一.简介

上一节介绍了硬间隔支持向量机,它可以在严格线性可分的数据集上工作的很好,但对于非严格线性可分的情况往往就表现很差了,比如:

In [1]:
import numpy as np
import matplotlib.pyplot as plt
import copy
import random
import os
os.chdir('../')
from ml_models import utils
from ml_models.svm import HardMarginSVM
%matplotlib inline

PS:请多试几次,生成含噪声点的数据

In [14]:
from sklearn.datasets import make_classification
data, target = make_classification(n_samples=100, n_features=2, n_classes=2, n_informative=1, n_redundant=0,
                                   n_repeated=0, n_clusters_per_class=1, class_sep=2.0)
plt.scatter(data[:,0],data[:,1],c=target)
Out[14]:
<matplotlib.collections.PathCollection at 0x202a6f55a58>
In [15]:
#训练
svm = HardMarginSVM()
svm.fit(data, target)
utils.plot_decision_function(data, target, svm, svm.support_vectors)

那怕仅含有一个异常点,对硬间隔支持向量机的训练影响就很大,我们希望它能具有一定的包容能力,容忍哪些放错的点,但又不能容忍过度,我们可以引入变量$\xi$和一个超参$C$来进行控制,原始的优化问题更新为如下:

$$ \min_{w,b,\xi} \frac{1}{2}w^Tw + C\sum_{i=1}^N\xi_i\\ s.t.y_i(w^Tx_i+b)\geq 1-\xi_i,i=1,2,...,N\\ \xi_i\geq0,i=1,2,...,N $$

这里$C$若越大,包容能力就越小,当取值很大时,就等价于硬间隔支持向量机,而$\xi$使得支持向量的间隔可以调整,不必像硬间隔那样,严格等于1

Lagrange函数

关于原问题的Lagrange函数:

$$ L(w,b,\xi,\alpha,\mu)=\frac{1}{2}w^Tw+C\sum_{i=1}^N\xi_i+\sum_{i=1}^N\alpha_i(1-\xi_i-y_i(w^Tx_i+b))-\sum_{i=1}^N\mu_i\xi_i\\ s.t.\mu_i\geq 0,\alpha_i\geq0,i=1,2,...,N $$

二.对偶问题

对偶问题的求解过程我就省略了,与硬间隔类似,我这里就直接写最终结果:

$$ \min_{\alpha} \frac{1}{2}\sum_{i=1}^N\sum_{j=1}^N\alpha_i\alpha_jy_iy_jx_i^Tx_j-\sum_{i=1}^N\alpha_i\\ s.t.\sum_{i=1}^N\alpha_iy_i=0,\\ 0\leq\alpha_i\leq C,i=1,2,...,N $$


可以发现与硬间隔的不同是$\alpha$加了一个上界的约束$C$

三.KKT条件

这里就直接写KKT条件看原优化变量与拉格朗日乘子之间的关系:

$$ \frac{\partial L}{\partial w}=0\Rightarrow w^*=\sum_{i=1}^N\alpha_i^*y_ix_i(关系1)\\ \frac{\partial L}{\partial b}=0\Rightarrow \alpha_i^*y_i=0(关系2)\\ \frac{\partial L}{\partial \xi}=0\Rightarrow C-\alpha_i^*-\mu_i^*=0(关系3)\\ \alpha_i^*(1-\xi_i^*-y_i({w^*}^Tx_i+b^*))=0(关系4)\\ \mu_i^*\xi_i^*=0(关系5)\\ y_i({w^*}^Tx_i+b^*)-1-\xi_i^*\geq0(关系6)\\ \xi_i^*\geq0(关系7)\\ \alpha_i^*\geq0(关系8)\\ \mu_i^*\geq0(关系9)\\ $$

四.$w^*,b^*$的求解

由KKT条件中的关系1,我们可以知道:

$$ w^*=\sum_{i=1}^N\alpha_i^*y_ix_i $$

对于$b^*$的求解,我们可以取某点,其$0<\alpha_k^*<C$,由关系3,4,5可以推得到:${w^*}^Tx_k+b^*=y_k$,所以:

$$ b^*=y_k-{w^*}^Tx_k $$

五.SMO求$\alpha^*$

好了,最终模型得求解落到了对$\alpha^*$得求解上,求解过程与硬间隔一样,无非就是就是对$\alpha$多加了一个约束:$\alpha_i^*<=C$,具体而言需要对$\alpha_2^{new}$的求解进行更新:

当$y_1\neq y_2$时:

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

当$y_1=y_2$时:

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

更新公式:

$$ \alpha_2^{new}=\left\{\begin{matrix} H & \alpha_2^{unc}> H\\ \alpha_2^{unc} & L \leq \alpha_2^{unc} \leq H\\ L & \alpha_2^{unc}<L \end{matrix}\right. $$

六.代码实现

In [16]:
"""
软间隔支持向量机的smo实现,放到ml_models.svm模块中
"""
class SoftMarginSVM(object):
    def __init__(self, epochs=100, C=1.0):
        self.w = None
        self.b = None
        self.alpha = None
        self.E = None
        self.epochs = epochs
        self.C = C
        # 记录支持向量
        self.support_vectors = None

    def init_params(self, X, y):
        """
        :param X: (n_samples,n_features)
        :param y: (n_samples,) y_i\in\{0,1\}
        :return:
        """
        n_samples, n_features = X.shape
        self.w = np.zeros(n_features)
        self.b = .0
        self.alpha = np.zeros(n_samples)
        self.E = np.zeros(n_samples)
        # 初始化E
        for i in range(0, n_samples):
            self.E[i] = np.dot(self.w, X[i, :]) + self.b - y[i]

    def _select_j(self, best_i):
        """
        选择j
        :param best_i:
        :return:
        """
        valid_j_list = [i for i in range(0, len(self.alpha)) if self.alpha[i] > 0 and i != best_i]
        best_j = -1
        # 优先选择使得|E_i-E_j|最大的j
        if len(valid_j_list) > 0:
            max_e = 0
            for j in valid_j_list:
                current_e = np.abs(self.E[best_i] - self.E[j])
                if current_e > max_e:
                    best_j = j
                    max_e = current_e
        else:
            # 随机选择
            l = list(range(len(self.alpha)))
            seq = l[: best_i] + l[best_i + 1:]
            best_j = random.choice(seq)
        return best_j

    def _meet_kkt(self, w, b, x_i, y_i, alpha_i):
        """
        判断是否满足KKT条件

        :param w:
        :param b:
        :param x_i:
        :param y_i:
        :return:
        """
        if alpha_i < self.C:
            return y_i * (np.dot(w, x_i) + b) >= 1
        else:
            return y_i * (np.dot(w, x_i) + b) <= 1

    def fit(self, X, y2, show_train_process=False):
        """

        :param X:
        :param y2:
        :param show_train_process: 显示训练过程
        :return:
        """
        y = copy.deepcopy(y2)
        y[y == 0] = -1
        # 初始化参数
        self.init_params(X, y)
        for _ in range(0, self.epochs):
            if_all_match_kkt = True
            for i in range(0, len(self.alpha)):
                x_i = X[i, :]
                y_i = y[i]
                alpha_i_old = self.alpha[i]
                E_i_old = self.E[i]
                # 外层循环:选择违反KKT条件的点i
                if not self._meet_kkt(self.w, self.b, x_i, y_i, alpha_i_old):
                    if_all_match_kkt = False
                    # 内层循环,选择使|Ei-Ej|最大的点j
                    best_j = self._select_j(i)

                    alpha_j_old = self.alpha[best_j]
                    x_j = X[best_j, :]
                    y_j = y[best_j]
                    E_j_old = self.E[best_j]

