# K-Nearest Neighbor from Scratch in Python

Posted by Kenzo Takahashi on Wed 06 January 2016

We are going to implement K-nearest neighbor(or k-NN for short) classifier from scratch in Python. k-NN is probably the easiest-to-implement ML algorithm. Besides, unlike other algorithms(e.g. Neural Network, Support Vector Machine), you do not need to know much math to understand it.

For this tutorial, I assume you know the followings:

• Python(list comprehension, basic OOP)
• Numpy
• Basic Linear Algebra
• Basic machine learning concepts

## Part 1: How k-NN Works

I will explain k-NN, but this is by no means a thorough introduction to the subject. It would be better if you already know how it works and just wants to implement it.

### The Big Picture

The idea of k-NN is pretty intuitive: If you want to know the class of a sample, look at its neighbors.

Let me give you an example. We are going to use the famous Iris flower data set for this tutorial. It has 4 features (Sepal length, Sepal width, Petal length, Petal width) and 3 classes(Setosa, Virginica, Versicolor). You want to predict the class of a flower(flower A) based on the 4 features. If a similar flower(flower B) is labeled as Setosa, it's quite likely that flower A is also Setosa. By similar, I mean they both have similar sepal length, sepal width, and so on. Flower A's most similar flower is called the nearest neighbor.

You can conclude flower A's class just by looking at its most similar flower(the nearest neighbor), but you can also look a little further and decide the class based on the 3, 5, or K nearest neighbors. If K is larger than 1, you simply take the majority. That's why K is usually an odd number in order to prevent tie.

### Scikit-learn Style

My implementation of k-NN closely follows the scikit-learn style. scikit-learn is the most widely used ML library for python. The nice thing about scikit-learn is its consistent API. It makes the design of your algorithm really easy. Take a look at the toy example.

from sklearn.neighbors import KNeighborsClassifier
import numpy as np
X_train = np.array([[-1, -1], [-2, -1], [-3, -2], [1, 1], [2, 1], [3, 2]])
y_train = np.array([1,1,1,0,0,0])
neighbor = KNeighborsClassifier(n_neighbors=3)
neighbor.fit(X_train, y_train)
X_test = np.array([[1, 0], [-2, -2]])
print(neighbor.predict(X_test))


Output:

[0,1]


Scikit-learn calls ML algorithms estimators. Every estimator is a class. You first initialize the class with parameters specific to that estimator. Then you call the fit method with a set of samples(X_train) and target classes(y_train). As you can see, X_train is a 2D array, or matrix. In this example, we have 6 samples, each of which consist of 2 features. y_train is an array of target class corresponding to X_train. You might be wondering why X is in uppercase and y is in lowercase. In Linear Algebra, we usually denote matrices by uppercase letters and vectors by lowercase letters, so it's good to stick to the convention. Then you call the predict method with a test set(X_test). The output is a list of predicted classes.

### Calculating k-NN by Hand

Let's calculate k-NN by hand using the above example. In my experience, hand calculation really helps me understand the algorithm. Unlike most algorithms, k-NN does nothing at fit. All the work happens at predict. The first test sample is [1,0]. To measure the similarity, we simply calculate the difference for each feature and add them up. This is called Manhattan distance. The Manhattan distance between vector $$p$$ and $$q$$ can be written as follows:

$$d(p, q) = \displaystyle\sum_{i=1}^{n} |p_i - q_i|$$

The first training sample is [-1, -1]. So the Manhattan distance is

$$|-1 - 1| + |-1 - 0| = 2 + 1 = 3$$

Notice the absolute value notation. If you don't take the absolute value, positive value and negative value cancel out.

Similarly, the Manhattan distances of the rest of the training data are 4, 6, 1, 2, 4, respectively. K = 3 in this example, so we pick the 3 nearest neighbors. The smallest value means the nearest, so the nearest neighbor is [1,1] with distance = 1. Its corresponding class is 0. The second and third nearest neighbors are [2,1], [-1,-1], whose classes are 0, 1. We have two 0's and one 1 so the prediction is 0.

