Can someone give an example of cosine similarity, in a very simple, graphical way?

Question

Cosine Similarity article on Wikipedia

Can you show the vectors here (in a list or something) and then do the math, and let us see how it works?

I\'m a begin

Answer 1

Two Vectors A and B exists in a 2D space or 3D space, the angle between those vectors is cos similarity.

If the angle is more (can reach max 180 degree) which is Cos 180=-1 and the minimum angle is 0 degree. cos 0 =1 implies the vectors are aligned to each other and hence the vectors are similar.

cos 90=0 (which is sufficient to conclude that the vectors A and B are not similar at all and since distance cant be negative, the cosine values will lie from 0 to 1. Hence, more angle implies implies reducing similarity (visualising also it makes sense)

Answer 2

For simplicity I am reducing the vector a and b:

Let :
    a : [1, 1, 0]
    b : [1, 0, 1]

Then cosine similarity (Theta):

 (Theta) = (1*1 + 1*0 + 0*1)/sqrt((1^2 + 1^2))* sqrt((1^2 + 1^2)) = 1/2 = 0.5

then inverse of cos 0.5 is 60 degrees.

Answer 3

This Python code is my quick and dirty attempt to implement the algorithm:

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))

Answer 4

import java.util.HashMap;
import java.util.HashSet;
import java.util.Map;
import java.util.Set;

/**
 * 
* @author Xiao Ma
* mail : 409791952@qq.com
*
*/
  public class SimilarityUtil {

public static double consineTextSimilarity(String[] left, String[] right) {
    Map<String, Integer> leftWordCountMap = new HashMap<String, Integer>();
    Map<String, Integer> rightWordCountMap = new HashMap<String, Integer>();
    Set<String> uniqueSet = new HashSet<String>();
    Integer temp = null;
    for (String leftWord : left) {
        temp = leftWordCountMap.get(leftWord);
        if (temp == null) {
            leftWordCountMap.put(leftWord, 1);
            uniqueSet.add(leftWord);
        } else {
            leftWordCountMap.put(leftWord, temp + 1);
        }
    }
    for (String rightWord : right) {
        temp = rightWordCountMap.get(rightWord);
        if (temp == null) {
            rightWordCountMap.put(rightWord, 1);
            uniqueSet.add(rightWord);
        } else {
            rightWordCountMap.put(rightWord, temp + 1);
        }
    }
    int[] leftVector = new int[uniqueSet.size()];
    int[] rightVector = new int[uniqueSet.size()];
    int index = 0;
    Integer tempCount = 0;
    for (String uniqueWord : uniqueSet) {
        tempCount = leftWordCountMap.get(uniqueWord);
        leftVector[index] = tempCount == null ? 0 : tempCount;
        tempCount = rightWordCountMap.get(uniqueWord);
        rightVector[index] = tempCount == null ? 0 : tempCount;
        index++;
    }
    return consineVectorSimilarity(leftVector, rightVector);
}

/**
 * The resulting similarity ranges from −1 meaning exactly opposite, to 1
 * meaning exactly the same, with 0 usually indicating independence, and
 * in-between values indicating intermediate similarity or dissimilarity.
 * 
 * For text matching, the attribute vectors A and B are usually the term
 * frequency vectors of the documents. The cosine similarity can be seen as
 * a method of normalizing document length during comparison.
 * 
 * In the case of information retrieval, the cosine similarity of two
 * documents will range from 0 to 1, since the term frequencies (tf-idf
 * weights) cannot be negative. The angle between two term frequency vectors
 * cannot be greater than 90°.
 * 
 * @param leftVector
 * @param rightVector
 * @return
 */
private static double consineVectorSimilarity(int[] leftVector,
        int[] rightVector) {
    if (leftVector.length != rightVector.length)
        return 1;
    double dotProduct = 0;
    double leftNorm = 0;
    double rightNorm = 0;
    for (int i = 0; i < leftVector.length; i++) {
        dotProduct += leftVector[i] * rightVector[i];
        leftNorm += leftVector[i] * leftVector[i];
        rightNorm += rightVector[i] * rightVector[i];
    }

    double result = dotProduct
            / (Math.sqrt(leftNorm) * Math.sqrt(rightNorm));
    return result;
}

public static void main(String[] args) {
    String left[] = { "Julie", "loves", "me", "more", "than", "Linda",
            "loves", "me" };
    String right[] = { "Jane", "likes", "me", "more", "than", "Julie",
            "loves", "me" };
    System.out.println(consineTextSimilarity(left,right));
}
}

