數據集以及畫圖部分代碼使用的 https://zhiyuanliplus.github.io/SVM-SMO
import numpy as np
import pandas as pd
import matplotlib.pyplot as plt
# -- coding: utf-8 --
# 沒有使用核函數
def kij(data_x):
return np.dot(data_x, data_x.T)
def gxi(index, alpha_, y, kij_, b):
return np.sum(alpha_ * y * (kij_[:, index].reshape(y.shape[0], 1))) + b
def gx(length, alpha_, y, kij_, b):
g = []
for i in range(length):
g.append(gxi(i, alpha_, y, kij_, b))
return g
def e(g_, y):
return g_ - y
# 判斷是否滿足Kkt條件，不滿足的話，求出違反的絕對誤差
def satisfy_kkt(index, alpha_, eps_, g_, y_, C_, variable_absolute_error):
val = y_[index] * g_[index]
if alpha_[index] == 0:
if val >= 1 - eps_:
return True
else:
variable_absolute_error[index] = abs(1 - eps_ - val)
return False
if 0 < alpha_[index] < C_:
if 1 - eps_ <= val <= 1 + eps_:
return True
else:
variable_absolute_error[index] = max(abs(1 - eps_ - val), abs(val - 1 - eps_))
return False
if alpha_[index] == C_:
if val <= 1 + eps_:
return True
else:
variable_absolute_error[index] = abs(val - 1 - eps)
return False
def draw(alpha, bet, data, label):
plt.xlabel(u"x1")
plt.xlim(0, 100)
plt.ylabel(u"x2")
for i in range(len(label)):
if label[i] > 0:
plt.plot(data[i][0], data[i][1], 'or')
else:
plt.plot(data[i][0], data[i][1], 'og')
w1 = 0.0
w2 = 0.0
for i in range(len(label)):
w1 += alpha[i] * label[i] * data[i][0]
w2 += alpha[i] * label[i] * data[i][1]
w = float(- w1 / w2)
b = float(- bet / w2)
r = float(1 / w2)
lp_x1 = list([10, 90])
lp_x2 = []
lp_x2up = []
lp_x2down = []
for x1 in lp_x1:
lp_x2.append(w * x1 + b)
lp_x2up.append(w * x1 + b + r)
lp_x2down.append(w * x1 + b - r)
lp_x2 = list(lp_x2)
lp_x2up = list(lp_x2up)
lp_x2down = list(lp_x2down)
plt.plot(lp_x1, lp_x2, 'b')
plt.plot(lp_x1, lp_x2up, 'b--')
plt.plot(lp_x1, lp_x2down, 'b--')
plt.show()
def smo(X, Y, C, eps, max_iter):
Kij = kij(X)
N = X.shape[0] # 有多少個樣本
# 初始值
alpha = np.zeros(len(X)).reshape(X.shape[0], 1) # 每個alpha
b = 0.0
G = np.array(gx(N, alpha_=alpha, y=Y, kij_=Kij, b=b)).reshape(N, 1)
G.reshape(N, 1)
E = e(G, Y)
visit_j = {}
visit_i = {}
loop = 0
while loop < max_iter:
# 選擇第一個變量
# 先找到所有違反KKT條件的樣本點
viable_indexes = [] # 所有可選擇的樣本
viable_indexes_alpha_less_c = [] # 所有可選擇樣本中alpha > 0 且 < C的
viable_indexes_and_absolute_error = {} # 違反KKT的數量以及違反的嚴重程度，用絕對值表示
for i in range(N):
if not satisfy_kkt(i, alpha, eps, G, Y, C, viable_indexes_and_absolute_error) and i not in visit_i:
viable_indexes.append(i)
if 0 < alpha[i] < C:
viable_indexes_alpha_less_c.append(i)
if len(viable_indexes) == 0: # 找到最優解了，退出
break
# 所有可選擇樣本中 alpha= 0 或 alpha = C的
viable_indexes_extra = [index for index in viable_indexes if index not in viable_indexes_alpha_less_c]
i = -1
