Implementation of K Nearest Neighbors

Last Updated : 09 Nov, 2022

Prerequisite: K nearest neighbors

Introduction

Say we are given a data set of items, each having numerically valued features (like Height, Weight, Age, etc). If the count of features is n, we can represent the items as points in an n-dimensional grid. Given a new item, we can calculate the distance from the item to every other item in the set. We pick the k closest neighbors and we see where most of these neighbors are classified in. We classify the new item there.
So the problem becomes how we can calculate the distances between items. The solution to this depends on the data set. If the values are real we usually use the Euclidean distance. If the values are categorical or binary, we usually use the Hamming distance.
Algorithm:

Given a new item:     1. Find distances between new item and all other items     2. Pick k shorter distances     3. Pick the most common class in these k distances     4. That class is where we will classify the new item

Reading Data

Let our input file be in the following format:

Height, Weight, Age, Class 1.70, 65, 20, Programmer 1.90, 85, 33, Builder 1.78, 76, 31, Builder 1.73, 74, 24, Programmer 1.81, 75, 35, Builder 1.73, 70, 75, Scientist 1.80, 71, 63, Scientist 1.75, 69, 25, Programmer

Each item is a line and under “Class” we see where the item is classified in. The values under the feature names (“Height” etc.) are the value the item has for that feature. All the values and features are separated by commas.
Place these data files in the working directory data2 and data. Choose one and paste the contents as-is into a text file named data.
We will read from the file (named “data.txt”) and we will split the input by lines:

f = open('data.txt', 'r'); lines = f.read().splitlines(); f.close();

The first line of the file holds the feature names, with the keyword “Class” at the end. We want to store the feature names into a list:

# Split the first line by commas, # remove the first element and  # save the rest into a list. The # list now holds the feature  # names of the data set. features = lines[0].split(', ')[:-1];

Then we move on to the data set itself. We will save the items into a list, named items, whose elements are dictionaries (one for each item). The keys to these item-dictionaries are the feature names, plus “Class” to hold the item class. In the end, we want to shuffle the items in the list (this is a safety measure, in case the items are in a weird order).

Python3

   items = [];
 
for i in range(1, len(lines)):
     
    line = lines[i].split(', ');
 
    itemFeatures = {"Class" : line[-1]};
 
    # Iterate through the features
    for j in range(len(features)):
     
        # Get the feature at index j
        f = features[j]; 
            
        # The first item in the line
        # is the class, skip it
        v = float(line[j]);
         
        # Add feature to dict
        itemFeatures[f] = v; 
     
    # Append temp dict to items
    items.append(itemFeatures); 
     
shuffle(items);
 
  

Classifying the data

With the data stored into items, we now start building our classifier. For the classifier, we will create a new function, Classify. It will take as input the item we want to classify, the items list, and k, the number of the closest neighbors.
If k is greater than the length of the data set, we do not go ahead with the classifying, as we cannot have more closest neighbors than the total amount of items in the data set. (alternatively, we could set k as the items length instead of returning an error message)

if(k > len(Items)):                  # k is larger than list         # length, abort         return "k larger than list length";

We want to calculate the distance between the item to be classified and all the items in the training set, in the end keeping the k shortest distances. To keep the current closest neighbors we use a list, called neighbors. Each element in the least holds two values, one for the distance from the item to be classified and another for the class the neighbor is in. We will calculate distance via the generalized Euclidean formula (for n dimensions). Then, we will pick the class that appears most of the time in neighbors and that will be our pick. In code:

Python3

   def Classify(nItem, k, Items):
    if(k > len(Items)):
         
        # k is larger than list
        # length, abort
        return "k larger than list length";
     
    # Hold nearest neighbors.
    # First item is distance, 
    # second class
    neighbors = [];
 
    for item in Items:
       
        # Find Euclidean Distance
        distance = EuclideanDistance(nItem, item);
 
        # Update neighbors, either adding
        # the current item in neighbors 
        # or not.
        neighbors = UpdateNeighbors(neighbors, item, distance, k);
 
    # Count the number of each
    # class in neighbors
    count = CalculateNeighborsClass(neighbors, k);
 
    # Find the max in count, aka the
    # class with the most appearances.
    return FindMax(count);
 
  

The external functions we need to implement are EuclideanDistance, UpdateNeighbors, CalculateNeighborsClass, and FindMax.