Answer 5

Simple JAVA code to calculate cosine similarity

/**
   * Method to calculate cosine similarity of vectors
   * 1 - exactly similar (angle between them is 0)
   * 0 - orthogonal vectors (angle between them is 90)
   * @param vector1 - vector in the form [a1, a2, a3, ..... an]
   * @param vector2 - vector in the form [b1, b2, b3, ..... bn]
   * @return - the cosine similarity of vectors (ranges from 0 to 1)
   */
  private double cosineSimilarity(List<Double> vector1, List<Double> vector2) {

    double dotProduct = 0.0;
    double normA = 0.0;
    double normB = 0.0;
    for (int i = 0; i < vector1.size(); i++) {
      dotProduct += vector1.get(i) * vector2.get(i);
      normA += Math.pow(vector1.get(i), 2);
      normB += Math.pow(vector2.get(i), 2);
    }
    return dotProduct / (Math.sqrt(normA) * Math.sqrt(normB));
  }

Answer 6

I'm guessing you are more interested in getting some insight into "why" the cosine similarity works (why it provides a good indication of similarity), rather than "how" it is calculated (the specific operations used for the calculation). If your interest is in the latter, see the reference indicated by Daniel in this post, as well as a related SO Question.

To explain both the how and even more so the why, it is useful, at first, to simplify the problem and to work only in two dimensions. Once you get this in 2D, it is easier to think of it in three dimensions, and of course harder to imagine in many more dimensions, but by then we can use linear algebra to do the numeric calculations and also to help us think in terms of lines / vectors / "planes" / "spheres" in n dimensions, even though we can't draw these.

So, in two dimensions: with regards to text similarity this means that we would focus on two distinct terms, say the words "London" and "Paris", and we'd count how many times each of these words is found in each of the two documents we wish to compare. This gives us, for each document, a point in the the x-y plane. For example, if Doc1 had Paris once, and London four times, a point at (1,4) would present this document (with regards to this diminutive evaluation of documents). Or, speaking in terms of vectors, this Doc1 document would be an arrow going from the origin to point (1,4). With this image in mind, let's think about what it means for two documents to be similar and how this relates to the vectors.

VERY similar documents (again with regards to this limited set of dimensions) would have the very same number of references to Paris, AND the very same number of references to London, or maybe, they could have the same ratio of these references. A Document, Doc2, with 2 refs to Paris and 8 refs to London, would also be very similar, only with maybe a longer text or somehow more repetitive of the cities' names, but in the same proportion. Maybe both documents are guides about London, only making passing references to Paris (and how uncool that city is ;-) Just kidding!!!.

Now, less similar documents may also include references to both cities, but in different proportions. Maybe Doc2 would only cite Paris once and London seven times.

Back to our x-y plane, if we draw these hypothetical documents, we see that when they are VERY similar, their vectors overlap (though some vectors may be longer), and as they start to have less in common, these vectors start to diverge, to have a wider angle between them.

By measuring the angle between the vectors, we can get a good idea of their similarity, and to make things even easier, by taking the Cosine of this angle, we have a nice 0 to 1 or -1 to 1 value that is indicative of this similarity, depending on what and how we account for. The smaller the angle, the bigger (closer to 1) the cosine value, and also the higher the similarity.

At the extreme, if Doc1 only cites Paris and Doc2 only cites London, the documents have absolutely nothing in common. Doc1 would have its vector on the x-axis, Doc2 on the y-axis, the angle 90 degrees, Cosine 0. In this case we'd say that these documents are orthogonal to one another.

Adding dimensions:
With this intuitive feel for similarity expressed as a small angle (or large cosine), we can now imagine things in 3 dimensions, say by bringing the word "Amsterdam" into the mix, and visualize quite well how a document with two references to each would have a vector going in a particular direction, and we can see how this direction would compare to a document citing Paris and London three times each, but not Amsterdam, etc. As said, we can try and imagine the this fancy space for 10 or 100 cities. It's hard to draw, but easy to conceptualize.

I'll wrap up just by saying a few words about the formula itself. As I've said, other references provide good information about the calculations.

First in two dimensions. The formula for the Cosine of the angle between two vectors is derived from the trigonometric difference (between angle a and angle b):

cos(a - b) = (cos(a) * cos(b)) + (sin (a) * sin(b))

This formula looks very similar to the dot product formula:

Vect1 . Vect2 =  (x1 * x2) + (y1 * y2)

where cos(a) corresponds to the x value and sin(a) the y value, for the first vector, etc. The only problem, is that x, y, etc. are not exactly the cos and sin values, for these values need to be read on the unit circle. That's where the denominator of the formula kicks in: by dividing by the product of the length of these vectors, the x and y coordinates become normalized.