CODEWITHSUNDEEP
pythonpython-3.xlist-comprehensionword-frequency
tim_frost
tim_frost

Reputation: 145

Forming Bigrams of words in list of sentences and counting bigrams using python

I need: 1. to form bigram pairs and store them in list 2. find sum of id in which there аrе top 3 bigram with highest frequency

I have a list of sentences:

[['22574999', 'your message communication sent']
, ['22582857', 'your message be delivered']
, ['22585166', 'message has be delivered']
, ['22585424', 'message originated communication sent']]

Here is what I did:

for row in messages: 
    sstrm = list(row)
    bigrams=[b for l in sstrm for b in zip(l.split(" ")[:1], l.split(" ")[1:])]
    print(sstrm[0],bigrams)

which yields:

22574999 [('your', 'message')]
22582857 [('[your', 'message')]
22585166 [('message', 'has')]
22585424 [('message', 'originated')]

What I want is:

22574999 [('your', 'message'),('communication','sent')]
22582857 [('[your', 'message'),('be','delivered')]
22585166 [('message', 'has'),('be','delivered')]
22585424 [('message', 'originated'),('communication','sent')]

I would like to get the following result RESULT:

top 3 bigrams with highest frequency:

('your', 'message') :2 
('communication','sent'):2    
('be','delivered'):2

sum of id in which there аре top 3 bigrams with highest frequency:

('your', 'message'):2           Is included (22574999,22582857)     
('communication','sent'):2      Is included(22574999,22585424)
('be','delivered'):2            Is included (22582857,22585166)

Thanks for your help!

Upvotes: 1

Views: 3338

Answers (2)

Daniele Bacarella
Daniele Bacarella

Reputation: 464

First thing I'd like to point out is that bigrams are sequences of two adjacent elements.

For instance, the bigrams of "the fox jumped over the lazy dog" are:

[("the", "fox"),("fox", "jumped"),("jumped", "over"),("over", "the"),("the", "lazy"),("lazy", "dog")]

This problem can be modeled using an inverted index, where the bigrams are the postings and the set of ids are the posting lists.

def bigrams(line):
    tokens = line.split(" ")
    return [(tokens[i], tokens[i+1]) for i in range(0, len(tokens)-1)]


if __name__ == "__main__":
    messages = [['22574999', 'your message communication sent'], ['22582857', 'your message be delivered'], ['22585166', 'message has be delivered'], ['22585424', 'message originated communication sent']]
    bigrams_set = set()

    for row in messages:
        l_bigrams = bigrams(row[1])
        for bigram in l_bigrams:
            bigrams_set.add(bigram)

    inverted_idx = dict((b,[]) for b in bigrams_set)

    for row in messages:
        l_bigrams = bigrams(row[1])
        for bigram in l_bigrams:
            inverted_idx[bigram].append(row[0])

    freq_bigrams = dict((b,len(ids)) for b,ids in inverted_idx.items())
    import operator
    top3_bigrams = sorted(freq_bigrams.iteritems(), key=operator.itemgetter(1), reverse=True)[:3]

Output

[(('communication', 'sent'), 2), (('your', 'message'), 2), (('be', 'delivered'), 2)]

Although this code can be optimized by a great deal, it gives you the idea.

Upvotes: 1

Jundiaius
Jundiaius

Reputation: 7718

You have an error in this line:

bigrams=[b for l in sstrm for b in zip(l.split(" ")[:1], l.split(" ")[1:])]

In your first argument in the zip you're stopping at the first element of the list with the [:1]. You want to get every element but the last, which corresponds to [:-1].

So the line should be like that:

bigrams=[b for l in sstrm for b in zip(l.split(" ")[:-1], l.split(" ")[1:])]

Upvotes: 0

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