Finding Euclidean Distance

The generalized Euclidean formula for two vectors x and y is this:

distance = sqrt{(x_{1}-y_{1})^2 + (x_{2}-y_{2})^2 + ... + (x_{n}-y_{n})^2}

In code:

Python3

   def EuclideanDistance(x, y):
     
    # The sum of the squared 
    # differences of the elements
    S = 0; 
     
    for key in x.keys():
        S += math.pow(x[key]-y[key], 2);
 
    # The square root of the sum
    return math.sqrt(S);
 
  

Updating Neighbors

We have our neighbors list (which should at most have a length of k) and we want to add an item to the list with a given distance. First, we will check if neighbors have a length of k. If it has less, we add the item to it regardless of the distance (as we need to fill the list up to k before we start rejecting items). If not, we will check if the item has a shorter distance than the item with the max distance in the list. If that is true, we will replace the item with max distance with the new item.
To find the max distance item more quickly, we will keep the list sorted in ascending order. So, the last item in the list will have the max distance. We will replace it with a new item and we will sort it again.
To speed this process up, we can implement an Insertion Sort where we insert new items in the list without having to sort the entire list. The code for this though is rather long and, although simple, will bog the tutorial down.

Python3

   def UpdateNeighbors(neighbors, item, distance, k):
     
    if(len(neighbors) > distance):
             
            # If yes, replace the last
            # element with new item
            neighbors[-1] = [distance, item["Class"]];
            neighbors = sorted(neighbors);
 
    return neighbors;

CalculateNeighborsClass

Here we will calculate the class that appears most often in neighbors. For that, we will use another dictionary, called count, where the keys are the class names appearing in neighbors. If a key doesn’t exist, we will add it, otherwise, we will increment its value.

Python3

   def CalculateNeighborsClass(neighbors, k):
    count = {};
     
    for i in range(k):
         
        if(neighbors[i][1] not in count):
         
            # The class at the ith index
            # is not in the count dict.
            # Initialize it to 1.
            count[neighbors[i][1]] = 1;
        else:
             
            # Found another item of class 
            # c[i]. Increment its counter.
            count[neighbors[i][1]] += 1;
 
    return count;
 
  

FindMax

We will input to this function the dictionary count we build in CalculateNeighborsClass and we will return its max.

Python3

   def FindMax(countList):
     
    # Hold the max
    maximum = -1;
     
    # Hold the classification
    classification = ""; 
     
    for key in countList.keys():
       
        if(countList[key] > maximum):
            maximum = countList[key];
            classification = key;
 
    return classification, maximum;

Conclusion

With that, this kNN tutorial is finished.
You can now classify new items, setting k as you see fit. Usually, for k an odd number is used, but that is not necessary. To classify a new item, you need to create a dictionary with keys the feature names, and the values that characterize the item. An example of classification:

newItem = {'Height' : 1.74, 'Weight' : 67, 'Age' : 22}; print Classify(newItem, 3, items);

The complete code of the above approach is given below:-

Python3

   # Python Program to illustrate
# KNN algorithm
 
# For pow and sqrt
import math 
from random import shuffle
 
###_Reading_### def ReadData(fileName):
 
    # Read the file, splitting by lines
    f = open(fileName, 'r')
    lines = f.read().splitlines()
    f.close()
 
    # Split the first line by commas, 
    # remove the first element and save
    # the rest into a list. The list 
    # holds the feature names of the 
    # data set.
    features = lines[0].split(', ')[:-1]
 
    items = []
 
    for i in range(1, len(lines)):
         
        line = lines[i].split(', ')
 
        itemFeatures = {'Class': line[-1]}
 
        for j in range(len(features)):
             
