countvectorizer 是否与 use_idf=false 的 tfidfvectorizer 相同? [英] Is a countvectorizer the same as tfidfvectorizer with use_idf=false?
问题描述
正如标题所述:countvectorizer
是否与 tfidfvectorizer
相同且带有 use_idf=false ?如果不是为什么不呢?
那么这是否也意味着在此处添加 tfidftransformer
是多余的?
vect = CountVectorizer(min_df=1)tweets_vector = vect.fit_transform(语料库)tf_transformer = TfidfTransformer(use_idf=False).fit(tweets_vector)tweets_vector_tf = tf_transformer.transform(tweets_vector)
不,它们不一样.TfidfVectorizer
对其结果进行归一化,即其输出中的每个向量都具有范数 1:
这样做是为了使行上的点积是余弦相似度.当给定选项 sublinear_tf=True
时,TfidfVectorizer
也可以使用对数折扣频率.
要使 TfidfVectorizer
表现得像 CountVectorizer
,请为其提供构造函数选项 use_idf=False, normalize=None
.
As the title states: Is a countvectorizer
the same as tfidfvectorizer
with use_idf=false ? If not why not ?
So does this also mean that adding the tfidftransformer
here is redundant ?
vect = CountVectorizer(min_df=1)
tweets_vector = vect.fit_transform(corpus)
tf_transformer = TfidfTransformer(use_idf=False).fit(tweets_vector)
tweets_vector_tf = tf_transformer.transform(tweets_vector)
No, they're not the same. TfidfVectorizer
normalizes its results, i.e. each vector in its output has norm 1:
>>> CountVectorizer().fit_transform(["foo bar baz", "foo bar quux"]).A
array([[1, 1, 1, 0],
[1, 0, 1, 1]])
>>> TfidfVectorizer(use_idf=False).fit_transform(["foo bar baz", "foo bar quux"]).A
array([[ 0.57735027, 0.57735027, 0.57735027, 0. ],
[ 0.57735027, 0. , 0.57735027, 0.57735027]])
This is done so that dot-products on the rows are cosine similarities. Also TfidfVectorizer
can use logarithmically discounted frequencies when given the option sublinear_tf=True
.
To make TfidfVectorizer
behave as CountVectorizer
, give it the constructor options use_idf=False, normalize=None
.
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