I want to calculate the cosine similarity between two lists, let’s say for example list 1 which is dataSetI and list 2 which is dataSetII.

Let’s say dataSetI is [3, 45, 7, 2] and dataSetII is [2, 54, 13, 15]. The length of the lists are always equal. I want to report cosine similarity as a number between 0 and 1.

dataSetI = [3, 45, 7, 2]
dataSetII = [2, 54, 13, 15]

def cosine_similarity(list1, list2):
  # How to?
  pass

print(cosine_similarity(dataSetI, dataSetII))

You should try SciPy. It has a bunch of useful scientific routines for example, “routines for computing integrals numerically, solving differential equations, optimization, and sparse matrices.” It uses the superfast optimized NumPy for its number crunching. See here for installing.

Note that spatial.distance.cosine computes the distance, and not the similarity. So, you must subtract the value from 1 to get the similarity.

from scipy import spatial

dataSetI = [3, 45, 7, 2]
dataSetII = [2, 54, 13, 15]
result = 1 - spatial.distance.cosine(dataSetI, dataSetII)

another version based on numpy only

from numpy import dot
from numpy.linalg import norm

cos_sim = dot(a, b)/(norm(a)*norm(b))

You can use cosine_similarity function form sklearn.metrics.pairwise docs

In [23]: from sklearn.metrics.pairwise import cosine_similarity

In [24]: cosine_similarity([[1, 0, -1]], [[-1,-1, 0]])
Out[24]: array([[-0.5]])

I don’t suppose performance matters much here, but I can’t resist. The zip() function completely recopies both vectors (more of a matrix transpose, actually) just to get the data in “Pythonic” order. It would be interesting to time the nuts-and-bolts implementation:

import math
def cosine_similarity(v1,v2):
    "compute cosine similarity of v1 to v2: (v1 dot v2)/{||v1||*||v2||)"
    sumxx, sumxy, sumyy = 0, 0, 0
    for i in range(len(v1)):
        x = v1[i]; y = v2[i]
        sumxx += x*x
        sumyy += y*y
        sumxy += x*y
    return sumxy/math.sqrt(sumxx*sumyy)

v1,v2 = [3, 45, 7, 2], [2, 54, 13, 15]
print(v1, v2, cosine_similarity(v1,v2))

Output: [3, 45, 7, 2] [2, 54, 13, 15] 0.972284251712

That goes through the C-like noise of extracting elements one-at-a-time, but does no bulk array copying and gets everything important done in a single for loop, and uses a single square root.

ETA: Updated print call to be a function. (The original was Python 2.7, not 3.3. The current runs under Python 2.7 with a from __future__ import print_function statement.) The output is the same, either way.

CPYthon 2.7.3 on 3.0GHz Core 2 Duo:

>>> timeit.timeit("cosine_similarity(v1,v2)",setup="from __main__ import cosine_similarity, v1, v2")
2.4261788514654654
>>> timeit.timeit("cosine_measure(v1,v2)",setup="from __main__ import cosine_measure, v1, v2")
8.794677709375264

So, the unpythonic way is about 3.6 times faster in this case.

without using any imports

math.sqrt(x)

can be replaced with

x** .5

without using numpy.dot() you have to create your own dot function using list comprehension:

def dot(A,B): 
    return (sum(a*b for a,b in zip(A,B)))

and then its just a simple matter of applying the cosine similarity formula:

def cosine_similarity(a,b):
    return dot(a,b) / ( (dot(a,a) **.5) * (dot(b,b) ** .5) )

I did a benchmark based on several answers in the question and the following snippet is believed to be the best choice:

def dot_product2(v1, v2):
    return sum(map(operator.mul, v1, v2))


def vector_cos5(v1, v2):
    prod = dot_product2(v1, v2)
    len1 = math.sqrt(dot_product2(v1, v1))
    len2 = math.sqrt(dot_product2(v2, v2))
    return prod / (len1 * len2)

The result makes me surprised that the implementation based on scipy is not the fastest one. I profiled and find that cosine in scipy takes a lot of time to cast a vector from python list to numpy array.

enter image description here

import math
from itertools import izip

def dot_product(v1, v2):
    return sum(map(lambda x: x[0] * x[1], izip(v1, v2)))

def cosine_measure(v1, v2):
    prod = dot_product(v1, v2)
    len1 = math.sqrt(dot_product(v1, v1))
    len2 = math.sqrt(dot_product(v2, v2))
    return prod / (len1 * len2)

