1new
    a=alpha[i]+traindata.iloc[i,2]*traindata.iloc[j,2]*(alpha[j]-alpha2new(traindata, i,j,alpha,b))
    return a
def bnew(traindata, i,j,alpha,b):#更新b
    ei=Ei(i,traindata,alpha,b)
    ej=Ei(j,traindata,alpha,b)
    yi=traindata.iloc[i,2]
    yj=traindata.iloc[j,2]
    alphai=alpha1new(traindata, i, j, alpha, b)
    alphaj=alpha2new(traindata, i,j,alpha,b)
    b1=b-ei-yi*(alphai-alpha[i])*(traindata.iloc[i,0]**2+traindata.iloc[i,1]**2)-yj*(alphaj-alpha[j])*(traindata.iloc[i,0]*traindata.iloc[j,0]+traindata.iloc[i,1]*traindata.iloc[j,1])
    b2=b-ej-yi*(alphai-alpha[i])*(traindata.iloc[i,0]*traindata.iloc[j,0]+traindata.iloc[i,1]*traindata.iloc[j,1])-yj*(alphaj-alpha[j])*(traindata.iloc[j,0]**2+traindata.iloc[j,1]**2)
    if alphai>0:
        return b1
    elif alphaj>0:
        return b2
    else:
        return (b1+b2)/2

#上面的所有函数只需要加载一下即可，重点在下面的实际运行

alpha = np.zeros(len(traindata))  # 初始化alpha向量，零向量，长度为数据的长度
alphav = alpha.copy()#alphav相当于经过一次更新后的alpha向量，目的在于下面运行的时候会和更新前的alpha做比较
b = 0  # 初始化b为0
for i in range(len(alpha)):  # 对所有违反kkt的alpha做更新，最后输出alpha为更新后的alpha，中间的b也是更新好的
    if (alpha[i]==0 and traindata.iloc[i,2]*ui(i,traindata,alpha,b)>=1)+(alpha[i]>0 and traindata.iloc[i,2]*ui(i,traindata,alpha,b)<=1)==0:
        j = alpha2(i, traindata, alpha, b)
        t = alpha2new(traindata, i, j, alpha, b)
        if t >= 0:
            alphav[j] = t
            alphav[i] = alpha1new(traindata, i, j, alpha, b)
            if alphav[i]!=alpha[i]:
                b = bnew(traindata, i, j, alpha, b)
    alpha = alphav
#下面是用来临时检验alpha和b的结果是否正确，方法是计算出w，初始化一组x，将w，b带入计算y，看y的结果是否合理，可以和原始数据作图对比
    w=np.zeros(2)#初始化w
    for i in range(len(alpha)):
        w=w+alpha[i]*traindata.iloc[i,2]*np.array(traindata.iloc[i,:2])
    w=list(w)
    x=[1,10]#初始化x
    y=[((-1)*b-w[0]*plot_x[0])/w[1],(b-w[0]*plot_x[1])/w[1]]#计算y