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我正在尝试LinearSVC使用scikit-learn. 我试过同时使用 0.10 和 0.14 版本。使用代码:

from sklearn.svm import LinearSVC, SVC
from numpy import *

data = array([[ 1007.,  1076.],
              [ 1017.,  1009.],
              [ 2021.,  2029.],
              [ 2060.,  2085.]])
groups = array([1, 1, 2, 2])

svc = LinearSVC()
svc.fit(data, groups)
svc.predict(data)

我得到输出:

array([2, 2, 2, 2])

但是,如果我将分类器替换为

svc = SVC(kernel='linear')

然后我得到结果

array([ 1.,  1.,  2.,  2.])

哪个是对的。有谁知道为什么使用LinearSVC会破坏这个简单的问题?

4

1 回答 1

13

底层算法LinearSVC对其输入中的极值非常敏感:

>>> svc = LinearSVC(verbose=1)
>>> svc.fit(data, groups)
[LibLinear]....................................................................................................
optimization finished, #iter = 1000

WARNING: reaching max number of iterations
Using -s 2 may be faster (also see FAQ)

Objective value = -0.001256
nSV = 4
LinearSVC(C=1.0, class_weight=None, dual=True, fit_intercept=True,
     intercept_scaling=1, loss='l2', multi_class='ovr', penalty='l2',
     random_state=None, tol=0.0001, verbose=1)

(警告指的是 LibLinear FAQ,因为 scikit-learnLinearSVC是基于该库的。)

您应该在拟合之前进行标准化:

>>> from sklearn.preprocessing import scale
>>> data = scale(data)
>>> svc.fit(data, groups)
[LibLinear]...
optimization finished, #iter = 39
Objective value = -0.240988
nSV = 4
LinearSVC(C=1.0, class_weight=None, dual=True, fit_intercept=True,
     intercept_scaling=1, loss='l2', multi_class='ovr', penalty='l2',
     random_state=None, tol=0.0001, verbose=1)
>>> svc.predict(data)
array([1, 1, 2, 2])
于 2013-12-17T01:52:15.153 回答