SVM之Python实现

时间：2019-03-07 19:10:04 收藏：0 阅读：161

SVM Python实现

Python实现SVM的理论知识

SVM原始最优化问题:

\[ min_{w,b,\xi}{1\over{2}}{||w||}^2 + C\sum_{i=1}^m\xi^{(i)} \]

\[ s.t. \ \ y^{(i)}(w^{T}x^{(i)} + b), i=1,2,...,m \\xi^{(i)} \ge 0, i=1,2,...m \]

原始问题转为对偶问题

\[ min_{\alpha}{1\over{2}}\sum_{i=1}^m\sum_{j=1}^{m}\alpha^{(i)}\alpha^{(j)}y^{(i)}y^{(j)}K(x^{(i)},x^{(j)})-\sum_{i=1}^m\alpha^{(i)} \]

\[ s.t. \ \ \sum_{i=1}^m\alpha^{(i)}y^{(i)}=0 \0 \le \alpha^{(i)} \le C, i=1,2,...,m \]

对于第2点, 设\(\alpha^*=(\alpha^*_{1},\alpha^*_{2}, ..., \alpha^*_{m})\), 若存在\(\alpha^*\)的一个分量\(\alpha*_{j}, 0 \lt \alpha^*_{j} \lt C\), 则原始问题的解\(w^*,b^*\)为

\[ w^* = \sum_{i=1}^m\alpha^*_{i}y^{(i)}x^{(i)} \b^* = y^{(j)} - \sum_{i=1}^my^{(i)}\alpha^{(i)}K(x^{(i)}, x^{(j)}) \]

SMO算法中使用到的公式(公式用使用下标1和2不是指在样本中的第1个样本和第2个样本, 而是指第1个参数和第2个参数, 在编程中我们使用i和j替代, i和j是在X输入样本中的样本下标)
- 计算\(E^{(i)}\) -> 在calc_E函数中
\[ E^{(i)} = g(x^{(i)})-y^{(i)} \ 其中g(x^{(i)}) = \sum_{i=1}^m\alpha^{(i)}y^{(i)}K(x^{(i)},x^{(j)}) + b \ 所以E^{(i)} = (\sum_{i=1}^m\alpha^{(i)}y^{(i)}K(x^{(i)},x^{(j)}) + b) - y^{(i)} \ \ \ where \ \ i = 1,2 \]
- 计算\(\alpha^{new,unc}_2\) -> 在iter函数中
\[ \alpha^{new,unc}_2 = \alpha^{old}_2 + {{y_2}(E^{(i)} - E^{(j)})\over{\eta}} \ 其中\eta = K_{11} + K_{22} - 2K_{12} \ 注意: K_{11}指的是在使用核函数映射之后的输入样本中的第i行与第j行样本, \ 同理K_{22}指的是在使用核函数映射之后的输入样本中的第i行与第j行样本... \ 注意: \eta不能为小于等于0, 如果出现这种情况, 则在迭代函数中直接返回0 \]
- 裁剪\(\alpha^{new,unc}_1\) -> 在clip_alpha函数中
  - 如果\(y^{(1)}\ne y^{(2)}\)则
    \[ L=max(0,\alpha^{old}_2-\alpha^{old}_1) \ H=min(C, C + \alpha^{old}_2 - \alpha^{old}_1) \]
  - 如果\(y^{(1)}=y^{(2)}\)则
    \[ L=max(0,\alpha^{old}_2+\alpha^{old}_1-C) \ H=min(C, \alpha^{old}_2+\alpha^{old}_1) \]
  - 注意: 得到的L与H不能相同, 如果相同则直接返回0
  - 定义函数clip_alpha(alpha, L, H)
```py
def calc_alpha(alpha, L, H):
if alpha > H:
alpha = H
elif alpha < L:
alpha = L
return alpha
```
  - 计算\(\alpha^{new}_1\)
  \[ \alpha^{new}_1=\alpha^{old}_1+y^{(1)}y^{(2)}(\alpha^{old}_2-\alpha^{new}_2) \]
  - 计算出\(\alpha^{new}_1\)之后, 比较\(abs(\alpha^{new}_1-\alpha^{old}_1)\)与我们规定的精度的值(一般零点几), 如果小于精度则直接返回0
- 选择第2个\(\alpha_2\), 选择依据就是让\(abs(^{(i)}-E^{(j)})\)最大, 那个\(j\)就是我们期望的值, 从而\(\alpha_2=\alpha_j\), 定义函数select_j(Ei, i, model)
  py # model封装了整个SVM公式中的参数(C, xi)与数据(X, y) # 其中model还有一个E属性, E为一个mx2的矩阵, 初始情况下都为0, 如果E对应的alpha被访问处理过了, 就会在赋予model.E[i, :] = [1, Ei] def select_j(Ei, i, model): j = 0 max_delta_E = 0 Ej = 0 # 查找所有已经被访问过了样本 nonzeros_indice = nonzeros(model.E[:, 0])[0] if len(nonzeros_indice) > 1: # 在for循环中查找使得abs(Ei-Ej)最大的Ej和j for index in nonzeros_indice: # 选择的两个alpha不能来自同一个样本 if index == i: continue E_temp = calc_E(i, model) delta_E = abs(E_temp - Ei) if delta_E > max_delta_E: delta_E = max_delta_E j = index Ej = E_temp else: j = i while j = i: Ej = int(random.uniform(0, model.m)) return j, Ej
- \(\alpha_1\)是否违反了KKT条件
```
if (yi * Ei > toler and alphai > 0) or (yi * Ei < -toler and alphaj < 0):
    # 违反了
    pass
else:
    # 没有违反KKT, 直接返回0
    pass
```
使用SMO算法对对偶问题求解, 求出\(\alpha\), 从而得出\(w,b\), 大致思路如下
- 初始化\(\alpha\)向量为\(m\times1\), 元素为0的向量, 一开始\(\alpha^{(1)}\)的选择没有之前的依据, 所以使用从第一个\(alpha\)开始选取
- 如果选入的\(\alpha\)没有违反KKT条件则跳过, 迭代下一个\(\alpha\)
- 将选出的\(\alpha^{(1)}\)代入iter函数, 该函数的工作是根据当前\(\alpha^{(1)}\)选择出第二个\(\alpha^{(2)}\), 接着根据公式更新\(\alpha^{(2)},\alpha^{(1)}\), 公式在上面已经给出, 注意选出来的\(\alpha_2\)的在输入样本中不能与\(\alpha_1\)是同一个
- 迭代完所有\(\alpha\), 下一步就是找出满足支持向量条件的\(\alpha\), 即\(0 \le \alpha \le C\), 再将他们迭代

Python实现SVM的代码



#!/usr/bin/env python
from numpy import *


class Model(object):

    def __init__(self, X, y, C, toler, kernel_param):
        self.X = X
        self.y = y
        self.C = C
        self.toler = toler
        self.kernel_param = kernel_param
        self.m = shape(X)[0]
        self.mapped_data = mat(zeros((self.m, self.m)))
        for i in range(self.m):
            self.mapped_data[:, i] = gaussian_kernel(self.X, X[i, :], self.kernel_param)
        self.E = mat(zeros((self.m, 2)))
        self.alphas = mat(zeros((self.m, 1)))
        self.b = 0


def load_data(filename):
    X = []
    y = []
    with open(filename, 'r') as fd:
        for line in fd.readlines():
            nums = line.strip().split(',')
            X_temp = []
            for i in range(len(nums)):
                if i == len(nums) - 1:
                    y.append(float(nums[i]))
                else:
                    X_temp.append(float(nums[i]))
            X.append(X_temp)
    return mat(X), mat(y)

def gaussian_kernel(X, l, kernel_param): 
    sigma = kernel_param 
    m = shape(X)[0]
    mapped_data = mat(zeros((m, 1)))
    for i in range(m):
        mapped_data[i] = exp(-sum((X[i, :] - l).T * (X[i, :] - l) / (2 * sigma ** 2)))
    return mapped_data

