Reputation: 53
I'm having a problem in understanding how we got the Tf-Idf in the following program:
I have tried calculating the value of a
in the document 2 ('And_this_is_the_third_one.'
) using the concept given on the site, but my value of 'a' using the above concept is
1/26*log(4/1)
((count of occurrence of 'a' character)/(no of characters in the given document)*log( # Docs/ # Docs in which given character occurred))
= 0.023156
But output is returned as 0.2203 as can be seen in the output.
from sklearn.feature_extraction.text import TfidfVectorizer
corpus = ['This_is_the_first_document.', 'This_document_is_the_second_document.', 'And_this_is_the_third_one.', 'Is_this_the_first_document?', ]
vectorizer = TfidfVectorizer(min_df=0.0, analyzer="char")
X = vectorizer.fit_transform(corpus)
print(vectorizer.get_feature_names())
print(vectorizer.vocabulary_)
m = X.todense()
print(m)
I expected the output to be 0.023156 using the concept explained above.
The output is:
['.', '?', '_', 'a', 'c', 'd', 'e', 'f', 'h', 'i', 'm', 'n', 'o', 'r', 's', 't', 'u']
{'t': 15, 'h': 8, 'i': 9, 's': 14, '_': 2, 'e': 6, 'f': 7, 'r': 13, 'd': 5, 'o': 12, 'c': 4, 'u': 16, 'm': 10, 'n': 11, '.': 0, 'a': 3, '?': 1}
[[0.14540332 0. 0.47550697 0. 0.14540332 0.11887674
0.23775349 0.17960203 0.23775349 0.35663023 0.14540332 0.11887674
0.11887674 0.14540332 0.35663023 0.47550697 0.14540332]
[0.10814145 0. 0.44206359 0. 0.32442434 0.26523816
0.35365088 0. 0.17682544 0.17682544 0.21628289 0.26523816
0.26523816 0. 0.26523816 0.35365088 0.21628289]
[0.14061506 0. 0.57481012 0.22030066 0. 0.22992405
0.22992405 0. 0.34488607 0.34488607 0. 0.22992405
0.11496202 0.14061506 0.22992405 0.34488607 0. ]
[0. 0.2243785 0.46836004 0. 0.14321789 0.11709001
0.23418002 0.17690259 0.23418002 0.35127003 0.14321789 0.11709001
0.11709001 0.14321789 0.35127003 0.46836004 0.14321789]]
Upvotes: 5
Views: 5276
Reputation: 16966
The TfidfVectorizer()
has smoothing added to the document counts and l2
normalization been applied on top tf-idf vector, as mentioned in the documentation.
(count of occurrence of the character)/(no of characters in the given document) *
log (1 + # Docs / 1 + # Docs in which the given character is present) +1 )
This Normalization is l2
by default, but you can change or remove this step by using the parameter norm
. Similarly, smoothing can be
To understand how does the exact score is computed, I am going to fit a CountVectorizer()
to know the counts of each character in every document.
countVectorizer = CountVectorizer(analyzer='char')
tf = countVectorizer.fit_transform(corpus)
tf_df = pd.DataFrame(tf.toarray(),
columns= countVectorizer.get_feature_names())
tf_df
#output:
. ? _ a c d e f h i m n o r s t u
0 1 0 4 0 1 1 2 1 2 3 1 1 1 1 3 4 1
1 1 0 5 0 3 3 4 0 2 2 2 3 3 0 3 4 2
2 1 0 5 1 0 2 2 0 3 3 0 2 1 1 2 3 0
3 0 1 4 0 1 1 2 1 2 3 1 1 1 1 3 4 1
Let us apply the tf-idf weighting based on sklearn implementation now for the second document now!
v=[]
doc_id = 2
# number of documents in the corpus + smoothing
n_d = 1+ tf_df.shape[0]
for char in tf_df.columns:
# calculate tf - count of this char in the doc / total number chars in the doc
tf = tf_df.loc[doc_id,char]/tf_df.loc[doc_id,:].sum()
# number of documents containing this char with smoothing
df_d_t = 1+ sum(tf_df.loc[:,char]>0)
# now calculate the idf with smoothing
idf = (np.log (n_d/df_d_t) + 1 )
# calculate the score now
v.append (tf*idf)
from sklearn.preprocessing import normalize
# normalize the vector with l2 norm and create a dataframe with feature_names
pd.DataFrame(normalize([v], norm='l2'), columns=vectorizer.get_feature_names())
#output:
. ? _ a c d e f h i m n o r s t u
0.140615 0.0 0.57481 0.220301 0.0 0.229924 0.229924 0.0 0.344886 0.344886 0.0 0.229924 0.114962 0.140615 0.229924 0.344886 0.0
you could find that the score for char a
matches with the TfidfVectorizer()
output!!!
Upvotes: 4