                    # 进行更新
                    # 1.首先获取无裁剪的最优alpha_2
                    eta = np.dot(x_i - x_j, x_i - x_j)
                    # 如果x_i和x_j很接近,则跳过
                    if eta < 1e-3:
                        continue
                    alpha_j_unc = alpha_j_old + y_j * (E_i_old - E_j_old) / eta
                    # 2.裁剪并得到new alpha_2
                    if y_i == y_j:
                        L = max(0., alpha_i_old + alpha_j_old - self.C)
                        H = min(self.C, alpha_i_old + alpha_j_old)
                    else:
                        L = max(0, alpha_j_old - alpha_i_old)
                        H = min(self.C, self.C + alpha_j_old - alpha_i_old)

                    if alpha_j_unc < L:
                        alpha_j_new = L
                    elif alpha_j_unc > H:
                        alpha_j_new = H
                    else:
                        alpha_j_new = alpha_j_unc

                    # 如果变化不够大则跳过
                    if np.abs(alpha_j_new - alpha_j_old) < 1e-5:
                        continue
                    # 3.得到alpha_1_new
                    alpha_i_new = alpha_i_old + y_i * y_j * (alpha_j_old - alpha_j_new)
                    # 4.更新w
                    self.w = self.w + (alpha_i_new - alpha_i_old) * y_i * x_i + (alpha_j_new - alpha_j_old) * y_j * x_j
                    # 5.更新alpha_1,alpha_2
                    self.alpha[i] = alpha_i_new
                    self.alpha[best_j] = alpha_j_new
                    # 6.更新b
                    b_i_new = y_i - np.dot(self.w, x_i)
                    b_j_new = y_j - np.dot(self.w, x_j)
                    if self.C > alpha_i_new > 0:
                        self.b = b_i_new
                    elif self.C > alpha_j_new > 0:
                        self.b = b_j_new
                    else:
                        self.b = (b_i_new + b_j_new) / 2.0
                    # 7.更新E
                    for k in range(0, len(self.E)):
                        self.E[k] = np.dot(self.w, X[k, :]) + self.b - y[k]
                    # 显示训练过程
                    if show_train_process is True:
                        utils.plot_decision_function(X, y2, self, [i, best_j])
                        utils.plt.pause(0.1)
                        utils.plt.clf()

            # 如果所有的点都满足KKT条件,则中止
            if if_all_match_kkt is True:
                break
        # 计算支持向量
        self.support_vectors = np.where(self.alpha > 1e-3)[0]
        # 显示最终结果
        if show_train_process is True:
            utils.plot_decision_function(X, y2, self, self.support_vectors)
            utils.plt.show()

    def get_params(self):
        """
        输出原始的系数
        :return: w
        """

        return self.w, self.b

    def predict_proba(self, x):
        """
        :param x:ndarray格式数据: m x n
        :return: m x 1
        """
        return utils.sigmoid(x.dot(self.w) + self.b)

    def predict(self, x):
        """
        :param x:ndarray格式数据: m x n
        :return: m x 1
        """
        proba = self.predict_proba(x)
        return (proba >= 0.5).astype(int)
In [20]:
svm = SoftMarginSVM(C=3.0)
svm.fit(data, target)
utils.plot_decision_function(data, target, svm, svm.support_vectors)

通过控制C可以调节宽容度,设置一个大的C可以取得和硬间隔一样的效果

In [22]:
svm = SoftMarginSVM(C=1000000)
svm.fit(data, target)
utils.plot_decision_function(data, target, svm, svm.support_vectors)

有时,太过宽容也不一定好

In [23]:
svm = SoftMarginSVM(C=0.01)
svm.fit(data, target)
utils.plot_decision_function(data, target, svm, svm.support_vectors)

七.支持向量

软间隔支持向量机的支持向量复杂一些,因为对于$\alpha>0$有许多种情况,如下图所示,大概可以分为4类:

(1)$0<\alpha_i<C,\xi_i=0$:位于间隔边界上;
(2)$\alpha_i=C,0<\xi_i<1$:分类正确,位于间隔边界与分离超平面之间;
(3)$\alpha_i=C,\xi_i=1$:位于分离超平面上;
(4)$\alpha_i=C,\xi_i>1$:位于错误分类的一侧

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