CODEWITHSUNDEEP
pythonloopstrigonometry
ChargingMuffin
ChargingMuffin

Reputation: 37

How to take two dictionary values to find cosine similarity in Python?

I need to find the cosine distance similarity from two vectors using the "user ratings" stored in the user_dict dictionary.

The ratings were imported from a CSV file then changed to a dictionary with the User as the Key with the values of each users ratings. My question is how do I loop through the dictionary to take two user's ratings and get the similarity using the cosine distance function?

The loop would need to not compare the same user together or compare the same users in a different order? (E.g. user 5 vs user 3 and user 3 vs user 5)

Upvotes: 0

Views: 891

Answers (2)

bigbounty
bigbounty

Reputation: 17408

from scipy import spatial

d = {'U1': [3, 4, 2, 5, 0, 4, 1, 3, 0, 0, 4],
'U2': [2, 3, 1, 0, 3, 0, 2, 0, 0, 3, 0], 
'U3': [0, 4, 0, 5, 0, 4, 0, 3, 0, 2, 4], 
'U4': [0, 0, 2, 1, 4, 3, 2, 0, 0, 2, 0], 
'U5': [0, 0, 0, 5, 0, 4, 0, 3, 0, 0, 4],
'U6': [2, 3, 4, 0, 3, 0, 3, 0, 3, 4, 0],
'U6': [2, 3, 4, 0, 3, 0, 3, 0, 3, 4, 0], 
'U7': [0, 4, 3, 5, 0, 5, 0, 0, 0, 0, 4], 
'U8': [4, 3, 0, 3, 4, 2, 2, 0, 2, 3, 2],
'U9': [0, 2, 0, 3, 1, 0, 1, 0, 0, 2, 0], 
'U10': [0, 3, 0, 4, 3, 3, 0, 3, 0, 4, 4],
'U11': [2, 2, 1, 2, 1, 0, 2, 0, 1, 0, 2],
'U12': [0, 4, 4, 5, 0, 0, 0, 3, 0, 4, 5],
'U13': [3, 3, 0, 2, 2, 3, 2, 0, 2, 0, 3],
'U14': [0, 3, 4, 5, 0, 5, 0, 0, 0, 4, 0],
'U15': [2, 0, 0, 3, 0, 2, 2, 3, 0, 0, 3], 
'U16': [4, 4, 0, 4, 3, 4, 0, 3, 0, 3, 0], 
'U17': [0, 2, 0, 3, 1, 0, 2, 0, 1, 0, 3],
'U18': [2, 3, 1, 0, 3, 2, 3, 2, 0, 2, 0], 
'U19': [0, 5, 0, 4, 0, 3, 0, 4, 0, 0, 5],
'U20': [0, 0, 3, 0, 3, 0, 4, 0, 2, 0, 0],
'U21': [3, 0, 2, 4, 2, 3, 0, 4, 2, 3, 3], 
'U22': [4, 4, 0, 5, 3, 5, 0, 4, 0, 3, 0],
'U23': [3, 0, 0, 0, 3, 0, 2, 0, 0, 4, 0], 
'U24': [4, 0, 3, 0, 3, 0, 3, 0, 0, 2, 2], 
'U25': [0, 5, 0, 3, 3, 4, 0, 3, 3, 4, 4]}

all_keys = list(d.keys())

for i in range(len(all_keys)):
    for j in range(i+1,len(all_keys)):
        print(f"Cosine similaity between {all_keys[i]} and {all_keys[j]} is {1 - spatial.distance.cosine(d[all_keys[i]], d[all_keys[j]])}")

OR

using pandas

import pandas as pd
from sklearn.metrics.pairwise import cosine_similarity

d = {'U1': [3, 4, 2, 5, 0, 4, 1, 3, 0, 0, 4],
'U2': [2, 3, 1, 0, 3, 0, 2, 0, 0, 3, 0], 
'U3': [0, 4, 0, 5, 0, 4, 0, 3, 0, 2, 4], 
'U4': [0, 0, 2, 1, 4, 3, 2, 0, 0, 2, 0], 
'U5': [0, 0, 0, 5, 0, 4, 0, 3, 0, 0, 4],
'U6': [2, 3, 4, 0, 3, 0, 3, 0, 3, 4, 0],
'U6': [2, 3, 4, 0, 3, 0, 3, 0, 3, 4, 0], 
'U7': [0, 4, 3, 5, 0, 5, 0, 0, 0, 0, 4], 
'U8': [4, 3, 0, 3, 4, 2, 2, 0, 2, 3, 2],
'U9': [0, 2, 0, 3, 1, 0, 1, 0, 0, 2, 0], 
'U10': [0, 3, 0, 4, 3, 3, 0, 3, 0, 4, 4],
'U11': [2, 2, 1, 2, 1, 0, 2, 0, 1, 0, 2],
'U12': [0, 4, 4, 5, 0, 0, 0, 3, 0, 4, 5],
'U13': [3, 3, 0, 2, 2, 3, 2, 0, 2, 0, 3],
'U14': [0, 3, 4, 5, 0, 5, 0, 0, 0, 4, 0],
'U15': [2, 0, 0, 3, 0, 2, 2, 3, 0, 0, 3], 
'U16': [4, 4, 0, 4, 3, 4, 0, 3, 0, 3, 0], 
'U17': [0, 2, 0, 3, 1, 0, 2, 0, 1, 0, 3],
'U18': [2, 3, 1, 0, 3, 2, 3, 2, 0, 2, 0], 
'U19': [0, 5, 0, 4, 0, 3, 0, 4, 0, 0, 5],
'U20': [0, 0, 3, 0, 3, 0, 4, 0, 2, 0, 0],
'U21': [3, 0, 2, 4, 2, 3, 0, 4, 2, 3, 3], 
'U22': [4, 4, 0, 5, 3, 5, 0, 4, 0, 3, 0],
'U23': [3, 0, 0, 0, 3, 0, 2, 0, 0, 4, 0], 
'U24': [4, 0, 3, 0, 3, 0, 3, 0, 0, 2, 2], 
'U25': [0, 5, 0, 3, 3, 4, 0, 3, 3, 4, 4]}

