Question

我是NLP的新手，请说明如何使用fit_transform转换TFIDF值。

以下用于计算IDF的公式工作正常，日志（文档总数+ 1 /出现的术语数+ 1）+ 1

EG：文档1中“ This”一词的IDF值（“ this is a string”为1.91629073

应用fit_transform后，所有术语的值都会更改，用于转换的公式\逻辑是什么

TFID = TF * IDF

EG：文档1中“ This”一词的TFIDF值（“ this is a string”）为0.61366674

如何得出这个值0.61366674？

from sklearn.feature_extraction.text import TfidfVectorizer
import pandas as pd

d = pd.Series(['This is a string','This is another string',
               'TFIDF Computation Calculation','TFIDF is the product of TF and IDF'])


df = pd.DataFrame(d)

tfidf_vectorizer = TfidfVectorizer()

tfidf = tfidf_vectorizer.fit_transform(df[0])


print (tfidf_vectorizer.idf_)

#output
#[1.91629073 1.91629073 1.91629073 1.91629073 1.91629073 1.22314355 1.91629073 
#1.91629073 1.51082562 1.91629073 1.51082562 1.91629073 1.51082562]

##-------------------------------------------------

##how the above values are getting transformed here 

##-------------------------------------------------


print (tfidf.toarray())


#[[0.         0.         0.         0.         0.         0.49681612  0.         
#0.         0.61366674 0.         0.         0.     0.61366674]
# [0.         0.61422608 0.         0.         0.         0.39205255
#  0.         0.         0.4842629  0.         0.         0.  0.4842629 ]
# [0.         0.         0.61761437 0.61761437 0.         0.
#  0.         0.         0.         0.         0.48693426 0.  0.        ]
# [0.37718389 0.         0.         0.         0.37718389 0.24075159
#  0.37718389 0.37718389 0.         0.37718389 0.29737611 0.37718389  0.        ]]

Answer 1

这是标准的TF-IDF向量，因为根据documentation，默认情况下为norm='l2'。因此，在tfidf.toarray()的输出中，级别0的每个元素/数组的行代表一个文档，级别1 /列的每个元素代表一个唯一的单词，每个文档的矢量元素的平方和等于1 ，您可以通过打印print([sum([word ** 2 for word in doc]) for doc in tfidf.toarray()])进行检查。

范数：“ l1”，“ l2”或无，可选（默认值=“ l2”）每个输出行将具有单位范数，或者：*'l2'：矢量元素的平方和为1。两个之间的余弦相似度向量是应用l2范数时的点积。 *‘l1’：向量元素的绝对值之和为1。请参阅 preprocessing.normalize

print(tfidf) #the same values you find in tfidf.toarray() but more readable
output: ([index of document on array lvl 0 / row], [index of unique word on array lvl 1 / column]) normed TF-IDF value
(0, 12) 0.6136667440107333  #1st word in 1st sentence: 'This'
(0, 5)  0.4968161174826459  #'is'
(0, 8)  0.6136667440107333  #'string', see that word 'a' is missing
(1, 12) 0.48426290003607125 #'This'
(1, 5)  0.3920525532545391  #'is'
(1, 8)  0.48426290003607125 #'string'
(1, 1)  0.6142260844216119  #'another'
(2, 10) 0.48693426407352264 #'TFIDF'
(2, 3)  0.6176143709756019  #'Computation'
(2, 2)  0.6176143709756019  #'Calculation'
(3, 5)  0.2407515909314943  #'is'
(3, 10) 0.2973761110467491  #'TFIDF'
(3, 11) 0.37718388973255157 #'the'
(3, 7)  0.37718388973255157 #'product'
(3, 6)  0.37718388973255157 #'of'
(3, 9)  0.37718388973255157 #'TF'
(3, 0)  0.37718388973255157 #'and'
(3, 4)  0.37718388973255157 #'IDF'

因为它是标准的TF-IDF值，所以矢量元素的平方和将等于1。对于索引为0的第一个文档，矢量元素的平方和将等于1：sum([0.6136667440107333 ** 2, 0.4968161174826459 ** 2, 0.6136667440107333 ** 2])

您可以通过设置norm=None来关闭此转换。

print(TfidfVectorizer(norm=None).fit_transform(df[0])) #the same values you find in TfidfVectorizer(norm=None).fit_transform(df[0]).toarray(), but more readable
output: ([index of document on array lvl 0 / row], [index of unique word on array lvl 1 / column]) TF-IDF value
(0, 12) 1.5108256237659907 #1st word in 1st sentence: 'This'
(0, 5)  1.2231435513142097 #'is'
(0, 8)  1.5108256237659907 #'string', see that word 'a' is missing
(1, 12) 1.5108256237659907 #'This'
(1, 5)  1.2231435513142097 #'is'
(1, 8)  1.5108256237659907 #'string'
(1, 1)  1.916290731874155  #'another'
(2, 10) 1.5108256237659907 #'TFIDF'
(2, 3)  1.916290731874155  #'Computation'
(2, 2)  1.916290731874155  #'Calculation'
(3, 5)  1.2231435513142097 #'is'
(3, 10) 1.5108256237659907 #'TFIDF'
(3, 11) 1.916290731874155  #'the'
(3, 7)  1.916290731874155  #'product'
(3, 6)  1.916290731874155  #'of'
(3, 9)  1.916290731874155  #'TF'
(3, 0)  1.916290731874155  #'and'
(3, 4)  1.916290731874155  #'IDF'

因为每个单词在每个文档中只出现一次，所以TF-IDF值是每个单词的IDF值乘以1：

tfidf_vectorizer = TfidfVectorizer(norm=None)
tfidf = tfidf_vectorizer.fit_transform(df[0])
print(tfidf_vectorizer.idf_)
output: Smoothed IDF-values
[1.91629073 1.91629073 1.91629073 1.91629073 1.91629073 1.22314355
 1.91629073 1.91629073 1.51082562 1.91629073 1.51082562 1.91629073
 1.51082562]

我希望以上对您有所帮助。

不幸的是，我无法复制该转换，因为

两个向量之间的余弦相似度是l2时的点积规范已应用。

似乎是附加步骤。因为当您使用默认设置norm='l2'时，TF-IDF值会因每个文档中的单词数而有偏差，所以我只需使用norm=None来关闭此设置。我发现，您不能简单地通过使用以下方式进行转换：

tfidf_norm_calculated = [
    [(word/sum(doc))**0.5 for word in doc]
    for doc in TfidfVectorizer(norm=None).fit_transform(df[0]).toarray()]
print(tfidf_norm_calculated)
print('Sum of squares of vector elements is 1: ', [sum([word**2 for word in doc]) for doc in tfidf_norm_calculated])
print('Compare to:', TfidfVectorizer().fit_transform(df[0]).toarray())

TFIDF值如何转换

1 个答案: