python - sklearn (scikit-learn) 逻辑回归包——为分类设置训练系数。

Question

所以我阅读了 scikit-learn 包 webpate:

http://scikit-learn.sourceforge.net/dev/modules/generated/sklearn.linear_model.LogisticRegression.html

我可以使用逻辑回归来拟合数据，在获得 LogisticRegression 实例后，我可以使用它对新数据点进行分类。到目前为止，一切都很好。

有没有办法设置 LogisticRegression() 实例的系数？因为我得到训练好的系数后，我想用同样的API对新的数据点进行分类。

或者也许其他人推荐了另一个具有更好 API 的 python 机器学习包？

谢谢

Answer 1

系数是估算器对象的属性——您在实例化 Logistic 回归类时创建的——因此您可以通过正常的 Python 方式访问它们:

>>> import numpy as NP
>>> from sklearn import datasets
>>> from sklearn import datasets as DS
>>> digits = DS.load_digits()
>>> D = digits.data
>>> T = digits.target

>>> # instantiate an estimator instance (classifier) of the Logistic Reg class
>>> clf = LR()
>>> # train the classifier
>>> clf.fit( D[:-1], T[:-1] )
    LogisticRegression(C=1.0, dual=False, fit_intercept=True, 
      intercept_scaling=1, penalty='l2', tol=0.0001)

>>> # attributes are accessed in the normal python way
>>> dx = clf.__dict__
>>> dx.keys()
    ['loss', 'C', 'dual', 'fit_intercept', 'class_weight_label', 'label_', 
     'penalty', 'multi_class', 'raw_coef_', 'tol', 'class_weight', 
     'intercept_scaling']

这就是获取系数的方法，但如果您只想使用这些系数进行预测，更直接的方法是使用估算器的预测 方法:

>>> # instantiate the L/R classifier, passing in norm used for penalty term 
>>> # and regularization strength
>>> clf = LR(C=.2, penalty='l1')
>>> clf
    LogisticRegression(C=0.2, dual=False, fit_intercept=True, 
      intercept_scaling=1, penalty='l1', tol=0.0001)

>>> # select some "training" instances from the original data
>>> # [of course the model should not have been trained on these instances]
>>> test = NP.random.randint(0, 151, 5)
>>> d = D[test,:]     # random selected data points w/o class labels
>>> t = T[test,:]     # the class labels that correspond to the points in d

>>> # generate model predictions for these 5 data points
>>> v = clf.predict(d)
>>> v
    array([0, 0, 2, 0, 2], dtype=int32)
>>> # how well did the model do?
>>> percent_correct = 100*NP.sum(t==v)/t.shape[0]
>>> percent_correct
    100