Cosine Similarity

Question

I was reading and came across this formula:

The formula is for cosine similarity. I thought this looked interesting and I created a numpy array that has user_id as row and item_id as column. For instance, let M be this matrix:

M = [[2,3,4,1,0],[0,0,0,0,5],[5,4,3,0,0],[1,1,1,1,1]]

Here the entries inside the matrix are ratings the people u has given to item i based on row u and column i. I want to calculate this cosine similarity for this matrix between items (rows). This should yield a 5 x 5 matrix I believe. I tried to do

df = pd.DataFrame(M)
item_mean_subtracted = df.sub(df.mean(axis=0), axis=1)
similarity_matrix = item_mean_subtracted.fillna(0).corr(method="pearson").values

However, this does not seem right.

Tonechas · Accepted Answer · 2017-03-19T19:53:05.267

4

Here's a possible implementation of the adjusted cosine similarity:

import numpy as np
from scipy.spatial.distance import pdist, squareform

M = np.asarray([[2, 3, 4, 1, 0], 
                [0, 0, 0, 0, 5], 
                [5, 4, 3, 0, 0], 
                [1, 1, 1, 1, 1]])

M_u = M.mean(axis=1)
item_mean_subtracted = M - M_u[:, None]
similarity_matrix = 1 - squareform(pdist(item_mean_subtracted.T, 'cosine'))

Remarks:

I'm taking advantage of NumPy broadcasting to subtract the mean.
If M is a sparse matrix, you could do something like ths: M.toarray().
From the docs:

Y = pdist(X, 'cosine')
Computes the cosine distance between vectors u and v,
1 − u⋅v / (||u||₂||v||₂)
where ||∗||₂ is the 2-norm of its argument *, and u⋅v is the dot product of u and v.
Array transposition is performed through the T method.

Demo:

In [277]: M_u
Out[277]: array([ 2. ,  1. ,  2.4,  1. ])

In [278]: item_mean_subtracted
Out[278]: 
array([[ 0. ,  1. ,  2. , -1. , -2. ],
       [-1. , -1. , -1. , -1. ,  4. ],
       [ 2.6,  1.6,  0.6, -2.4, -2.4],
       [ 0. ,  0. ,  0. ,  0. ,  0. ]])

In [279]: np.set_printoptions(precision=2)

In [280]: similarity_matrix
Out[280]: 
array([[ 1.  ,  0.87,  0.4 , -0.68, -0.72],
       [ 0.87,  1.  ,  0.8 , -0.65, -0.91],
       [ 0.4 ,  0.8 ,  1.  , -0.38, -0.8 ],
       [-0.68, -0.65, -0.38,  1.  ,  0.27],
       [-0.72, -0.91, -0.8 ,  0.27,  1.  ]])

edited Mar 19 '17 at 19:53

answered Mar 18 '17 at 17:45

Tonechas

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I believe the adjusted cosine similarity values should be between -1 and 1 right? – Mike El Jackson Mar 18 '17 at 20:56
Yes, absolutely. I misread the formula. Please take a look again at my edited answer, I think that now the adjusted cosine similarity is correctly implemented. – Tonechas Mar 19 '17 at 02:23
does this group the users for like u in U that the formula asks for? – Mike El Jackson Mar 19 '17 at 15:18
so like for each user (row in original matrix) we need to find the other users who rated the current item we are looking at – Mike El Jackson Mar 19 '17 at 15:23
Yes, it does. Notice that the shape of similarity matrix is (I, I), being I the number of items. – Tonechas Mar 19 '17 at 15:23
ok so like if we are looking at item 2 then it finds all users who have a nonzero entry for item2 and computes the mean there for $\bar{R}_u$ – Mike El Jackson Mar 19 '17 at 15:28
Can you explain how this does this, I can't tell how it does – Mike El Jackson Mar 19 '17 at 15:39
Let us [continue this discussion in chat](http://chat.stackoverflow.com/rooms/138481/discussion-between-mike-el-jackson-and-tonechas). – Mike El Jackson Mar 19 '17 at 19:33

Cosine Similarity

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