在使用scikit学习最有用的功能时遇到问题吗? [英] Problems obtaining most informative features with scikit learn?

问题描述

我正在尝试从文本语料库中获取最多信息.从这个得到很好回答的问题我知道可以按以下步骤完成此任务:

Im triying to obtain the most informative features from a textual corpus. From this well answered question I know that this task could be done as follows:

def most_informative_feature_for_class(vectorizer, classifier, classlabel, n=10):
    labelid = list(classifier.classes_).index(classlabel)
    feature_names = vectorizer.get_feature_names()
    topn = sorted(zip(classifier.coef_[labelid], feature_names))[-n:]

    for coef, feat in topn:
        print classlabel, feat, coef

然后:

most_informative_feature_for_class(tfidf_vect, clf, 5)

对于此类同学:

X = tfidf_vect.fit_transform(df['content'].values)
y = df['label'].values


from sklearn import cross_validation
X_train, X_test, y_train, y_test = cross_validation.train_test_split(X,
                                                    y, test_size=0.33)
clf = SVC(kernel='linear', C=1)
clf.fit(X, y)
prediction = clf.predict(X_test)

问题是most_informative_feature_for_class的输出:

5 a_base_de_bien bastante   (0, 2451)   -0.210683496368
  (0, 3533) -0.173621065386
  (0, 8034) -0.135543062425
  (0, 10346)    -0.173621065386
  (0, 15231)    -0.154148294738
  (0, 18261)    -0.158890483047
  (0, 21083)    -0.297476572586
  (0, 434)  -0.0596263855375
  (0, 446)  -0.0753492277856
  (0, 769)  -0.0753492277856
  (0, 1118) -0.0753492277856
  (0, 1439) -0.0753492277856
  (0, 1605) -0.0753492277856
  (0, 1755) -0.0637950312345
  (0, 3504) -0.0753492277856
  (0, 3511) -0.115802483001
  (0, 4382) -0.0668983049212
  (0, 5247) -0.315713152154
  (0, 5396) -0.0753492277856
  (0, 5753) -0.0716096348446
  (0, 6507) -0.130661516772
  (0, 7978) -0.0753492277856
  (0, 8296) -0.144739048504
  (0, 8740) -0.0753492277856
  (0, 8906) -0.0753492277856
  : :
  (0, 23282)    0.418623443832
  (0, 4100) 0.385906085143
  (0, 15735)    0.207958503155
  (0, 16620)    0.385906085143
  (0, 19974)    0.0936828782325
  (0, 20304)    0.385906085143
  (0, 21721)    0.385906085143
  (0, 22308)    0.301270427482
  (0, 14903)    0.314164150621
  (0, 16904)    0.0653764031957
  (0, 20805)    0.0597723455204
  (0, 21878)    0.403750815828
  (0, 22582)    0.0226150073272
  (0, 6532) 0.525138162099
  (0, 6670) 0.525138162099
  (0, 10341)    0.525138162099
  (0, 13627)    0.278332617058
  (0, 1600) 0.326774799211
  (0, 2074) 0.310556919237
  (0, 5262) 0.176400451433
  (0, 6373) 0.290124806858
  (0, 8593) 0.290124806858
  (0, 12002)    0.282832270298
  (0, 15008)    0.290124806858
  (0, 19207)    0.326774799211

它不返回标签或单词.为什么会发生这种情况，如何打印单词和标签?你们是因为我正在使用熊猫读取数据而发生这种情况吗?我尝试的另一件事是以下内容，形成此问题:

It is not returning the label nor the words. Why this is happening and how can I print the words and the labels?. Do you guys this is happening since I am using pandas to read the data?. Another thing I tried is the following, form this question:

def print_top10(vectorizer, clf, class_labels):
    """Prints features with the highest coefficient values, per class"""
    feature_names = vectorizer.get_feature_names()
    for i, class_label in enumerate(class_labels):
        top10 = np.argsort(clf.coef_[i])[-10:]
        print("%s: %s" % (class_label,
              " ".join(feature_names[j] for j in top10)))


print_top10(tfidf_vect,clf,y)

但是我得到了这个追溯:

But I get this traceback:

回溯(最近通话最近一次):

Traceback (most recent call last):

  File "/Users/user/PycharmProjects/TESIS_FINAL/Classification/Supervised_learning/Final/experimentos/RBF/SVM_con_rbf.py", line 237, in <module>
    print_top10(tfidf_vect,clf,5)
  File "/Users/user/PycharmProjects/TESIS_FINAL/Classification/Supervised_learning/Final/experimentos/RBF/SVM_con_rbf.py", line 231, in print_top10
    for i, class_label in enumerate(class_labels):
TypeError: 'int' object is not iterable

关于如何解决这个问题的任何想法，以便获得具有最高系数值的特征?.