df = pd.DataFrame(d)
cos_df = pd.DataFrame(cosine_similarity(df.T), columns = df.columns)
cos_df.insert(0,"Columns",df.columns)
print(cos_df)

Output:

  Columns        U1        U2        U3        U4  ...       U21       U22       U23       U24       U25
0       U1  1.000000  0.374228  0.902462  0.380803  ...  0.787351  0.805479  0.182123  0.414455  0.742959
1       U2  0.374228  1.000000  0.323498  0.648886  ...  0.428580  0.588035  0.838144  0.746816  0.574696
2       U3  0.902462  0.323498  1.000000  0.367348  ...  0.747476  0.790950  0.139942  0.181195  0.867595
3       U4  0.380803  0.648886  0.367348  1.000000  ...  0.562244  0.572351  0.631579  0.636035  0.543830
4       U5  0.829156  0.000000  0.876038  0.339457  ...  0.770675  0.651439  0.000000  0.137890  0.660241
5       U6  0.348816  0.864242  0.254164  0.650011  ...  0.500694  0.448630  0.707365  0.759113  0.553116
6       U7  0.888018  0.262071  0.870404  0.442141  ...  0.621170  0.642383  0.000000  0.249542  0.712893
7       U8  0.671751  0.808290  0.610121  0.655610  ...  0.735867  0.793363  0.749269  0.711438  0.774202
8       U9  0.561951  0.650011  0.667940  0.483810  ...  0.512989  0.681623  0.483810  0.321246  0.659221
9      U10  0.768376  0.545545  0.905945  0.584094  ...  0.817316  0.810441  0.442495  0.381958  0.930116
10     U11  0.766131  0.625543  0.584602  0.405906  ...  0.606128  0.561442  0.439732  0.700749  0.599162
11     U12  0.769604  0.451144  0.813118  0.329333  ...  0.724166  0.583435  0.250921  0.406111  0.740772
12     U13  0.806747  0.577813  0.687871  0.517409  ...  0.666687  0.708161  0.427425  0.582552  0.757112
13     U14  0.695436  0.436785  0.734756  0.612195  ...  0.644610  0.720248  0.272087  0.293578  0.662689
14     U15  0.849837  0.213504  0.759751  0.337691  ...  0.805629  0.669039  0.259762  0.448449  0.582825
15     U16  0.781028  0.663914  0.757364  0.578184  ...  0.785252  0.992774  0.561179  0.455047  0.783178
16     U17  0.713653  0.409462  0.713247  0.337227  ...  0.528221  0.456211  0.214599  0.396942  0.669745
17     U18  0.569298  0.879408  0.487692  0.733674  ...  0.573070  0.741858  0.709218  0.696631  0.664230
18     U19  0.898717  0.262071  0.949531  0.221071  ...  0.656330  0.691049  0.000000  0.146789  0.813301
19     U20  0.165567  0.540738  0.000000  0.684211  ...  0.290191  0.135557  0.447368  0.681466  0.233070
20     U21  0.787351  0.428580  0.747476  0.562244  ...  1.000000  0.809693  0.489696  0.563602  0.771035
21     U22  0.805479  0.588035  0.790950  0.572351  ...  0.809693  1.000000  0.497042  0.403039  0.782601
22     U23  0.182123  0.838144  0.139942  0.631579  ...  0.489696  0.497042  1.000000  0.795044  0.388450
23     U24  0.414455  0.746816  0.181195  0.636035  ...  0.563602  0.403039  0.795044  1.000000  0.335306
24     U25  0.742959  0.574696  0.867595  0.543830  ...  0.771035  0.782601  0.388450  0.335306  1.000000

[25 rows x 26 columns]

Upvotes: 2

Riccardo Bucco
Riccardo Bucco

Reputation: 15384

You can use itertools.combinations for that (dct is your input dictionary):

from itertools import combinations

for k1, k2 in combinations(dct.keys(), 2):
    # compute cosine similarity between dct[k1] and dct[k2]
    ...

Upvotes: 1

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