以@Bill Bell为例，在[R]中有两种方法

a = c(2,1,0,2,0,1,1,1)

b = c(2,1,1,1,1,0,1,1)

d = (a %*% b) / (sqrt(sum(a^2)) * sqrt(sum(b^2)))

或者利用crossprod()方法的性能…

e = crossprod(a, b) / (sqrt(crossprod(a, a)) * sqrt(crossprod(b, b)))

两个向量A和B存在于二维空间或三维空间中，它们之间的夹角为cos相似度。

如果角度更大(可以达到最大180度)，即Cos 180=-1，最小角度为0度。cos0 =1意味着向量是对齐的，因此向量是相似的。

cos 90=0(这足以得出向量A和B根本不相似，因为距离不能为负，余弦值将在0到1之间。因此，更多的角度意味着降低相似性(视觉化也有意义)

这段Python代码是我实现算法的快速而肮脏的尝试:

import math
from collections import Counter

def build_vector(iterable1, iterable2):
    counter1 = Counter(iterable1)
    counter2 = Counter(iterable2)
    all_items = set(counter1.keys()).union(set(counter2.keys()))
    vector1 = [counter1[k] for k in all_items]
    vector2 = [counter2[k] for k in all_items]
    return vector1, vector2

def cosim(v1, v2):
    dot_product = sum(n1 * n2 for n1, n2 in zip(v1, v2) )
    magnitude1 = math.sqrt(sum(n ** 2 for n in v1))
    magnitude2 = math.sqrt(sum(n ** 2 for n in v2))
    return dot_product / (magnitude1 * magnitude2)


l1 = "Julie loves me more than Linda loves me".split()
l2 = "Jane likes me more than Julie loves me or".split()


v1, v2 = build_vector(l1, l2)
print(cosim(v1, v2))

这是我在c#中的实现。

using System;

namespace CosineSimilarity
{
    class Program
    {
        static void Main()
        {
            int[] vecA = {1, 2, 3, 4, 5};
            int[] vecB = {6, 7, 7, 9, 10};

            var cosSimilarity = CalculateCosineSimilarity(vecA, vecB);

            Console.WriteLine(cosSimilarity);
            Console.Read();
        }

        private static double CalculateCosineSimilarity(int[] vecA, int[] vecB)
        {
            var dotProduct = DotProduct(vecA, vecB);
            var magnitudeOfA = Magnitude(vecA);
            var magnitudeOfB = Magnitude(vecB);

            return dotProduct/(magnitudeOfA*magnitudeOfB);
        }

        private static double DotProduct(int[] vecA, int[] vecB)
        {
            // I'm not validating inputs here for simplicity.            
            double dotProduct = 0;
            for (var i = 0; i < vecA.Length; i++)
            {
                dotProduct += (vecA[i] * vecB[i]);
            }

            return dotProduct;
        }

        // Magnitude of the vector is the square root of the dot product of the vector with itself.
        private static double Magnitude(int[] vector)
        {
            return Math.Sqrt(DotProduct(vector, vector));
        }
    }
}

以@Bill Bell为例，在[R]中有两种方法

a = c(2,1,0,2,0,1,1,1)

b = c(2,1,1,1,1,0,1,1)

d = (a %*% b) / (sqrt(sum(a^2)) * sqrt(sum(b^2)))

或者利用crossprod()方法的性能…

e = crossprod(a, b) / (sqrt(crossprod(a, a)) * sqrt(crossprod(b, b)))

下面是一个简单的计算余弦相似度的Python代码:

import math

def dot_prod(v1, v2):
    ret = 0
    for i in range(len(v1)):
        ret += v1[i] * v2[i]
    return ret

def magnitude(v):
    ret = 0
    for i in v:
        ret += i**2
    return math.sqrt(ret)

def cos_sim(v1, v2):
    return (dot_prod(v1, v2)) / (magnitude(v1) * magnitude(v2))