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我现在正在使用 nltk 学习 naivebayes 分类器。

在文档（http://www.nltk.org/book/ch06.html）1.3文档分类中，有一个featureset示例。

featuresets = [(document_features(d), c) for (d,c) in documents]
train_set, test_set = featuresets[100:], featuresets[:100]
classifier = nltk.NaiveBayesClassifier.train(train_set)

all_words = nltk.FreqDist(w.lower() for w in movie_reviews.words())
word_features = list(all_words)[:2000] [1]

def document_features(document): [2]
    document_words = set(document) [3]
    features = {}
    for word in word_features:
        features['contains({})'.format(word)] = (word in document_words)
    return features

所以特征集形式的例子是 {('contains(waste)': False, 'contains(lot)': False, ...},'neg')...}

但我想将字典形式从'contains(waste)': False更改为'contains(waste)': 2。我认为那种形式（'contains（waste）'：2）很好地解释了文件，因为它可以计算世界的频率。所以特征集是{('contains(waste)': 2, 'contains(lot)': 5, ...},'neg')...}

但我担心'contains(waste)': 2和'contains(waste)': 1是否与 naivebayesclassifier 完全不同。然后它无法解释'contains(waste)': 2和'contains(waste)': 1的相似性。

{'contains(lot)': 1 和 'contains(waste)': 1}和{'contains(waste)': 2 和 'contains(waste)': 1}可以与程序相同。

nltk.naivebayesclassifier 能理解单词的频率吗？

这是我使用的代码

def split_and_count_word(data):
    #belongs_to : Main
    #Role : make featuresets from korean words using konlpy.
    #Parameter : dictionary data(dict of contents ex.{'politic':{'parliament': [content,content]}..})
    #Return : list featuresets([{'word':True',...},'politic'] == featureset + category)

    featuresets = []
    twitter = konlpy.tag.Twitter()#Korean word splitter

    for big_cat in data:

        for small_cat in data[big_cat]:
            #save category name needed in featuresets 
            category = str(big_cat[0:3])+'/'+str(small_cat)
            count = 0; print(small_cat)

            for one_news in data[big_cat][small_cat]:
                count+=1; if count%100==0: print(count,end=' ')                
                #one_news is list in list so open it!
                doc = one_news
                #split word as using konlpy
                list_of_splited_word = twitter.morphs(doc[:-63])#delete useless sentences. 
                #get word length is higher than two and get list of splited words
                list_of_up_two_word = [word for word in list_of_splited_word if len(word)>1]
                dict_of_featuresets = make_featuresets(list_of_up_two_word)
                #save 
                featuresets.append((dict_of_featuresets,category))

    return featuresets


def make_featuresets(data):
    #belongs_to : split_and_count_word
    #Role : make featuresets
    #Parameter : list list_of_up_two_word(ex.['비누','떨어','지다']
    #Return : dictionary {word : True for word in data}

    #PROBLEM :(
    #cannot consider the freqency of word
    return {word : True for word in data}

def naive_train(featuresets):
    #belongs_to : Main
    #Role : Learning by naive bayes rule
    #Parameter : list featuresets([{'word':True',...},'pol/pal'])
    #Return : object classifier(nltk naivebayesclassifier object),
    #         list test_set(the featuresets that are randomly selected)

    random.shuffle(featuresets)
    train_set, test_set = featuresets[1000:], featuresets[:1000]
    classifier = naivebayes.NaiveBayesClassifier.train(train_set)

    return classifier,test_set

featuresets = split_and_count_word(data)
classifier,test_set = naive_train(featuresets)
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1 回答 1

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nltk 的朴素贝叶斯分类器将特征值视为逻辑上不同的。值不限于True和False，但它们从不被视为数量。如果您有特征f=2和f=3，它们将被视为不同的值。将数量添加到此类模型的唯一方法是将它们分类到“桶”中，例如f=1，f="few"(2-5)、f="several"(6-10)、f="many"(11 或更多)。（注意：如果你走这条路，有一些算法可以为桶选择好的值范围。）即使这样，模型也不“知道”“很少”介于“一个”和“几个”之间。你需要一个不同的机器学习工具来直接处理数量。

于 2016-11-13T20:21:43.447 回答