## Part 2: A Simple k-NN

Now that you understand the algorithm, it's time to write some code! I'm using python3. If you want to use python2, add this line at the beginning of your file and everything will work fine.

from __future__ import division


### The Initial Code

__init__, fit, predict are pretty easy, so we start with them.

class KNeighborsClassifier(object):
def __init__(self, n_neighbors=5):
self.n_neighbors = n_neighbors

def fit(self, X, y):
self.X = X
self.y = y
return self

def _predict_one(self, test):
return 1

def predict(self, X):
return [self._predict_one(i) for i in X]


For now, __init__ takes one parameter, n_neighbors. We just save it to the instance variable. As I said before, fit doesn't do anything in k-NN. Again, we just save X and y to the instance variables for later use. predict in scikit-learn predicts not just 1 instance, but multiple instances. So we have _predict_one for predicting 1 instance. predict calls _predict_one for each instance and puts them in a list. By the way, I am using the term data, sample, and instance loosely. They pretty much mean the same thing here. For now, _predict_one always returns 1, but let's test the code.

import numpy as np
X = np.array([[-1, -1], [-2, -1], [-3, -2], [1, 1], [2, 1], [3, 2]])
y = np.array([1,1,1,0,0,0])
neighbor = KNeighborsClassifier()
neighbor.fit(X, y)
print(neighbor.predict(np.array([[1, 0], [-2, -2]])))


Output:

[1,1]


It is quite common to initialize the class and fit in 1 line.

neighbor = KNeighborsClassifier().fit(X, y)


Now let's work on _predict_one.

### Manhattan Distance

#### Exercise 1

The first thing you have to do is calculate distance. The method _distance takes two numpy arrays data1, data2, and returns the Manhattan distance between the two. This shouldn't be that hard, so I want you to write it by yourself. Dont' worry, I will show you my solution in a moment. Hint: This can be done in 1 line of code:

def _distance(self, data1, data2):
"""returns Manhattan distance"""

neighbor = KNeighborsClassifier()
print(neighbor._distance(np.array([-1, -1]),np.array([1, -2])))


#### Solution

Here is my solution:

def _distance(self, data1, data2):
return sum(abs(data1 - data2))


Here, I'm taking advantage of numpy array to use element-wise operations(- and abs). If you gave lists instead, you would get TypeError:

TypeError: unsupported operand type(s) for -: 'list' and 'list'


By the way, Scikit-learn accepts any array-like objects in fit and predict, so it could be a list, or it could be a numpy array. But it converts it to numpy array under the hood because it makes calculation simpler. On the other hand, my code only accepts numpy array because I don't want to make it complicated. If you are going to use it just for yourself, it shouldn't be a big deal.

### Getting K nearest neighbors

Now that you have distance, we can get K nearest neighbors. To do that, first we naively sort the samples. If you have 1000 examples and just want get, say, 3 smallest values, it is unnecessary to sort them all. But our goal is to understand how k-NN works, not write efficient code:

def _predict_one(self, test):
distances = sorted((self._distance(x, test), y) for x, y in zip(self.X, self.y))
return distances


It's important to sort a pair of distance and a target class. Otherwise you won't be able to find a corresponding class value. When sorting tuples or lists instead of numbers, sorted function sorts by the first element. The first element of our tuple is distance, so we are good. I'm breaking _predict_one method temporarily; it no longer returns a number. But that's OK. Let's test it:

X = np.array([[-1, -1], [-2, -1], [-3, -2], [1, 1], [2, 1], [3, 2]])
y = np.array([1,1,1,0,0,0])
neighbor = KNeighborsClassifier().fit(X, y)
print(neighbor._predict_one(np.array([1, 0])))


Output:

[(1, 0), (2, 0), (3, 1), (4, 0), (4, 1), (6, 1)]


It works! It matches what we calculated by hand earlier.

As a side note, when applying a function to a list/dictionary comprehension, you don't need square brackets or curly braces. Using them would make it ugly and hard to read:

distances = sorted([(self._distance(x, test), y) for x, y in zip(self.X, self.y)])


Once we sorted the distances, we can take the first k elements:

def _predict_one(self, test):
distances = sorted((self._distance(x, test), y) for x, y in zip(self.X, self.y))
neighbors = distances[:self.n_neighbors]
return neighbors


### Uniform Weights

#### Exercise 2

After we take the k nearest neighbors, we no longer care about their distance. What we care is how many of them belong to each class. So _compute_weights method changes distances to 1. This is fairly straightforward. Again, this can be done is 1 line of code:

def _compute_weights(self, distances):
"""computes uniform weights"""

neighbor = KNeighborsClassifier()
print(neighbor._compute_weights(np.array([(1, 0), (2, 0), (3, 1)])))


Output:

[(1, 0), (1, 0), (1, 1)]


#### Solution

Here is my solution:

def _compute_weights(self, distances):
return [(1, y) for d, y in distances]


Now we call the method with neighbors as the argument:

def _predict_one(self, test):
distances = sorted((self._distance(x, test), y) for x, y in zip(self.X, self.y))
neighbors = distances[:self.n_neighbors]
weights = self._compute_weights(neighbors)
return weights


You can shorten the code a little bit as long as it's readable:

weights = self._compute_weights(distances[:self.n_neighbors])


### Predicting 1 Instance

At this point, our k-NN class looks like this:

import numpy as np

class KNeighborsClassifier(object):
def __init__(self, n_neighbors=5):
self.n_neighbors = n_neighbors

def fit(self, X, y):
self.X = X
self.y = y
return self

def _distance(self, data1, data2):
return sum(abs(data1 - data2))

def _compute_weights(self, distances):
return [(1, y) for d, y in distances]

def _predict_one(self, test):
distances = sorted((self._distance(x, test), y) for x, y in zip(self.X, self.y))
weights = self._compute_weights(distances[:self.n_neighbors])
return weights

def predict(self, X):
return [self._predict_one(i) for i in X]


We need to group the weights by class so that we can count them. To do that, we can use defaultdict:

from collections import defaultdict

def _predict_one(self, test):
distances = sorted((self._distance(x, test), y) for x, y in zip(self.X, self.y))
weights = self._compute_weights(distances[:self.n_neighbors])
weights_by_class = defaultdict(list)
for d, c in weights:
weights_by_class[c].append(d)
return weights_by_class

X = np.array([[-1, -1], [-2, -1], [-3, -2], [1, 1], [2, 1], [3, 2]])
y = np.array([1,1,1,0,0,0])
neighbor = KNeighborsClassifier().fit(X, y)
print(neighbor._predict_one(np.array([1, 0])))


Output:

defaultdict(<class 'list'>, {0: [1, 1, 1], 1: [1, 1]})


#### Exercise 3

We are almost done. Given the weights_by_class, we can predict the class. As always, this can be done in 1 line. The toy example we've been using is binary classification, but your code should also work with multiclass classification:

def _predict_one(self, test):
distances = sorted((self._distance(x, test), y) for x, y in zip(self.X, self.y))
weights = self._compute_weights(distances[:self.n_neighbors])
weights_by_class = defaultdict(list)
for d, c in weights:
weights_by_class[c].append(d)

X = np.array([[-1, -1], [-2, -1], [-3, -2], [1, 1], [2, 1], [3, 2]])
y = np.array([1,1,1,0,0,0])
neighbor = KNeighborsClassifier().fit(X, y)
print(neighbor._predict_one(np.array([1, 0])))


This should print 0

#### Solution

Here is mine:

return max((sum(val), key) for key, val in weights_by_class.items())[1]


There is quite a lot going on in this line. Some people might prefer to break it down to multiple lines, which is totally fine(In fact, I often write code like this and make it shorter later):

counts = [(sum(val), key) for key, val in weights_by_class.items()]
majority = max(counts)
return majority[1]


First we count how many neighbors belong to each class using sum, then take the majority with max. Just like sorted, when dealing with tuples, max looks at the first element by default. Finally, you take the second element of the tuple, which is the predicted class.

That's it!

import numpy as np
from collections import defaultdict

class KNeighborsClassifier(object):
def __init__(self, n_neighbors=5):
self.n_neighbors = n_neighbors

def fit(self, X, y):
self.X = X
self.y = y
return self

def _distance(self, data1, data2):
return sum(abs(data1 - data2))

def _compute_weights(self, distances):
return [(1, y) for d, y in distances]

def _predict_one(self, test):
distances = sorted((self._distance(x, test), y) for x, y in zip(self.X, self.y))
weights = self._compute_weights(distances[:self.n_neighbors])
weights_by_class = defaultdict(list)
for d, c in weights:
weights_by_class[c].append(d)
return max((sum(val), key) for key, val in weights_by_class.items())[1]

def predict(self, X):
return [self._predict_one(i) for i in X]


Let's test it.