Any idea of how to solve this, in order to get the features with the highest coefficient values?.

推荐答案

要专门针对线性SVM解决此问题，我们首先必须了解sklearn中SVM的公式及其与MultinomialNB的区别.

To solve this specifically for linear SVM, we first have to understand the formulation of the SVM in sklearn and the differences that it has to MultinomialNB.

most_informative_feature_for_class用于MultinomialNB的原因是因为coef_的输出本质上是给定类的要素的对数概率(因此，由于天真的公式，其大小为[nclass, n_features])贝叶斯问题.但是，如果我们查看文档对于SVM，coef_并不是那么简单，相反，(线性)SVM的coef_是[n_classes * (n_classes -1)/2, n_features]，因为每个二进制模型都适合每个可能的类.

The reason why the most_informative_feature_for_class works for MultinomialNB is because the output of the coef_ is essentially the log probability of features given a class (and hence would be of size [nclass, n_features], due to the formulation of the naive bayes problem. But if we check the documentation for SVM, the coef_ is not that simple. Instead coef_ for (linear) SVM is [n_classes * (n_classes -1)/2, n_features] because each of the binary models are fitted to every possible class.

如果我们确实了解我们感兴趣的特定系数，则可以将函数更改为如下所示:

If we do possess some knowledge on which particular coefficient we're interested in, we could alter the function to look like the following:

def most_informative_feature_for_class_svm(vectorizer, classifier,  classlabel, n=10):
    labelid = ?? # this is the coef we're interested in. 
    feature_names = vectorizer.get_feature_names()
    svm_coef = classifier.coef_.toarray() 
    topn = sorted(zip(svm_coef[labelid], feature_names))[-n:]

    for coef, feat in topn:
        print feat, coef

这将按预期工作，并根据您要使用的系数向量打印出标签和前n个特征.

This would work as intended and print out the labels and the top n features according to the coefficient vector that you're after.

对于获得特定类的正确输出，这取决于假设和输出目标.我建议您通读SVM文档中的多类文档，以了解您的需求.

As for getting the correct output for a particular class, that would depend on the assumptions and what you aim to output. I suggest reading through the multi-class documentation within the SVM documentation to get a feel for what you're after.

因此，使用train.txt 文件(在此

So using the train.txt file which was described in this question, we can get some kind of output, though in this situation it isn't particularly descriptive or helpful to interpret. Hopefully this helps you.

import codecs, re, time
from itertools import chain

import numpy as np

from sklearn.feature_extraction.text import CountVectorizer
from sklearn.naive_bayes import MultinomialNB

trainfile = 'train.txt'

# Vectorizing data.
train = []
word_vectorizer = CountVectorizer(analyzer='word')
trainset = word_vectorizer.fit_transform(codecs.open(trainfile,'r','utf8'))
tags = ['bs','pt','es','sr']

# Training NB
mnb = MultinomialNB()
mnb.fit(trainset, tags)

from sklearn.svm import SVC
svcc = SVC(kernel='linear', C=1)
svcc.fit(trainset, tags)

def most_informative_feature_for_class(vectorizer, classifier, classlabel, n=10):
    labelid = list(classifier.classes_).index(classlabel)
    feature_names = vectorizer.get_feature_names()
    topn = sorted(zip(classifier.coef_[labelid], feature_names))[-n:]

    for coef, feat in topn:
        print classlabel, feat, coef

def most_informative_feature_for_class_svm(vectorizer, classifier,  n=10):
    labelid = 3 # this is the coef we're interested in. 
    feature_names = vectorizer.get_feature_names()
    svm_coef = classifier.coef_.toarray() 
    topn = sorted(zip(svm_coef[labelid], feature_names))[-n:]

    for coef, feat in topn:
        print feat, coef

most_informative_feature_for_class(word_vectorizer, mnb, 'pt')
print 
most_informative_feature_for_class_svm(word_vectorizer, svcc)

输出:

pt teve -4.63472898823
pt tive -4.63472898823
pt todas -4.63472898823
pt vida -4.63472898823
pt de -4.22926388012
pt foi -4.22926388012
pt mais -4.22926388012
pt me -4.22926388012
pt as -3.94158180767
pt que -3.94158180767

no 0.0204081632653
parecer 0.0204081632653
pone 0.0204081632653
por 0.0204081632653
relación 0.0204081632653
una 0.0204081632653
visto 0.0204081632653
ya 0.0204081632653
es 0.0408163265306
lo 0.0408163265306

这篇关于在使用scikit学习最有用的功能时遇到问题吗?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！