# Understanding and Calculating Cosine Similarity

This article is a continuation from the [previous article](https://www.azureguru.net/dot-product-and-cosine-similarity) where we explored the mathematical foundation of `dot product` and `cosine similarity` and calculated the values for `cos(θ)` and `θ` of a two dimensional vector using `cosine similarity`.

$$A = 3i + 5j \;\; \text{and } \; B = 4i +8j$$

The values obtained from the calculation for `cos(θ)` and `θ` were `0.9970` and `4.4°` respectively.

In this article we would dive into understanding the different classifications of cosine similarities and practical implementation of them.

It is important to understand that similarity is measured between a value of `+1` and `-1`.Smaller angles between the vectors results in larger cosine values. In the example of the [previous article](https://www.azureguru.net/dot-product-and-cosine-similarity) an angle of `4.4°` resulted in a cosine value of less than `1`.

### Similar Vectors

Similar vectors are vectors that have the same direction (they are parallel) but may differ in magnitude. The value of `θ` between the vectors is close `O` and the `cos(θ)` is close to `1`.The following schematic represents an Similar vector.

![](https://cdn.hashnode.com/res/hashnode/image/upload/v1725125343461/da4bc5be-2782-4100-865e-df5f6ed3e04d.png align="center")

Lets take an example of a Similar vector and see how to check if a given vector is classifies as a similar vector.

Consider the following three dimensional vector

$$A=4i +1j-3k\;\text{and}\;B=8i+2j-6k$$

To verify if these vectors are similar , it is necessary to check if `B` is a positive scalar multiple of `A` i.e.

$$B=kA$$

To find `k`

$$8i+2j-6k=k(4i +1j-3k)$$

Comparing the elements of the vector we get

$$B_i=2A_i\;\text{i.e}\;8=2*4$$

$$B_j=2A_j\;\text{i.e}\;2=2*1$$

$$B_k=2A_k\;\text{i.e}\;6=2*3$$

So `B` is a positive scalar multiple of `A` and hence the vectors are similar (i.e., they point in the same direction).

Lets calculate the angle between them to reconfirm if they are indeed similar using the `cosine` formula

$$cos(θ)= \frac{∣A∣∣B∣}{A⋅B} ​$$

The dot product of the vectors is

$$​A⋅B=(2)(4)+(3)(6)+(−1)(−2)$$

$$A⋅B=8+18+2=28$$

while its magnitude is

$$∣A∣= \sqrt{ (2)^2 +(3) ^2 +(−1)^2 } = \sqrt{4+9+1} ​ = \sqrt{14}$$

$$∣ 𝐵 ∣ = \sqrt{( 4 ) ^2 + ( 6 ) ^2 + ( − 2 ) ^ 2} = \sqrt{16+36+4} = \sqrt{56}$$

value of `cos(θ)` is

$$cos(θ)= \frac{28}{\sqrt{14 * 56}}=1$$

and value of `θ` is

$$θ=cos −1 (1)=0^∘$$

Given that the value `cos(θ)` is `1` and value of `θ` is `O`, this reconfirms that the vector is indeed a similar vector.

### Orthogonal Vectors

Orthogonal vectors are vectors where the angle between the vectors is almost `90°` and value of `cos(θ)` is `O` and the dot product of every pair of the vector is `O` which corresponds to an angle of exactly `90°` between them.

The following diagram represents an Orthogonal vector.

![](https://cdn.hashnode.com/res/hashnode/image/upload/v1725125156427/86bb4ec4-8e4c-4472-890d-2a91861cb3dd.png align="center")

Lets consider a three dimensional vector

$$A=i+2j+3k\;\text{and}\;B=4i+5j−\frac{14}{3}k$$

The value of the dot product is

$$A⋅B=1×4+2×5+3×(− \frac{14}3 ​ )$$

$$A⋅B=4+10−14=0$$

The dot product of the vector is `O`. Lets reconfirm it by calculating its `cos(θ)` and the `0` value.

The magnitude of `A` is

$$∣A∣= \sqrt{(1)^2 +(2)^2 +(3)^2}=\sqrt{14}$$

and magnitude of `B` is

$$∣B∣= \sqrt{ (4)^ 2 +(5)^ 2 +(− \frac{14}{3} ​ )^2}=\frac{\sqrt{565}}{3}$$

Calculate `cos(θ)`

$$cos(θ)= \frac{A⋅B}{∣A∣∣B∣} = \frac{0}{\sqrt{14 * \frac{565}{3}}} ​ =0$$

​ and the angle `θ` between the two vectors is

$$θ=cos^{−1} (0)=90^∘$$

This reconfirms that the vector is an orthogonal vector.

### Opposite Vectors

Opposite vectors are vector types that have the same magnitude but the directions are opposite and the angle `0` between the vectors is close to `180°` and value of`cos(θ)` is close to `-1`

The following diagram represents an opposite vector.

![](https://cdn.hashnode.com/res/hashnode/image/upload/v1725125247709/8325a0e5-5080-49ea-acad-2db9dde8572a.png align="center")

Consider the following vector

$$A=3i+4j−5k\;\;\text{and }\; B=−3i−4j+5k$$

To confirm if these are vectors are opposite vectors we have to check if `B=-A`.

$$B=−1×(3i+4j−5k)=−3i−4j+5k$$

Since `B` is negative of `A` its safe to say that `A` and `B` are opposite vectors.

