Return 'similar score' based on two dictionaries' similarity in Python?

Question

I know it's possible to return how similar two strings are by using the following function:

from difflib import SequenceMatcher
def similar(a, b):
    output=SequenceMatcher(None, a, b).ratio()
    return output

In [37]: similar("Hey, this is a test!","Hey, man, this is a test, man.")
Out[37]: 0.76
In [38]: similar("This should be one.","This should be one.")
Out[38]: 1.0

But is it possible to score two dictionaries based on the similarity of keys and their corresponding values? Not a number of in common keys, or what is in common, but a score from 0 to 1, like the example above with strings.

I'm trying to find the similarity score between ratings['Shane'] and ratings['Joe'] in this dictionary:

ratings={'Shane': {'127 Hours': 3.0, 'Avatar': 4.0, 'Nonstop': 5.0}, 'Joe': {'127 Hours': 5.0, 'Taken 3': 4.0, 'Avatar': 5.0, 'Nonstop': 3.0}}

I am using Python 2.7.10

Answer 1

import math

ratings={'Shane': {'127 Hours': 3.0, 'Avatar': 4.0, 'Nonstop': 5.0}, 'Joe': {'127 Hours': 5.0, 'Taken 3': 4.0, 'Avatar': 5.0, 'Nonstop': 3.0}}

def cosine_similarity(vec1,vec2):
        sum11, sum12, sum22 = 0, 0, 0
        for i in range(len(vec1)):
            x = vec1[i]; y = vec2[i]
            sum11 += x*x
            sum22 += y*y
            sum12 += x*y
        return sum12/math.sqrt(sum11*sum22)

list1 = list(ratings['Shane'].values())
list2 =  list(ratings['Joe'].values())

sim = cosine_similarity(list1,list2)
print(sim)

output

o/p : 0.9205746178983233

Updated When i use :

ratings={'Shane': {'127 Hours': 5.0, 'Avatar': 4.0, 'Nonstop': 5.0},
         'Joe': {'127 Hours': 5.0, 'Taken 3': 4.0, 'Avatar': 5.0, 'Nonstop': 3.0}}

output :0.9574271077563381

Update2: Normalized length and considered keys

from math import*


ratings={'Shane': {'127 Hours': 5.0, 'Avatar': 4.0, 'Nonstop': 5.0},
         'Joe': {'127 Hours': 5.0, 'Taken 3': 4.0, 'Avatar': 5.0, 'Nonstop': 3.0},
         'Bob': {'Panic Room':5.0,'Nonstop':5.0}}


def square_rooted(x):

    return round(sqrt(sum([a*a for a in x])),3)

def cosine_similarity(x,y):

    input1 = {}
    input2 = {}
    vector2 = []
    vector1 =[]

    if len(x) > len(y):
        input1 = x
        input2 = y
    else:
        input1 = y
        input2 = x


    vector1 = list(input1.values())

    for k in input1.keys():    # Normalizing input vectors. 
        if k in input2:
            vector2.append(float(input2[k])) #picking the values for the common keys from input 2
        else :
            vector2.append(float(0))


    numerator = sum(a*b for a,b in zip(vector2,vector1))
    denominator = square_rooted(vector1)*square_rooted(vector2)
    return round(numerator/float(denominator),3)


print("Similarity between Shane and Joe")
print (cosine_similarity(ratings['Shane'],ratings['Joe']))

print("Similarity between Joe and Bob")
print (cosine_similarity(ratings['Joe'],ratings['Bob']))

print("Similarity between Shane and Bob")
print (cosine_similarity(ratings['Shane'],ratings['Bob']))

output:

Similarity between Shane and Joe
0.887
Similarity between Joe and Bob
0.346
Similarity between Shane and Bob
0.615

Nice explanation between jaccurd and cosine : https://datascience.stackexchange.com/questions/5121/applications-and-differences-for-jaccard-similarity-and-cosine-similarity

i am using Python 3.4

NOTE: I have assigned 0 to missing values. But you can assign some proper values too. Refer : http://www.analyticsvidhya.com/blog/2015/02/7-steps-data-exploration-preparation-building-model-part-2/