You can round it after computing:

cosine = format(round(cosine_measure(v1, v2), 3))

If you want it really short, you can use this one-liner:

from math import sqrt
from itertools import izip

def cosine_measure(v1, v2):
    return (lambda (x, y, z): x / sqrt(y * z))(reduce(lambda x, y: (x[0] + y[0] * y[1], x[1] + y[0]**2, x[2] + y[1]**2), izip(v1, v2), (0, 0, 0)))

You can use this simple function to calculate the cosine similarity:

def cosine_similarity(a, b):
  return sum([i*j for i,j in zip(a, b)])/(math.sqrt(sum([i*i for i in a]))* math.sqrt(sum([i*i for i in b])))

Python code to calculate:

  • Cosine Distance
  • Cosine Similarity
  • Angular Distance
  • Angular Similarity

import math

from scipy import spatial


def calculate_cosine_distance(a, b):
    cosine_distance = float(spatial.distance.cosine(a, b))
    return cosine_distance


def calculate_cosine_similarity(a, b):
    cosine_similarity = 1 - calculate_cosine_distance(a, b)
    return cosine_similarity


def calculate_angular_distance(a, b):
    cosine_similarity = calculate_cosine_similarity(a, b)
    angular_distance = math.acos(cosine_similarity) / math.pi
    return angular_distance


def calculate_angular_similarity(a, b):
    angular_similarity = 1 - calculate_angular_distance(a, b)
    return angular_similarity

You can do this in Python using simple function:

def get_cosine(text1, text2):
  vec1 = text1
  vec2 = text2
  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 vec1.keys()])
  sum2 = sum([vec2[x]**2 for x in vec2.keys()])
  denominator = math.sqrt(sum1) * math.sqrt(sum2)
  if not denominator:
     return 0.0
  else:
     return round(float(numerator) / denominator, 3)
dataSet1 = [3, 45, 7, 2]
dataSet2 = [2, 54, 13, 15]
get_cosine(dataSet1, dataSet2)

Using numpy compare one list of numbers to multiple lists(matrix):

def cosine_similarity(vector,matrix):
   return ( np.sum(vector*matrix,axis=1) / ( np.sqrt(np.sum(matrix**2,axis=1)) * np.sqrt(np.sum(vector**2)) ) )[::-1]

If you happen to be using PyTorch already, you should go with their CosineSimilarity implementation.

Suppose you have two n-dimensional numpy.ndarrays, v1 and v2, i.e. their shapes are both (n,). Here’s how you get their cosine similarity:

import torch
import torch.nn as nn

cos = nn.CosineSimilarity()
cos(torch.tensor([v1]), torch.tensor([v2])).item()

Or suppose you have two numpy.ndarrays w1 and w2, whose shapes are both (m, n). The following gets you a list of cosine similarities, each being the cosine similarity between a row in w1 and the corresponding row in w2:

cos(torch.tensor(w1), torch.tensor(w2)).tolist()

Another version, if you have a scenario where you have list of vectors and a query vector and you want to compute the cosine similarity of query vector with all the vectors in the list, you can do it in one go in the below fashion:

>>> import numpy as np

>>> A      # list of vectors, shape -> m x n
array([[ 3, 45,  7,  2],
       [ 1, 23,  3,  4]])

>>> B      # query vector, shape -> 1 x n
array([ 2, 54, 13, 15])

>>> similarity_scores = A.dot(B)/ (np.linalg.norm(A, axis=1) * np.linalg.norm(B))

>>> similarity_scores
array([0.97228425, 0.99026919])

We can easily calculate cosine similarity with simple mathematics equations.
Cosine_similarity = 1- (dotproduct of vectors/(product of norm of the vectors)). We can define two functions each for calculations of dot product and norm.

def dprod(a,b):
    sum=0
    for i in range(len(a)):
        sum+=a[i]*b[i]
    return sum

def norm(a):

    norm=0
    for i in range(len(a)):
    norm+=a[i]**2
    return norm**0.5

    cosine_a_b = 1-(dprod(a,b)/(norm(a)*norm(b)))

All the answers are great for situations where you cannot use NumPy. If you can, here is another approach:

def cosine(x, y):
    dot_products = np.dot(x, y.T)
    norm_products = np.linalg.norm(x) * np.linalg.norm(y)
    return dot_products / (norm_products + EPSILON)

Also bear in mind about EPSILON = 1e-07 to secure the division.