python - 如何为 SVM One-Versus-All 绘制超平面?

标签 python machine-learning scikit-learn svm multiclass-classification

当 SVM-OVA 执行如下时,我试图绘制超平面:

import matplotlib.pyplot as plt
import numpy as np
from sklearn.svm import SVC
x = np.array([[1,1.1],[1,2],[2,1]])
y = np.array([0,100,250])
classifier = OneVsRestClassifier(SVC(kernel='linear'))
基于对这个问题的回答 Plot hyperplane Linear SVM python ,我写了以下代码:
fig, ax = plt.subplots()
# create a mesh to plot in
x_min, x_max = x[:, 0].min() - 1, x[:, 0].max() + 1
y_min, y_max = x[:, 1].min() - 1, x[:, 1].max() + 1
xx2, yy2 = np.meshgrid(np.arange(x_min, x_max, .2),np.arange(y_min, y_max, .2))
Z = classifier.predict(np.c_[xx2.ravel(), yy2.ravel()])
Z = Z.reshape(xx2.shape)
ax.contourf(xx2, yy2, Z, cmap=plt.cm.winter, alpha=0.3)
ax.scatter(x[:, 0], x[:, 1], c=y, cmap=plt.cm.winter, s=25)

# First line: class1 vs (class2 U class3)
w = classifier.coef_[0]
a = -w[0] / w[1]
xx = np.linspace(-5, 5)
yy = a * xx - (classifier.intercept_[0]) / w[1]
ax.plot(xx,yy)

# Second line: class2 vs (class1 U class3)
w = classifier.coef_[1]
a = -w[0] / w[1]
xx = np.linspace(-5, 5)
yy = a * xx - (classifier.intercept_[1]) / w[1]
ax.plot(xx,yy)

# Third line: class 3 vs (class2 U class1)
w = classifier.coef_[2]
a = -w[0] / w[1]
xx = np.linspace(-5, 5)
yy = a * xx - (classifier.intercept_[2]) / w[1]
ax.plot(xx,yy)
然而,这是我得到的:
enter image description here
线条显然是错误的:实际上,角度系数似乎是正确的,但截距却不是。特别是,橙色线在向下平移 0.5 倍时正确,绿色线向左平移 0.5 倍,蓝色线向上平移 1.5 倍。
我画线是错误的,还是分类器由于训练点很少而无法正常工作?

最佳答案

问题是 C SVC 的参数太小(默认情况下 1.0 )。根据 this post ,

Conversely, a very small value of C will cause the optimizer to look for a larger-margin separating hyperplane, even if that hyperplane misclassifies more points.


因此,解决方案是使用更大的 C ,例如 1e5
import matplotlib.pyplot as plt
import numpy as np
from sklearn.svm import SVC
from sklearn.multiclass import OneVsRestClassifier


x = np.array([[1,1.1],[1,2],[2,1]])
y = np.array([0,100,250])
classifier = OneVsRestClassifier(SVC(C=1e5,kernel='linear'))
classifier.fit(x,y)

fig, ax = plt.subplots()
# create a mesh to plot in
x_min, x_max = x[:, 0].min() - 1, x[:, 0].max() + 1
y_min, y_max = x[:, 1].min() - 1, x[:, 1].max() + 1
xx2, yy2 = np.meshgrid(np.arange(x_min, x_max, .2),np.arange(y_min, y_max, .2))
Z = classifier.predict(np.c_[xx2.ravel(), yy2.ravel()])
Z = Z.reshape(xx2.shape)
ax.contourf(xx2, yy2, Z, cmap=plt.cm.winter, alpha=0.3)
ax.scatter(x[:, 0], x[:, 1], c=y, cmap=plt.cm.winter, s=25)

def reconstruct(w,b):

    k = - w[0] / w[1]
    b = - b[0] / w[1]

    if k >= 0:
        x0 = max((y_min-b)/k,x_min)
        x1 = min((y_max-b)/k,x_max)
    else:
        x0 = max((y_max-b)/k,x_min)
        x1 = min((y_min-b)/k,x_max)
    if np.abs(x0) == np.inf: x0 = x_min
    if np.abs(x1) == np.inf: x1 = x_max
    
    xx = np.linspace(x0,x1)
    yy = k*xx+b

    return xx,yy

xx,yy = reconstruct(classifier.coef_[0],classifier.intercept_[0])
ax.plot(xx,yy,'r')
xx,yy = reconstruct(classifier.coef_[1],classifier.intercept_[1])
ax.plot(xx,yy,'g')
xx,yy = reconstruct(classifier.coef_[2],classifier.intercept_[2])
ax.plot(xx,yy,'b')
这一次,因为更大的C被采用,结果看起来更好
output

关于python - 如何为 SVM One-Versus-All 绘制超平面?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/65020970/

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