Calculate cosine similarity given 2 sentence strings

From Python: tf-idf-cosine: to find document similarity , it is possible to calculate document similarity using tf-idf cosine. Without importing external libraries, are that any ways to calculate cosine similarity between 2 strings?

s1 = "This is a foo bar sentence ."
s2 = "This sentence is similar to a foo bar sentence ."
s3 = "What is this string ? Totally not related to the other two lines ."

cosine_sim(s1, s2) # Should give high cosine similarity
cosine_sim(s1, s3) # Shouldn't give high cosine similarity value
cosine_sim(s2, s3) # Shouldn't give high cosine similarity value

Solution 1:

A simple pure-Python implementation would be:

import math
import re
from collections import Counter

WORD = re.compile(r"\w+")


def get_cosine(vec1, vec2):
    intersection = set(vec1.keys()) & set(vec2.keys())
    numerator = sum([vec1[x] * vec2[x] for x in intersection])

    sum1 = sum([vec1[x] ** 2 for x in list(vec1.keys())])
    sum2 = sum([vec2[x] ** 2 for x in list(vec2.keys())])
    denominator = math.sqrt(sum1) * math.sqrt(sum2)

    if not denominator:
        return 0.0
    else:
        return float(numerator) / denominator


def text_to_vector(text):
    words = WORD.findall(text)
    return Counter(words)


text1 = "This is a foo bar sentence ."
text2 = "This sentence is similar to a foo bar sentence ."

vector1 = text_to_vector(text1)
vector2 = text_to_vector(text2)

cosine = get_cosine(vector1, vector2)

print("Cosine:", cosine)

Prints:

Cosine: 0.861640436855

The cosine formula used here is described here.

This does not include weighting of the words by tf-idf, but in order to use tf-idf, you need to have a reasonably large corpus from which to estimate tfidf weights.

You can also develop it further, by using a more sophisticated way to extract words from a piece of text, stem or lemmatise it, etc.

Solution 2:

The short answer is "no, it is not possible to do that in a principled way that works even remotely well". It is an unsolved problem in natural language processing research and also happens to be the subject of my doctoral work. I'll very briefly summarize where we are and point you to a few publications:

Meaning of words

The most important assumption here is that it is possible to obtain a vector that represents each word in the sentence in quesion. This vector is usually chosen to capture the contexts the word can appear in. For example, if we only consider the three contexts "eat", "red" and "fluffy", the word "cat" might be represented as [98, 1, 87], because if you were to read a very very long piece of text (a few billion words is not uncommon by today's standard), the word "cat" would appear very often in the context of "fluffy" and "eat", but not that often in the context of "red". In the same way, "dog" might be represented as [87,2,34] and "umbrella" might be [1,13,0]. Imagening these vectors as points in 3D space, "cat" is clearly closer to "dog" than it is to "umbrella", therefore "cat" also means something more similar to "dog" than to an "umbrella".

This line of work has been investigated since the early 90s (e.g. this work by Greffenstette) and has yielded some surprisingly good results. For example, here is a few random entries in a thesaurus I built recently by having my computer read wikipedia:

theory -> analysis, concept, approach, idea, method
voice -> vocal, tone, sound, melody, singing
james -> william, john, thomas, robert, george, charles

These lists of similar words were obtained entirely without human intervention- you feed text in and come back a few hours later.

The problem with phrases

You might ask why we are not doing the same thing for longer phrases, such as "ginger foxes love fruit". It's because we do not have enough text. In order for us to reliably establish what X is similar to, we need to see many examples of X being used in context. When X is a single word like "voice", this is not too hard. However, as X gets longer, the chances of finding natural occurrences of X get exponentially slower. For comparison, Google has about 1B pages containing the word "fox" and not a single page containing "ginger foxes love fruit", despite the fact that it is a perfectly valid English sentence and we all understand what it means.

Composition

To tackle the problem of data sparsity, we want to perform composition, i.e. to take vectors for words, which are easy to obtain from real text, and to put the together in a way that captures their meaning. The bad news is nobody has been able to do that well so far.

The simplest and most obvious way is to add or multiply the individual word vectors together. This leads to undesirable side effect that "cats chase dogs" and "dogs chase cats" would mean the same to your system. Also, if you are multiplying, you have to be extra careful or every sentences will end up represented by [0,0,0,...,0], which defeats the point.

Further reading

I will not discuss the more sophisticated methods for composition that have been proposed so far. I suggest you read Katrin Erk's "Vector space models of word meaning and phrase meaning: a survey". This is a very good high-level survey to get you started. Unfortunately, is not freely available on the publisher's website, email the author directly to get a copy. In that paper you will find references to many more concrete methods. The more comprehensible ones are by Mitchel and Lapata (2008) and Baroni and Zamparelli (2010).

Edit after comment by @vpekar: The bottom line of this answer is to stress the fact that while naive methods do exist (e.g. addition, multiplication, surface similarity, etc), these are fundamentally flawed and in general one should not expect great performance from them.

Solution 3:

I have similar solution but might be useful for pandas

import math
import re
from collections import Counter
import pandas as pd

WORD = re.compile(r"\w+")


def get_cosine(vec1, vec2):
    intersection = set(vec1.keys()) & set(vec2.keys())
    numerator = sum([vec1[x] * vec2[x] for x in intersection])

    sum1 = sum([vec1[x] ** 2 for x in list(vec1.keys())])
    sum2 = sum([vec2[x] ** 2 for x in list(vec2.keys())])
    denominator = math.sqrt(sum1) * math.sqrt(sum2)

    if not denominator:
        return 0.0
    else:
        return float(numerator) / denominator


def text_to_vector(text):
    words = WORD.findall(text)
    return Counter(words)

df=pd.read_csv('/content/drive/article.csv')
df['vector1']=df['headline'].apply(lambda x: text_to_vector(x)) 
df['vector2']=df['snippet'].apply(lambda x: text_to_vector(x)) 
df['simscore']=df.apply(lambda x: get_cosine(x['vector1'],x['vector2']),axis=1)