从线性 SVM 绘制 3D 决策边界 [英] Plotting 3D Decision Boundary From Linear SVM

问题描述

我已经使用sklearn.svm.svc()拟合了3个要素数据集.我可以使用 matplotlib 和 Axes3D 为每个观察点绘制点.我想绘制决策边界以查看适合度.我尝试改编2D示例，以无济于事地绘制决策边界.我了解clf.coef_是垂直于决策边界的向量.我如何绘制它以查看它划分点的位置?

I've fit a 3 feature data set using sklearn.svm.svc(). I can plot the point for each observation using matplotlib and Axes3D. I want to plot the decision boundary to see the fit. I've tried adapting the 2D examples for plotting the decision boundary to no avail. I understand that clf.coef_ is a vector normal to the decision boundary. How can I plot this to see where it divides the points?

推荐答案

以下是玩具数据集上的一个示例.请注意，使用 matplotlib 进行 3D 绘图很时髦.有时，飞机后面的点可能看起来好像在飞机前面，因此您可能需要摆弄旋转图来确定发生了什么.

Here is an example on a toy dataset. Note that plotting in 3D is funky with matplotlib. Sometimes points that are behind the plane might appear as though they are in front of it, so you may have to fiddle with rotating the plot to ascertain what's going on.

import numpy as np
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D
from sklearn.svm import SVC

rs = np.random.RandomState(1234)

# Generate some fake data.
n_samples = 200
# X is the input features by row.
X = np.zeros((200,3))
X[:n_samples/2] = rs.multivariate_normal( np.ones(3), np.eye(3), size=n_samples/2)
X[n_samples/2:] = rs.multivariate_normal(-np.ones(3), np.eye(3), size=n_samples/2)
# Y is the class labels for each row of X.
Y = np.zeros(n_samples); Y[n_samples/2:] = 1

# Fit the data with an svm
svc = SVC(kernel='linear')
svc.fit(X,Y)

# The equation of the separating plane is given by all x in R^3 such that:
# np.dot(svc.coef_[0], x) + b = 0. We should solve for the last coordinate
# to plot the plane in terms of x and y.

z = lambda x,y: (-svc.intercept_[0]-svc.coef_[0][0]*x-svc.coef_[0][1]*y) / svc.coef_[0][2]

tmp = np.linspace(-2,2,51)
x,y = np.meshgrid(tmp,tmp)

# Plot stuff.
fig = plt.figure()
ax  = fig.add_subplot(111, projection='3d')
ax.plot_surface(x, y, z(x,y))
ax.plot3D(X[Y==0,0], X[Y==0,1], X[Y==0,2],'ob')
ax.plot3D(X[Y==1,0], X[Y==1,1], X[Y==1,2],'sr')
plt.show()

输出:

编辑(上面的注释中的关键数学线性代数语句):

EDIT (Key Mathematical Linear Algebra Statement In Comment Above):

# The equation of the separating plane is given by all x in R^3 such that:
# np.dot(coefficients, x_vector) + intercept_value = 0. 
# We should solve for the last coordinate: x_vector[2] == z
# to plot the plane in terms of x and y.

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