def clip_alpha(L, H, alpha):
    if alpha > H:
        alpha = H
    elif alpha < L:
        alpha = L
    return alpha

def calc_b(b1, b2):
    return (b1 + b2) / 2

def calc_E(i, model):
    yi = float(model.y[i])
    gxi = float(multiply(model.alphas, model.y).T * model.mapped_data[:, i] + model.b)
    Ei = gxi - yi
    return Ei

def select_j(Ei, i, model):
    nonzero_indices = nonzero(model.E[:, 0].A)[0]
    Ej = 0
    j = 0
    max_delta = 0
    if len(nonzero_indices) > 1:
        for index in nonzero_indices:
            if index == i:
                continue
            E_temp = calc_E(index, model)
            delta = abs(E_temp - Ei)
            if delta > max_delta:
                max_delta = delta
                Ej = E_temp
                j = index
    else:
        j = i
        while j == i:
            j = int(random.uniform(0, model.m))
        Ej = calc_E(j, model)
    return j, Ej

def iterate(i, model):
    yi = model.y[i]
    Ei = calc_E(i, model)
    model.E[i] = [1, Ei]
    # 如果alpahi不满足KKT条件, 则进行之后的操作, 选择alphaj, 更新alphai与alphaj, 还有b
    if (yi * Ei > model.toler and model.alphas[i] > 0) or (yi * Ei < -model.toler and model.alphas[i] < model.C):
        # alphai不满足KKT条件
        # 选择alphaj
        j, Ej = select_j(Ei, i, model)
        yj = model.y[j] 
        alpha1old = model.alphas[i].copy()
        alpha2old = model.alphas[j].copy()
        eta = model.mapped_data[i, i] + model.mapped_data[j, j] - 2 * model.mapped_data[i, j]   
        if eta <= 0:
            return 0
        alpha2new_unclip = alpha2old + yj * (Ei - Ej) / eta
        if yi == yj:
            L = max(0, alpha2old + alpha1old - model.C)
            H = min(model.C, alpha1old + alpha2old)
        else:
            L = max(0, alpha2old - alpha1old)
            H = min(model.C, model.C - alpha1old + alpha2old)
        if L == H:
            return 0
        alpha2new = clip_alpha(L, H, alpha2new_unclip)
        if abs(alpha2new - alpha2old) < 0.00001:
           return 0
        alpha1new = alpha1old + yi * yj * (alpha2old - alpha2new)
        b1new = -Ei - yi * model.mapped_data[i, i] * (alpha1new - alpha1old)                 - yj * model.mapped_data[j, i] * (alpha2new - alpha2old) + model.b
        b2new = -Ej - yi * model.mapped_data[i, j] * (alpha1new - alpha1old)                 - yj * model.mapped_data[j, j] * (alpha2new - alpha2old) + model.b
        model.b = calc_b(b1new, b2new)
        model.alphas[i] = alpha1new
        model.alphas[j] = alpha2new
        model.E[i] = [1, calc_E(i, model)]
        model.E[j] = [1, calc_E(j, model)]
        return 1
    return 0

def smo(X, y, C, toler, iter_num, kernel_param):
    model = Model(X, y.T, C, toler, kernel_param)
    changed_alphas = 0
    current_iter = 0
    for i in range(model.m):
        changed_alphas += iterate(i, model)
        print("iter:%d i:%d,pairs changed %d"
              %(current_iter, i, changed_alphas))
    current_iter += 1
    print('start...') 
    while current_iter < iter_num and changed_alphas > 0:
        changed_alphas = 0
        # 处理支持向量
        alphas_indice = nonzero((model.alphas.A > 0) * (model.alphas.A < C))[0]
        for i in alphas_indice:
            changed_alphas += iterate(i, model)
            print("iter:%d i:%d,pairs changed %d"
                  %(current_iter, i, changed_alphas))
        current_iter += 1
    return model.alphas, model.b

注意: 在测试SVM的时候, 我们也需要把测试集通过核函数转为m个特征的输入, 所以我们需要将训练集和测试集代入核函数中

原文：https://www.cnblogs.com/megachen/p/10491461.html