X_train = np.array([[-1, -1], [-2, -1], [-3, -2], [1, 1], [2, 1], [3, 2]])
y_train = np.array([1,1,1,0,0,0])
neighbor = KNeighborsClassifier(n_neighbors=3)
neighbor.fit(X_train, y_train)
X_test = np.array([[1, 0], [-2, -2]])
print(neighbor.predict(X_test))


The output looks as you expect:

[0,1]


## Part 3: Some Improvements

If you look at the scikit-learn documentation, you see that KNeighborsClassifier takes a lot of parameters and has many more methods.

### Distance-Weighted k-NN

The way we pick a winner from k nearest neighbors is simply take the majority. But what happens in this case?

X = np.array([[1, 1], [4, 4], [5, 5]])
y = np.array([1,0,0])
neighbor = KNeighborsClassifier(n_neighbors=3).fit(X, y)
print(neighbor._predict_one(np.array([0, 0])))


According to our code, it will produce 0 because we have 2 neighbors of class 0 whereas we have only 1 neighbor of class 1. But those two neighbors of class 0 are further away. Intuitively, 1 is more likely to be correct in this case. So we would like to take the distances into account, giving closer points more weight. To do that, we take the inverse of a distance($$1/d$$). There is also $$1/(d^2)$$, which implies that further points matter even less than just the inverse, which makes the choice of K matters less. I don't know which one is better, but scikit-learn chooses the former, so I will stick to that.

First we need to add a new parameter, weights. The default is uniform.

def __init__(self, n_neighbors=5, weights='uniform'):
self.n_neighbors = n_neighbors
self.weights = weights


#### Exercise 4

Let's make some change to _compute_weights. I only left the main part for the exercise.

def _compute_weights(self, distances):
if self.weights == 'uniform':
return [(1, y) for d, y in distances]
elif self.weights == 'distance':
raise ValueError("weights not recognized: should be 'uniform' or 'distance'")

neighbor = KNeighborsClassifier()
print(neighbor._compute_weights(np.array([(1, 0), (2, 0), (3, 1)])))


Output:

[(1.0, 0), (0.5, 0), (0.3333333333333333, 1)]


#### Solution

This should be fairly straightforward:

return [(1/d, y) for d, y in distances]


Now this is classified as 1 instead of 0:

X = np.array([[1, 1], [4, 4], [5, 5]])
y = np.array([1,0,0])
neighbor = KNeighborsClassifier(n_neighbors=3, weights='distance').fit(X, y)
print(neighbor._predict_one(np.array([0, 0])))


There is one problem though. What happens if some distance is 0? You will get division by 0 error. Distance = 0 means two samples have exactly the same value on every feature. This might happen sometimes, especially if the sample size is big. Or, if you fit and predict with the same X to measure the training set accuracy, you will definitely get the error on every single instance(If you don't know why, pause and think about it).

So when at least one distance is 0, we will forget about weighted distance and assign 1 to distance = 0 and 0 to everything else. In other words, we compute uniform weights on those samples whose distance is 0.

#### Exercise 5

Let's implement the improved version of _compute_weights:

def _compute_weights(self, distances):
if self.weights == 'uniform':
return [(1, y) for d, y in distances]
elif self.weights == 'distance':
raise ValueError("weights not recognized: should be 'uniform' or 'distance'")

neighbor = KNeighborsClassifier(weights='distance')
print(neighbor._compute_weights([(0, 1),(0, 1),(3, 0),(0, 0)]))


Output:

[(1, 1), (1, 1), (1, 0)]


#### Solution

Here is mine:

matches = [(1, y) for d, y in distances if d == 0]
return matches if matches else [(1/d, y) for d, y in distances]


First it collects all the samples whose distance is 0 and assigns 1. If the matches contains something, it returns it. Otherwise, it just returns what we had previously.

