支持向量机python实现(简易版)

支持向量机的理论这里就不介绍了,可参考《统计学习方法》这本书,书上已经把整个过程写得很详细。如果觉得看书太枯燥,花费时间多,可以参考这边博文https://blog.csdn.net/v_JULY_v/article/details/7624837,写得非常好,通俗易懂。下面,是我参考《机器学习实战》这本书上代码写得支持向量机实现代码。

from numpy import *
import numpy as np
import matplotlib.pyplot as plt

def load_set_data(file_name):
    data_mat = []
    label_mat = []
    n = 0

    fr = open(file_name)
    for line in fr.readlines():
        line_arr = line.strip().split("\t")
        data_mat.append([float(line_arr[0]), float(line_arr[1])])
        label_mat.append(float(line_arr[2]))
        n += 1

    return data_mat, label_mat, n

def select_j_rand(i, m):
    j = i
    while (j == i):
        j = int(random.uniform(0, m))

    return j

def clip_alpha(aj, H, L):
    if aj > H:
        aj = H
    if aj < L:
        aj = L

    return aj

def smo_simple(data_matin, class_labele, C, toler, max_iter):
    data_matrix = mat(data_matin)  #将输入列表转成矩阵
    label_mat = mat(class_labele).transpose()  #将训练数据转成列向量
    b = 0
    m,n = shape(data_matrix)
    alphas = mat(zeros((m, 1)))
    iter = 0
while(iter < max_iter):
        alpha_pirs_change = 0
        for i in range(m):
            fxi = float(multiply(alphas, label_mat).T * (data_matrix * (data_matrix[i, :].T))) + b
            ei = fxi - float(label_mat[i])
            #if ((label_mat[i] * ei < -toler) and (alphas[i] < C)) or \
            #    ((label_mat[i] * ei > toler) and \
            #    (alphas[i] > 0)):
            if((label_mat[i] * (fxi - 2 * b) <= 1 and alphas[i] < C)\
                or (label_mat[i] * (fxi - 2 * b) >= 1 and alphas[i] > 0)\
                or (label_mat[i] * (fxi - 2 * b) == 1 and (alphas[i] == 0 or alphas[i] == C))):
                j = select_j_rand(i, m)  #随机选择aj且i != j,相当于随机选择ai和aj
                fxj = float(multiply(alphas, label_mat).T * \
                    (data_matrix * data_matrix[j, :].T)) + b
                ej = fxj - float(label_mat[j])
                alpha_iold = alphas[i].copy()  #保存alphai更新前的值
                alpha_jold = alphas[j].copy()  #保存alphaj更新前的值
                #求解alphaj的上下边界
                if (label_mat[i] != label_mat[j]):
                    L = max(0, alpha_jold - alpha_iold)
                    H = min(C, C + alpha_jold + alpha_iold)
               else:
                    L = max(0, alpha_jold + alpha_iold - C)
                    H = min(C, alpha_iold + alpha_jold)
                if (L == H):
                    print("L == H")
                    continue
               eta = 2 * data_matrix[i, :] * data_matrix[j, :].T \
                    - data_matrix[i, :] * data_matrix[i, :].T - data_matrix[j, :] * data_matrix[j, :].T
                if (eta > 0):
                    print("eta > 0")
                    continue
                alphas[j] = alpha_jold - (label_mat[j] * (ei - ej) * 1.0 / eta)
                alphas[j] = clip_alpha(alphas[j], H, L)
                if (abs(alphas[j] - alpha_jold) < 0.00001):  
                    print("j not moving enough")
                    continue
                alphas[i] = alpha_iold + (label_mat[i] * label_mat[j] * (alpha_jold - alphas[j]))
                b1 = b - ei - label_mat[i] * (alphas[i] - alpha_iold) * (data_matrix[i, :] * data_matrix[i, :].T)\
                    - label_mat[j] * (alphas[j] - alpha_jold) * (data_matrix[i, :] * data_matrix[j, :].T)
                b2 = b - ej - label_mat[i] * (alphas[i] - alpha_iold) * (data_matrix[i, :] * data_matrix[j, :].T)\
                    - label_mat[j] * (alphas[j] - alpha_jold) * (data_matrix[j, :] * data_matrix[j, :].T) 
                if (alphas[i] > 0 and alphas[i] < C):
                    b = b1
                elif (alphas[j] > 0 and alphas[j] < C):
                    b = b2
                else:
                    b = (b1 + b2) / 2.0
                alpha_pirs_change += 1
                print("iter: %d i: %d, paris changed % d" % (iter, i, alpha_pirs_change))
        if (alpha_pirs_change == 0):
            iter += 1
        else:
            iter = 0
        print("iteration number: %d" % iter)
    return b,alphas

def show_experiment_plot(alphas, data_list_in, label_list_in, b, n):
    data_arr_in = array(data_list_in)
    label_arr_in = array(label_list_in)
    alphas_arr = alphas.getA()
    data_mat = mat(data_list_in)
    label_mat = mat(label_list_in).transpose()

