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%matplotlib inline

SVM Tie Breaking Example

Tie breaking is costly if decision_function_shape='ovr', and therefore it is not enabled by default. This example illustrates the effect of the break_ties parameter for a multiclass classification problem and decision_function_shape='ovr'.

The two plots differ only in the area in the middle where the classes are tied. If break_ties=False, all input in that area would be classified as one class, whereas if break_ties=True, the tie-breaking mechanism will create a non-convex decision boundary in that area.

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# Code source: Andreas Mueller, Adrin Jalali
# License: BSD 3 clause

import numpy as np
import matplotlib.pyplot as plt
from sklearn.svm import SVC
from sklearn.datasets import make_blobs

X, y = make_blobs(random_state=27)

fig, sub = plt.subplots(2, 1, figsize=(5, 8))
titles = ("break_ties = False",
          "break_ties = True")

for break_ties, title, ax in zip((False, True), titles, sub.flatten()):

    svm = SVC(kernel="linear", C=1, break_ties=break_ties,
              decision_function_shape='ovr').fit(X, y)

    xlim = [X[:, 0].min(), X[:, 0].max()]
    ylim = [X[:, 1].min(), X[:, 1].max()]

    xs = np.linspace(xlim[0], xlim[1], 1000)
    ys = np.linspace(ylim[0], ylim[1], 1000)
    xx, yy = np.meshgrid(xs, ys)

    pred = svm.predict(np.c_[xx.ravel(), yy.ravel()])

    colors = [ for i in [0, 4, 7]]

    points = ax.scatter(X[:, 0], X[:, 1], c=y, cmap="Accent")
    classes = [(0, 1), (0, 2), (1, 2)]
    line = np.linspace(X[:, 1].min() - 5, X[:, 1].max() + 5)
    ax.imshow(-pred.reshape(xx.shape), cmap="Accent", alpha=.2,
              extent=(xlim[0], xlim[1], ylim[1], ylim[0]))

    for coef, intercept, col in zip(svm.coef_, svm.intercept_, classes):
        line2 = -(line * coef[1] + intercept) / coef[0]
        ax.plot(line2, line, "-", c=colors[col[0]])
        ax.plot(line2, line, "--", c=colors[col[1]])