word-word co-occurrence matrix

问题:

I am looking for a module in sklearn that lets you derive the word-word co-ocurrence matrix. I can get the document-term matrix but not sure how to go about obtaining a word-word matrix of co-occurences.

回答1:

Here is my example solution using CountVectorizer in scikit-learn. And referring to this post, you can simply use matrix multiplication to get word-word co-occurrence matrix.

from sklearn.feature_extraction.text import CountVectorizer
docs = ['this this this book',
        'this cat good',
        'cat good shit']
count_model = CountVectorizer(ngram_range=(1,1)) # default unigram model
X = count_model.fit_transform(docs)
# X[X > 0] = 1 # run this line if you don't want extra within-text cooccurence (see below)
Xc = (X.T * X) # this is co-occurrence matrix in sparse csr format
Xc.setdiag(0) # sometimes you want to fill same word cooccurence to 0
print(Xc.todense()) # print out matrix in dense format

You can also refer to dictionary of words in count_model,

count_model.vocabulary_

Or, if you want to normalize by diagonal component (referred to answer in previous post).

import scipy.sparse as sp
Xc = (X.T * X)
g = sp.diags(1./Xc.diagonal())
Xc_norm = g * Xc # normalized co-occurence matrix

Extra to note @Federico Caccia answer, if you don't want co-occurrence that are spurious from the own text, set occurrence that is greater that 1 to 1 e.g.

X[X > 0] = 1 # do this line first before computing cooccurrence
Xc = (X.T * X)
...

回答2:

You can use the ngram_range parameter in the CountVectorizer or TfidfVectorizer

Code example:

bigram_vectorizer = CountVectorizer(ngram_range=(2, 2)) # by saying 2,2 you are telling you only want pairs of 2 words

In case you want to explicitly say which co-occurrences of words you want to count, use the vocabulary param, i.e: vocabulary = {'awesome unicorns':0, 'batman forever':1}

http://scikit-learn.org/stable/modules/generated/sklearn.feature_extraction.text.CountVectorizer.html

Self-explanatory and ready to use code with predefined word-word co-occurrences. In this case we are tracking for co-occurrences of awesome unicorns and batman forever:

from sklearn.feature_extraction.text import CountVectorizer
import numpy as np
samples = ['awesome unicorns are awesome','batman forever and ever','I love batman forever']
bigram_vectorizer = CountVectorizer(ngram_range=(1, 2), vocabulary = {'awesome unicorns':0, 'batman forever':1}) 
co_occurrences = bigram_vectorizer.fit_transform(samples)
print 'Printing sparse matrix:', co_occurrences
print 'Printing dense matrix (cols are vocabulary keys 0-> "awesome unicorns", 1-> "batman forever")', co_occurrences.todense()
sum_occ = np.sum(co_occurrences.todense(),axis=0)
print 'Sum of word-word occurrences:', sum_occ
print 'Pretty printig of co_occurrences count:', zip(bigram_vectorizer.get_feature_names(),np.array(sum_occ)[0].tolist())

Final output is ('awesome unicorns', 1), ('batman forever', 2), which corresponds exactly to our samples provided data.

回答3:

@titipata I think your solution is not a good metric because we are giving the same weight to real co-ocurrences and to occurrences that are just spurious. For example, if I have 5 texts and the words apple and house appears with this frecuency:

text1: apple:10, "house":1

text2: apple:10, "house":0

text3: apple:10, "house":0

text4: apple:10, "house":0

text5: apple:10, "house":0

The co-occurrence we are going to measure is 10*1+10*0+10*0+10*0+10*0=10, but is just spurious.

And, in this another important cases, like the following:

text1: apple:1, "banana":1

text2: apple:1, "banana":1

text3: apple:1, "banana":1

text4: apple:1, "banana":1

text5: apple:1, "banana":1

we are going to get just a co-occurrence of 1*1+1*1+1*1+1*1=5, when in fact that co-occurrence really important.

@Guiem Bosch In this case co-occurrences are measured only when the two words are contiguous.

I propose to use something the @titipa solution to compute the matrix:

Xc = (Y.T * Y) # this is co-occurrence matrix in sparse csr format

where, instead of using X, use a matrix Y with ones in positions greater than 0 and zeros in another positions.

Using this, in the first example we are going to have: co-occurrence:1*1+1*0+1*0+1*0+1*0=1 and in the second example: co-occurrence:1*1+1*1+1*1+1*1+1*0=5 which is what we are really looking for.