Sid*_*ney 14 machine-learning svm libsvm scikit-learn
我读了这篇关于scikit-learn SVC()和LinearSVC()scikit-learn 之间差异的帖子.
现在我有一个二进制分类问题的数据集(对于这样的问题,两个函数之间的一对一/一对一策略差异可以忽略.)
我想尝试在这两个函数给出相同结果的参数下.首先,当然,我们应该设置kernel='linear'为SVC()
但是,我无法从两个函数中得到相同的结果.我无法从文档中找到答案,有人可以帮我找到我想要的等效参数集吗?
更新:我从scikit-learn网站的一个例子中修改了以下代码,显然它们不一样:
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()
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结果: 输出前一代码的图
lej*_*lot 24
在数学意义上,你需要设置:
SVC(kernel='linear', **kwargs) # by default it uses RBF kernel
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和
LinearSVC(loss='hinge', **kwargs) # by default it uses squared hinge loss
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另一元件时,其不能被容易地固定在增加intercept_scaling中LinearSVC,如在该实施方式中偏压正规化(这是不正确的在SVC也不应在SVM是真实的-因而这不是SVM) -因此它们将从未完全相等(除非偏压因为他们假设有两个不同的模型,因此你的问题是0
1/2||w||^2 + C SUM xi_i1/2||[w b]||^2 + C SUM xi_i我个人认为LinearSVC是sklearn开发者的错误之一 - 这个类不是线性SVM.
增加拦截比例后(to 10.0)
但是,如果你将它放大太多 - 它也会失败,因为现在容差和迭代次数是至关重要的.
总结一下:LinearSVC不是线性SVM,如果不必要,请不要使用它.
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