scikit-learn 中的 SVC 和 LinearSVC 在什么参数下是等价的? [英] Under what parameters are SVC and LinearSVC in scikit-learn equivalent?

问题描述

我阅读了

但是，如果您将其放大太多 - 它也会失败，因为现在容差和迭代次数至关重要.

总结:LinearSVC 不是线性 SVM，非必要就不要使用.

I read this thread about the difference between SVC() and LinearSVC() in scikit-learn.

Now I have a data set of binary classification problem(For such a problem, the one-to-one/one-to-rest strategy difference between both functions could be ignore.)

I want to try under what parameters would these 2 functions give me the same result. First of all, of course, we should set kernel='linear' for SVC() However, I just could not get the same result from both functions. I could not find the answer from the documents, could anybody help me to find the equivalent parameter set I am looking for?

Updated: I modified the following code from an example of the scikit-learn website, and apparently they are not the same:

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

# import some data to play with
iris = datasets.load_iris()
X = iris.data[:, :2]  # we only take the first two features. We could
                      # avoid this ugly slicing by using a two-dim dataset
y = iris.target

for i in range(len(y)):
    if (y[i]==2):
        y[i] = 1

h = .02  # step size in the mesh

# we create an instance of SVM and fit out data. We do not scale our
# data since we want to plot the support vectors
C = 1.0  # SVM regularization parameter
svc = svm.SVC(kernel='linear', C=C).fit(X, y)
lin_svc = svm.LinearSVC(C=C, dual = True, loss = 'hinge').fit(X, y)

# 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
xx, yy = np.meshgrid(np.arange(x_min, x_max, h),
                     np.arange(y_min, y_max, h))

# title for the plots
titles = ['SVC with linear kernel',
          'LinearSVC (linear kernel)']

for i, clf in enumerate((svc, lin_svc)):
    # Plot the decision boundary. For that, we will assign a color to each
    # point in the mesh [x_min, m_max]x[y_min, y_max].
    plt.subplot(1, 2, i + 1)
    plt.subplots_adjust(wspace=0.4, hspace=0.4)

    Z = clf.predict(np.c_[xx.ravel(), yy.ravel()])

    # Put the result into a color plot
    Z = Z.reshape(xx.shape)
    plt.contourf(xx, yy, Z, cmap=plt.cm.Paired, alpha=0.8)

    # Plot also the training points
    plt.scatter(X[:, 0], X[:, 1], c=y, cmap=plt.cm.Paired)
    plt.xlabel('Sepal length')
    plt.ylabel('Sepal width')
    plt.xlim(xx.min(), xx.max())
    plt.ylim(yy.min(), yy.max())
    plt.xticks(())
    plt.yticks(())
    plt.title(titles[i])

plt.show()

Result: Output Figure from previous code

解决方案

In mathematical sense you need to set:

SVC(kernel='linear', **kwargs) # by default it uses RBF kernel

and

LinearSVC(loss='hinge', **kwargs) # by default it uses squared hinge loss

Another element, which cannot be easily fixed is increasing intercept_scaling in LinearSVC, as in this implementation bias is regularized (which is not true in SVC nor should be true in SVM - thus this is not SVM) - consequently they will never be exactly equal (unless bias=0 for your problem), as they assume two different models

• SVC : 1/2||w||^2 + C SUM xi_i
• LinearSVC: 1/2||[w b]||^2 + C SUM xi_i

Personally I consider LinearSVC one of the mistakes of sklearn developers - this class is simply not a linear SVM.

After increasing intercept scaling (to 10.0)

However, if you scale it up too much - it will also fail, as now tolerance and number of iterations are crucial.

To sum up: LinearSVC is not linear SVM, do not use it if do not have to.

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