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If we have two lists of strings:

A = "Hello how are you? The weather is fine. I'd like to go for a walk.".split()
B = "bank, weather, sun, moon, fun, hi".split(",")

The words in list A constitute my word vector basis. How can I calculate the cosine similarity scores of each word in B?

What I've done so far: I can calculate the cosine similarity of two whole lists with the following function:

def counter_cosine_similarity(c1, c2):
    terms = set(c1).union(c2)
    dotprod = sum(c1.get(k, 0) * c2.get(k, 0) for k in terms)
    magA = math.sqrt(sum(c1.get(k, 0)**2 for k in terms))
    magB = math.sqrt(sum(c2.get(k, 0)**2 for k in terms))
    return dotprod / (magA * magB)

But how do I have to integrate my vector basis and how can I calculate then the similarities between the terms in B?

JohnD
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  • What do you mean by "calculate the cosine similarity scores of each word in B"? As you see in the parameters for `counter_cosine_similarity`, that similariy relates to two vectors, so I assume you want this between two words. So do you want the similarity for each pair of words, one from `A` and one from `B`? – Rory Daulton Nov 05 '16 at 12:00

1 Answers1

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import math
from collections import Counter

ListA = "Hello how are you? The weather is fine. I'd like to go for a walk.".split()
ListB = "bank, weather, sun, moon, fun, hi".split(",")

def cosdis(v1, v2):
    common = v1[1].intersection(v2[1])
    return sum(v1[0][ch] * v2[0][ch] for ch in common) / v1[2] / v2[2]

def word2vec(word):
    cw = Counter(word)
    sw = set(cw)
    lw = math.sqrt(sum(c * c for c in cw.values()))
    return cw, sw, lw

def removePunctuations(str_input):
    ret = []
    punctuations = '''!()-[]{};:'"\,<>./?@#$%^&*_~'''
    for char in str_input:
        if char not in punctuations:
            ret.append(char)

    return "".join(ret)


for i in ListA:
    for j in ListB:
       print(cosdis(word2vec(removePunctuations(i)), word2vec(removePunctuations(j))))
Januka samaranyake
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