How to compute the similarity between two text documents?

I am looking at working on an NLP project, in any programming language (though Python will be my preference).

I want to take two documents and determine how similar they are.


Solution 1:

The common way of doing this is to transform the documents into TF-IDF vectors and then compute the cosine similarity between them. Any textbook on information retrieval (IR) covers this. See esp. Introduction to Information Retrieval, which is free and available online.

Computing Pairwise Similarities

TF-IDF (and similar text transformations) are implemented in the Python packages Gensim and scikit-learn. In the latter package, computing cosine similarities is as easy as

from sklearn.feature_extraction.text import TfidfVectorizer

documents = [open(f).read() for f in text_files]
tfidf = TfidfVectorizer().fit_transform(documents)
# no need to normalize, since Vectorizer will return normalized tf-idf
pairwise_similarity = tfidf * tfidf.T

or, if the documents are plain strings,

>>> corpus = ["I'd like an apple", 
...           "An apple a day keeps the doctor away", 
...           "Never compare an apple to an orange", 
...           "I prefer scikit-learn to Orange", 
...           "The scikit-learn docs are Orange and Blue"]                                                                                                                                                                                                   
>>> vect = TfidfVectorizer(min_df=1, stop_words="english")                                                                                                                                                                                                   
>>> tfidf = vect.fit_transform(corpus)                                                                                                                                                                                                                       
>>> pairwise_similarity = tfidf * tfidf.T 

though Gensim may have more options for this kind of task.

See also this question.

[Disclaimer: I was involved in the scikit-learn TF-IDF implementation.]

Interpreting the Results

From above, pairwise_similarity is a Scipy sparse matrix that is square in shape, with the number of rows and columns equal to the number of documents in the corpus.

>>> pairwise_similarity                                                                                                                                                                                                                                      
<5x5 sparse matrix of type '<class 'numpy.float64'>'
    with 17 stored elements in Compressed Sparse Row format>

You can convert the sparse array to a NumPy array via .toarray() or .A:

>>> pairwise_similarity.toarray()                                                                                                                                                                                                                            
array([[1.        , 0.17668795, 0.27056873, 0.        , 0.        ],
       [0.17668795, 1.        , 0.15439436, 0.        , 0.        ],
       [0.27056873, 0.15439436, 1.        , 0.19635649, 0.16815247],
       [0.        , 0.        , 0.19635649, 1.        , 0.54499756],
       [0.        , 0.        , 0.16815247, 0.54499756, 1.        ]])

Let's say we want to find the document most similar to the final document, "The scikit-learn docs are Orange and Blue". This document has index 4 in corpus. You can find the index of the most similar document by taking the argmax of that row, but first you'll need to mask the 1's, which represent the similarity of each document to itself. You can do the latter through np.fill_diagonal(), and the former through np.nanargmax():

>>> import numpy as np     
                                                                                                                                                                                                                                  
>>> arr = pairwise_similarity.toarray()     
>>> np.fill_diagonal(arr, np.nan)                                                                                                                                                                                                                            
                                                                                                                                                                                                                 
>>> input_doc = "The scikit-learn docs are Orange and Blue"                                                                                                                                                                                                  
>>> input_idx = corpus.index(input_doc)                                                                                                                                                                                                                      
>>> input_idx                                                                                                                                                                                                                                                
4

>>> result_idx = np.nanargmax(arr[input_idx])                                                                                                                                                                                                                
>>> corpus[result_idx]                                                                                                                                                                                                                                       
'I prefer scikit-learn to Orange'

Note: the purpose of using a sparse matrix is to save (a substantial amount of space) for a large corpus & vocabulary. Instead of converting to a NumPy array, you could do:

>>> n, _ = pairwise_similarity.shape                                                                                                                                                                                                                         
>>> pairwise_similarity[np.arange(n), np.arange(n)] = -1.0
>>> pairwise_similarity[input_idx].argmax()                                                                                                                                                                                                                  
3

Solution 2:

Identical to @larsman, but with some preprocessing

import nltk, string
from sklearn.feature_extraction.text import TfidfVectorizer

nltk.download('punkt') # if necessary...


stemmer = nltk.stem.porter.PorterStemmer()
remove_punctuation_map = dict((ord(char), None) for char in string.punctuation)

def stem_tokens(tokens):
    return [stemmer.stem(item) for item in tokens]

'''remove punctuation, lowercase, stem'''
def normalize(text):
    return stem_tokens(nltk.word_tokenize(text.lower().translate(remove_punctuation_map)))

vectorizer = TfidfVectorizer(tokenizer=normalize, stop_words='english')

def cosine_sim(text1, text2):
    tfidf = vectorizer.fit_transform([text1, text2])
    return ((tfidf * tfidf.T).A)[0,1]


print cosine_sim('a little bird', 'a little bird')
print cosine_sim('a little bird', 'a little bird chirps')
print cosine_sim('a little bird', 'a big dog barks')

Solution 3:

It's an old question, but I found this can be done easily with Spacy. Once the document is read, a simple api similarity can be used to find the cosine similarity between the document vectors.

Start by installing the package and downloading the model:

pip install spacy
python -m spacy download en_core_web_sm

Then use like so:

import spacy
nlp = spacy.load('en_core_web_sm')
doc1 = nlp(u'Hello hi there!')
doc2 = nlp(u'Hello hi there!')
doc3 = nlp(u'Hey whatsup?')

print (doc1.similarity(doc2)) # 0.999999954642
print (doc2.similarity(doc3)) # 0.699032527716
print (doc1.similarity(doc3)) # 0.699032527716