    i = 0
    weights = zeros((2, 1))
    while(i < n):
        if(label_arr_in[i] == -1):
            plt.plot(data_arr_in[i, 0], data_arr_in[i, 1], "ob")
        elif(label_arr_in[i] == 1):
            plt.plot(data_arr_in[i, 0], data_arr_in[i, 1], "or")
        if(alphas_arr[i] > 0):
            plt.plot(data_arr_in[i, 0], data_arr_in[i, 1], "oy")
            weights += multiply(alphas[i] * label_mat[i], data_mat[i, :].T)
        i += 1

    x = arange(-2, 12, 0.1)
    y = []
    for k in x:
        y.append(float(-b - weights[0] * k) / weights[1])

    plt.plot(x, y, '-g')
    plt.xlabel("X")
    plt.ylabel("Y")
    plt.show()

def main():
    data_list,label_list, n = load_set_data("test_set.txt")
    b,alphas = smo_simple(data_list, label_list, 0.6, 0.001, 40)
    b_data = array(b)[0][0]
    show_experiment_plot(alphas, data_list, label_list, b_data, n)

main()

代码中被注释的部分是《机器学习实战》这本书的写法,看了很久还是没有搞明白作者为什么这么写,如果哪位大神知道,还请说一下,大家好互相学习。根据KKT条件我自己重现实现了更新alphas的条件判断。
特征到结果的输出函数:
ui = w.x – b
其中w,x和ui均为向量

《支持向量机python实现(简易版)》 2.PNG

也就是说,当没有满足KKT条件时,则需要更新alphas的值。
实验数据:

3.542485    1.977398    -1
3.018896    2.556416    -1
7.551510    -1.580030   1
2.114999    -0.004466   -1
8.127113    1.274372    1
7.108772    -0.986906   1
8.610639    2.046708    1
2.326297    0.265213    -1
3.634009    1.730537    -1
0.341367    -0.894998   -1
3.125951    0.293251    -1
2.123252    -0.783563   -1
0.887835    -2.797792   -1
7.139979    -2.329896   1
1.696414    -1.212496   -1
8.117032    0.623493    1
8.497162    -0.266649   1
4.658191    3.507396    -1
8.197181    1.545132    1
1.208047    0.213100    -1
1.928486    -0.321870   -1
2.175808    -0.014527   -1
7.886608    0.461755    1
3.223038    -0.552392   -1
3.628502    2.190585    -1
7.407860    -0.121961   1
7.286357    0.251077    1
2.301095    -0.533988   -1
-0.232542   -0.547690   -1
3.457096    -0.082216   -1
3.023938    -0.057392   -1
8.015003    0.885325    1
8.991748    0.923154    1
7.916831    -1.781735   1
7.616862    -0.217958   1
2.450939    0.744967    -1
7.270337    -2.507834   1
1.749721    -0.961902   -1
1.803111    -0.176349   -1
8.804461    3.044301    1
1.231257    -0.568573   -1
2.074915    1.410550    -1
-0.743036   -1.736103   -1
3.536555    3.964960    -1
8.410143    0.025606    1
7.382988    -0.478764   1
6.960661    -0.245353   1
8.234460    0.701868    1
8.168618    -0.903835   1
1.534187    -0.622492   -1
9.229518    2.066088    1
7.886242    0.191813    1
2.893743    -1.643468   -1
1.870457    -1.040420   -1
5.286862    -2.358286   1
6.080573    0.418886    1
2.544314    1.714165    -1
6.016004    -3.753712   1
0.926310    -0.564359   -1
0.870296    -0.109952   -1
2.369345    1.375695    -1
1.363782    -0.254082   -1
7.279460    -0.189572   1
1.896005    0.515080    -1
8.102154    -0.603875   1
2.529893    0.662657    -1
1.963874    -0.365233   -1
8.132048    0.785914    1
8.245938    0.372366    1
6.543888    0.433164    1
-0.236713   -5.766721   -1
8.112593    0.295839    1
9.803425    1.495167    1
1.497407    -0.552916   -1
1.336267    -1.632889   -1
9.205805    -0.586480   1
1.966279    -1.840439   -1
8.398012    1.584918    1
7.239953    -1.764292   1
7.556201    0.241185    1
9.015509    0.345019    1
8.266085    -0.230977   1
8.545620    2.788799    1
9.295969    1.346332    1
2.404234    0.570278    -1
2.037772    0.021919    -1
1.727631    -0.453143   -1
1.979395    -0.050773   -1
8.092288    -1.372433   1
1.667645    0.239204    -1
9.854303    1.365116    1
7.921057    -1.327587   1
8.500757    1.492372    1
1.339746    -0.291183   -1
3.107511    0.758367    -1
2.609525    0.902979    -1
3.263585    1.367898    -1
2.912122    -0.202359   -1
1.731786    0.589096    -1
2.387003    1.573131    -1

实验结果如下:

《支持向量机python实现(简易版)》 1.PNG

图中黄色的点是支持向量的(alphas[i] > 0对应的(xi, yi))。实验结果表明。该算法能正确分类,效果良好。

    原文作者:幸福洋溢的季节
    原文地址: https://www.jianshu.com/p/5ec5162fe8f3
    本文转自网络文章,转载此文章仅为分享知识,如有侵权,请联系博主进行删除。
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