2-6 感知机 - 对偶形式 - 梯度下降法的算法过程

Posted 2022-03-12 windmissing

输入：
训练数据集$T=(x_1,y_1),(x_1,y_1),\cdots ,(x_n,y_n)$，其中
$$\begin{gathered} x_i\in R^n \\ {y}_i\in y=-1,+1, \\ i=1,2,\cdots ,n \end{gathered}$$
学习率为$\eta$
输出：
a, b
感知机模型$f(x)=sign({\sum }_{j=1}^m{a}_j{y}_j{x}_j\cdot x+b)$
过程：

1. 选取初值a_0 = 0, b_0 = 0
2. 在训练集中选取数据$(x_i,y_i)$
3. 如果$y_i({\sum }_{j=1}^m{a}_j{y}_j{x}_j\cdot x_i+b)\le 0$，则
   $$\begin{array}{l}{a}_i\leftarrow a_i+\eta \\ b\leftarrow b+\eta y_i\end{array}$$
4. 转至2，直至训练集中没有错误分类点

对偶形式中训练数据仅以内积的形式出现。为了方便，可以预先将训练集中的特征向量的内积计算出来，并以矩阵的形式存储。这个矩阵就所谓的Gram矩阵（Gram matrix）
$$G=[x_i\cdot x_j{]}_{m\times m}$$

代码：
https://github.com/windmissing/LiHang-TongJiXueXiFangFa/blob/master/Chapter2/perceptron-2.ipynb

def calcGramMaxtrix(X):
    m = X.shape[0]
    gram = np.zeros((m, m))
    for i in range(m):
        for j in range(i, m):
            gram[i, j] = X[i].dot(X[j])
            gram[j, i] = X[i].dot(X[j])
    return gram

def calcI(X, y, a, b, i, gram):
    #print (X.shape, y.shape, a. shape)
    sum = 0
    for j in range(X.shape[0]):
        sum += a[j] *y[j] * gram[j, i]
    return (sum + b)*y[i]
        
# 感知机原始形式
def perceptron(X, y, eta):
    a, b = np.zeros(X.shape[0]),0
    gram = calcGramMaxtrix(X)
    isFinished = False
    while not isFinished:
        isFinished = True
        for i in range(X.shape[0]):
            if calcI(X, y, a, b, i, gram) <= 0:
                isFinished = False
                a[i] += eta
                b += eta * y[i]
    def f(x):
        sum = 0
        for j in range(X.shape[0]):
            sum += a[j] *y[j] * X[j].dot(x)
        return sum + b
    return a, b, f