To reconfirm it we can calculate the value of `θ` and `cos(θ)`

The value of the dot product is

$$A⋅B=(7)(−7)+(−2)(2)+(4)(−4)$$

$$A⋅B=−49−4−16=−69$$

and the magnitude of `A` is

$$∣A∣= \sqrt{(7)^2 +(−2)^2 +(4)^2} ​ = 49+4+16 ​ = 69 ​$$

and the magnitude of `B` is

$$∣B∣= \sqrt{ (−7) 2 +(2) 2 +(−4) 2} ​ = 49+4+16 ​ = 69$$

Calculate `cos(θ)`

$$cos(θ)= \frac{A⋅B}{∣A∣∣B∣} = \frac{-69}{\sqrt{69} *\sqrt{69} } ​ =-1$$

​ and the angle `θ` between the two vectors is

$$θ=cos −1 (−1)=180^∘$$

The vectors have have an angle of `180°` between them and `cos(θ)` is `-1`.This confirms that the vectors are indeed opposite vectors.

### Lets Test

We will use a python script to test the cosine similarity. The vectors are the same vectors that were used to manually compute the cosine similarity in the [previous article](https://www.azureguru.net/dot-product-and-cosine-similarity).The calculated value was `0.99`.

```python
import numpy as np

#numpy arrays
v1 = np.array((3,5))
v2 = np.array((4,8))

dot_product = np.dot(v1, v2)

x_magnitude = np.sqrt(np.sum(v1**2)) 
y_magnitude = np.sqrt(np.sum(v2**2))
    
cos_similarity = dot_product / (x_magnitude * y_magnitude)

print(cos_similarity)
```

the output is

![](https://cdn.hashnode.com/res/hashnode/image/upload/v1725139048715/25413cc8-a724-4589-8e11-2c456b0f40da.png align="left")

We could also use the `Scikit-learn’s` inbuilt `cosine_similarity` function

```python
import numpy as np
from sklearn.metrics.pairwise import cosine_similarity

A = np.array([[3, 5]])
B = np.array([[4, 8]])

cos_similarity = cosine_similarity(A, B)
print(cos_similarity)
```

The output is similar to the previous output

![](https://cdn.hashnode.com/res/hashnode/image/upload/v1725139207836/20416fe6-297e-492d-9a75-f122f8fab145.png align="left")

Lets test it on some phrases. The phrases used below are the same phrases that was used in this [article](https://www.azureguru.net/euclidean-distance-similarity-metric) to calculate the Euclidean distance between the vectors.

```python
import numpy as np

def ret_cos_similarity(v1, v2):

    dot_product = np.dot(v1, v2)
    x_magnitude = np.sqrt(np.sum(v1**2)) 
    y_magnitude = np.sqrt(np.sum(v2**2))
     
    cos_similarity = dot_product / (x_magnitude * y_magnitude)
 return cos_similarity


corpus = [ 'Paris is capital of France',
           'Boeing and Airbus are two companies that build aircrafts',
           'Rome is capital of Italy'  ]

from sklearn.feature_extraction.text import CountVectorizer

V = CountVectorizer().fit_transform(corpus).toarray()

v1=(V[0, :])
v2=(V[1, :])
v3=(V[2, :])

print('Similarity between: ')
print('\tPhrase 1 and Phrase 2: ',ret_cos_similarity(v1,v2))
print('\tPhrase 2 and Phrase 3: ', ret_cos_similarity(v2,v3))
print('\tPhrase 1 and Phrase 3: ', ret_cos_similarity(v1,v3))
```

I created a custom cosine similarity function `ret_cos_similarity`

![](https://cdn.hashnode.com/res/hashnode/image/upload/v1725141133418/d6e281cb-443b-4094-b513-6ff57a7bd8d6.png align="left")

As expected `Phrase1` and `Phrase2` & `Phrase2` and `Phrase3` are `orthogonal vectors` given the calculated value of `cos(θ)` is `O` while `Phrase1` and `Phrase3` are `similar vectors` as the value of `cos(θ)` is closer to 1.

You could also use `Scikit-learn’s` inbuilt `cosine_similarity` function

```python
import numpy as np
from scipy.spatial import distance
from sklearn.metrics.pairwise import cosine_similarity

corpus = [ 'Paris is capital of France',
           'Boeing and Airbus are two companies that build aircrafts',
           'Rome is capital of Italy'  ]

from sklearn.feature_extraction.text import CountVectorizer

V = CountVectorizer().fit_transform(corpus).toarray()

v1=(V[0, :] )
v2=(V[1, :] )
v3=(V[2, :] )


print('Similarity between: ')
print('\tPhrase 1 and Phrase 2: ',cosine_similarity((v1,v2)))
print('\tPhrase 2 and Phrase 3: ', cosine_similarity((v2,v3)))
print('\tPhrase 1 and Phrase 3: ', cosine_similarity((v1,v3)))
```

The output is in form of a matrix which isnt that intuitive.

![](https://cdn.hashnode.com/res/hashnode/image/upload/v1725141601835/e04fc431-ef55-4b91-a6e7-236021990def.png align="left")

### Closing Notes

Cosine similarity is a powerful and widely-used metric for measuring the similarity between vectors in high-dimensional spaces. It is one of the most widely used metric for vector analysis. Hope this article has helped to provide some insights into the mathematical intricacies of cosine similarity and its practical usage.

Thank you for reading !!!