# 先選擇間隔邊界上的支持向量點
if len(viable_indexes_alpha_less_c) > 0:
most_obey = -1
for index in viable_indexes_alpha_less_c:
if most_obey < viable_indexes_and_absolute_error[index] and index not in visit_i:
most_obey = viable_indexes_and_absolute_error[index]
i = index
else:
most_obey = -1
for index in viable_indexes_extra:
if most_obey < viable_indexes_and_absolute_error[index] and index not in visit_i:
most_obey = viable_indexes_and_absolute_error[index]
i = index
# 到這裏以後，i肯定不爲-1
j = -1
# 選擇|E1 - Ej|最大的那個j
max_absolute_error = -1
for index in viable_indexes:
if i == index:
continue
if max_absolute_error < abs(E[i] - E[index]) and index not in visit_j:
max_absolute_error = abs(E[i] - E[index])
j = index
# 找不到j，重新選擇i
if j == -1:
visit_j.clear()
visit_i[i] = 1
continue
# 假設已經選擇到了j
alpha1_old = alpha[i].copy() # 這裏一定要用copy..因爲後面alpha[i]的值會改變，它變了，alpha1_old也隨之會變,找了好多原因
alpha2_old = alpha[j].copy()
alpha2_new_uncut = alpha2_old + Y[j] * (E[i] - E[j]) / (Kij[i][i] + Kij[j][j] - 2 * Kij[i][j])
if Y[i] != Y[j]:
L = max(0, alpha2_old - alpha1_old)
H = min(C, C + alpha2_old - alpha1_old)
else:
L = max(0, alpha2_old + alpha1_old - C)
H = min(C, alpha2_old + alpha1_old)
# 剪輯切割
if alpha2_new_uncut > H:
alpha2_new = H
elif L <= alpha2_new_uncut <= H:
alpha2_new = alpha2_new_uncut
else:
alpha2_new = L
# 變化不大，重新選擇j
if abs(alpha2_new - alpha2_old) < 0.0001:
visit_j[j] = 1
continue
alpha1_new = alpha1_old + Y[i] * Y[j] * (alpha2_old - alpha2_new)
if alpha1_new < 0:
visit_j[j] = 1
continue
# 更新值
alpha[i] = alpha1_new
alpha[j] = alpha2_new
b1_new = -E[i] - Y[i] * Kij[i][i] * (alpha1_new - alpha1_old) - Y[j] * Kij[i][j] * (alpha2_new - alpha2_old) + b
b2_new = -E[j] - Y[i] * Kij[i][j] * (alpha1_new - alpha1_old) - Y[j] * Kij[j][j] * (alpha2_new - alpha2_old) + b
if 0 < alpha1_new < C:
b = b1_new
elif 0 < alpha2_new < C:
b = b2_new
else:
b = (b1_new + b2_new) / 2
# 更新值
G = np.array(gx(N, alpha_=alpha, y=Y, kij_=Kij, b=b)).reshape(N, 1)
Y = Y.reshape(N, 1)
E = e(G, Y)
print("iter ", loop)
print("i:%d from %f to %f" % (i, float(alpha1_old), alpha1_new))
print("j:%d from %f to %f" % (j, float(alpha2_old), alpha2_new))
visit_j.clear()
visit_i.clear()
loop = loop + 1
# print(alpha, b)
return alpha, b
if __name__ == '__main__':
data = pd.read_csv("data.csv", header=None)
X = np.array(data.values[:, : -1])
Y = np.array(data.values[:, -1])
Y = Y.reshape(X.shape[0], 1)
C = 1
eps = 1e-3 # 誤差值
max_iter = 10000 # 最大迭代次數
alpha, bb = smo(X, Y, C, eps, max_iter)
print(alpha)
print(bb)
draw(alpha, bb, X, Y)
# 注意np.array (n,) 和 (n ,1)是不一樣的，（n , 1) - (n, ） = (n, n) 一定要把(n, )轉化reshape爲(n, 1)
輸出結果表明：當迭代到6587次時，所有變量的解都滿足KKT條件。
效果圖如下：