            # Get the feature at index j
            f = features[j]  
 
            # Convert feature value to float
            v = float(line[j]) 
             
             # Add feature value to dict
            itemFeatures[f] = v
         
        items.append(itemFeatures)
 
    shuffle(items)
 
    return items
 
 
###_Auxiliary Function_### def EuclideanDistance(x, y):
     
    # The sum of the squared differences
    # of the elements
    S = 0 
     
    for key in x.keys():
        S += math.pow(x[key] - y[key], 2)
 
    # The square root of the sum
    return math.sqrt(S)
 
def CalculateNeighborsClass(neighbors, k):
    count = {}
 
    for i in range(k):
        if neighbors[i][1] not in count:
 
            # The class at the ith index is
            # not in the count dict. 
            # Initialize it to 1.
            count[neighbors[i][1]] = 1
        else:
 
            # Found another item of class 
            # c[i]. Increment its counter.
            count[neighbors[i][1]] += 1
 
    return count
 
def FindMax(Dict):
 
    # Find max in dictionary, return 
    # max value and max index
    maximum = -1
    classification = ''
 
    for key in Dict.keys():
         
        if Dict[key] > maximum:
            maximum = Dict[key]
            classification = key
 
    return (classification, maximum)
 
 
###_Core Functions_### def Classify(nItem, k, Items):
 
    # Hold nearest neighbours. First item
    # is distance, second class
    neighbors = []
 
    for item in Items:
 
        # Find Euclidean Distance
        distance = EuclideanDistance(nItem, item)
 
        # Update neighbors, either adding the
        # current item in neighbors or not.
        neighbors = UpdateNeighbors(neighbors, item, distance, k)
 
    # Count the number of each class 
    # in neighbors
    count = CalculateNeighborsClass(neighbors, k)
 
    # Find the max in count, aka the
    # class with the most appearances
    return FindMax(count)
 
 
def UpdateNeighbors(neighbors, item, distance,
                                          k, ):
    if len(neighbors) < k:
 
        # List is not full, add 
        # new item and sort
        neighbors.append([distance, item['Class']])
        neighbors = sorted(neighbors)
    else:
 
        # List is full Check if new 
        # item should be entered
        if neighbors[-1][0] > distance:
 
            # If yes, replace the 
            # last element with new item
            neighbors[-1] = [distance, item['Class']]
            neighbors = sorted(neighbors)
 
    return neighbors
 
###_Evaluation Functions_### def K_FoldValidation(K, k, Items):
     
    if K > len(Items):
        return -1
 
    # The number of correct classifications
    correct = 0 
     
    # The total number of classifications
    total = len(Items) * (K - 1)  
     
    # The length of a fold
    l = int(len(Items) / K)  
 
    for i in range(K):
 
        # Split data into training set
        # and test set
        trainingSet = Items[i * l:(i + 1) * l]
        testSet = Items[:i * l] + Items[(i + 1) * l:]
 
        for item in testSet:
            itemClass = item['Class']
 
            itemFeatures = {}
 
            # Get feature values
            for key in item:
                if key != 'Class':
 
                    # If key isn't "Class", add 
                    # it to itemFeatures
                    itemFeatures[key] = item[key]
 
            # Categorize item based on
            # its feature values
            guess = Classify(itemFeatures, k, trainingSet)[0]
 
            if guess == itemClass:
 
                # Guessed correctly
                correct += 1
 
    accuracy = correct / float(total)
    return accuracy
 
 
def Evaluate(K, k, items, iterations):
 
    # Run algorithm the number of
    # iterations, pick average
    accuracy = 0
     
    for i in range(iterations):
        shuffle(items)
        accuracy += K_FoldValidation(K, k, items)
 
    print accuracy / float(iterations)
 
 
###_Main_### def main():
    items = ReadData('data.txt')
 
    Evaluate(5, 5, items, 100)
 
if __name__ == '__main__':
    main()
 
  

Output:

0.9375

The output can vary from machine to machine. The code includes a Fold Validation function, but it is unrelated to the algorithm, it is there for calculating the accuracy of the algorithm.

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MrDupin

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