Plot hyperplane Linear SVM python

Question

I am trying to plot the hyperplane for the model I trained with LinearSVC and sklearn. Note that I am working with natural languages; before fitting the model I extracted features with CountVectorizer and TfidfTransformer.

Here the classifier:

from sklearn.svm import LinearSVC
from sklearn import svm

clf = LinearSVC(C=0.2).fit(X_train_tf, y_train)

Then I tried to plot as suggested on the Scikit-learn website:

# get the separating hyperplane
w = clf.coef_[0]
a = -w[0] / w[1]
xx = np.linspace(-5, 5)
yy = a * xx - (clf.intercept_[0]) / w[1]

# plot the parallels to the separating hyperplane that pass through the
# support vectors
b = clf.support_vectors_[0]
yy_down = a * xx + (b[1] - a * b[0])
b = clf.support_vectors_[-1]
yy_up = a * xx + (b[1] - a * b[0])

# plot the line, the points, and the nearest vectors to the plane
plt.plot(xx, yy, 'k-')
plt.plot(xx, yy_down, 'k--')
plt.plot(xx, yy_up, 'k--')

plt.scatter(clf.support_vectors_[:, 0], clf.support_vectors_[:, 1],
            s=80, facecolors='none')
plt.scatter(X[:, 0], X[:, 1], c=Y, cmap=plt.cm.Paired)

plt.axis('tight')
plt.show()

This example uses svm.SVC(kernel='linear'), while my classifier is LinearSVC. Therefore, I get this error:

AttributeError                            Traceback (most recent call last)
<ipython-input-39-6e231c530d87> in <module>()
      7 # plot the parallels to the separating hyperplane that pass through the
      8 # support vectors
----> 9 b = clf.support_vectors_[0]
     1 yy_down = a * xx + (b[1] - a * b[0])
     11 b = clf.support_vectors_[-1]

AttributeError: 'LinearSVC' object has no attribute 'support_vectors_'

How can I successfully plot the hyperplan of my LinearSVC classifier?

I know that in that example they use another kernel for linear SVM, but I would like to plot my classifier — Alex
– Alex, Commented Oct 1, 2017 at 9:35
Well the question reads like you are looking for a the reason of the error, but as the documentation clearly states: "Note that LinearSVC does not accept keyword kernel, as this is assumed to be linear. It also lacks some of the members of SVC and NuSVC, like support_." I don't know much about the usage of LinearSVC, but I guess you could rephrase the question to make it clear that you want a similar output as SVC, but use LinearSVC instead, which does not have a 'support_vectors_' attribute. This way people don't need to dig deep to find the obvious. — ImportanceOfBeingErnest
– ImportanceOfBeingErnest, Commented Oct 1, 2017 at 9:44

ImportanceOfBeingErnest · Accepted Answer · 2017-10-01 11:16:08Z

12

What about leaving the support_ out, which is not defined for a LinearSVC?

import numpy as np
import matplotlib.pyplot as plt
from sklearn import svm

np.random.seed(0)
X = np.r_[np.random.randn(20, 2) - [2, 2], np.random.randn(20, 2) + [2, 2]]
Y = [0] * 20 + [1] * 20

fig, ax = plt.subplots()
clf2 = svm.LinearSVC(C=1).fit(X, Y)

# get the separating hyperplane
w = clf2.coef_[0]
a = -w[0] / w[1]
xx = np.linspace(-5, 5)
yy = a * xx - (clf2.intercept_[0]) / w[1]

# 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 = clf2.predict(np.c_[xx2.ravel(), yy2.ravel()])

Z = Z.reshape(xx2.shape)
ax.contourf(xx2, yy2, Z, cmap=plt.cm.coolwarm, alpha=0.3)
ax.scatter(X[:, 0], X[:, 1], c=Y, cmap=plt.cm.coolwarm, s=25)
ax.plot(xx,yy)

ax.axis([x_min, x_max,y_min, y_max])
plt.show()

answered Oct 1, 2017 at 11:16

ImportanceOfBeingErnest

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6 Comments

Alex Over a year ago

It fails at this line: ----> 8 x_min, x_max = X[:, 0].min() - 1, X[:, 0].max() + 1 Error: TypeError: unhashable type: 'slice'

Alex Over a year ago

Yes, I am running the exact code. I use python 3.6, sklearn 0.19, numpy 1.13

Alex Over a year ago

I tried to restart the kernel and now it works. But now how can I plot the hyperplane of the model I have trained?

ImportanceOfBeingErnest Over a year ago

What exactly do you mean by hyperplane? Isn't that what's shown in the answer?

Alex Over a year ago

Yes it's exactly that. But in this graph it shows randomly generated points. I want to plot those which my model has classified

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