### Euclidean Distance

We used Manhattan distance to calculate the distance between the two points. Another popular distance is Euclidean Distance. To calculate it, we square the difference for each feature, add them up, and take the square root of it.

$$d(p, q) = \sqrt{\displaystyle\sum_{i=1}^{n} (p_i - q_i)^2}$$

Let's add Euclidean distance to our code. The first thing we need to do is add a new parameter p to __init__:

def __init__(self, n_neighbors=5, p=2):
self.n_neighbors = n_neighbors
self.weights = weights
self.p = p


When p = 1, Manhattan distance is used, and when p = 2, Euclidean distance. The default is 2. You might think why we use numbers instead of something like 'manhattan' and 'euclidean' as we did on weights. The reason for this is that Manhattan distance and Euclidean distance are the special case of Minkowski distance. For arbitrary p, Minkowski distance is used in scikit-learn. It's beyond the scope of this tutorial, and we are going to only consider p = 1 or 2.

#### Exercise 6

Just like Manhattan distance, if you follow the formula, you will be fine. Hint: sqrt in math module does not work on numpy array. You can use numpy's own sqrt instead.

def _distance(self, data1, data2):
"""1: Manhattan, 2: Euclidean"""
if self.p == 1:
return sum(abs(data1 - data2))
elif self.p == 2:
raise ValueError("p not recognized: should be 1 or 2")

neighbor = KNeighborsClassifier(p=2)
print(neighbor._distance(np.array([-1, -1]),np.array([1, -2])))


The output should be $$2.2360679775$$, which is $$\sqrt{5}$$

#### Solution

Again, we can take advantage of numpy array:

return np.sqrt(sum((data1 - data2)**2))


### Score method

#### Exercise 7

We have one more method to write which is score. It's just the mean accuracy of the given test data. It calls predict, compares the output to y, and returns what fraction of them got right. Hint: You don't need list comprehension for this:

def score(self, X, y):

X_train = np.array([[-1, -1], [-2, -1], [-3, -2], [1, 1], [2, 1], [3, 2]])
y_train = np.array([1,1,1,0,0,0])
neighbor = KNeighborsClassifier().fit(X_train, y_train)
X_test = np.array([[1, 0], [-2, -2]])
y_test = np.array([0, 0])
print(neighbor.score(X_test, y_test))


The output should be 0.5

#### Solution

Thr obvious solution is this:

return sum(1 for p, t in zip(self.predict(X), y) if p == t) / len(y)


After I published this post, I realized that there is a much better way using the element-wise operation:

return sum(self.predict(X) == y) / len(y)


Now we have the improved version of k-NN.

import numpy as np
from collections import defaultdict

class KNeighborsClassifier(object):
def __init__(self, n_neighbors=5, weights='uniform', p=2):
self.n_neighbors = n_neighbors
self.weights = weights
self.p = p

def fit(self, X, y):
self.X = X
self.y = y
return self

def _distance(self, data1, data2):
"""1: Manhattan, 2: Euclidean"""
if self.p == 1:
return sum(abs(data1 - data2))
elif self.p == 2:
return np.sqrt(sum((data1 - data2)**2))
raise ValueError("p not recognized: should be 1 or 2")

def _compute_weights(self, distances):
if self.weights == 'uniform':
return [(1, y) for d, y in distances]
elif self.weights == 'distance':
matches = [(1, y) for d, y in distances if d == 0]
return matches if matches else [(1/d, y) for d, y in distances]
raise ValueError("weights not recognized: should be 'uniform' or 'distance'")

def _predict_one(self, test):
distances = sorted((self._distance(x, test), y) for x, y in zip(self.X, self.y))
weights = self._compute_weights(distances[:self.n_neighbors])
weights_by_class = defaultdict(list)
for d, c in weights:
weights_by_class[c].append(d)
return max((sum(val), key) for key, val in weights_by_class.items())[1]

def predict(self, X):
return [self._predict_one(x) for x in X]

def score(self, X, y):
return sum(1 for p, t in zip(self.predict(X), y) if p == t) / len(y)


### Iris Flower dataset

Let's try our k-NN on a more realistic example. You can import a handful of classic datasets directly from scikit-learn's datasets module. We are going to use the iris flower dataset.

from sklearn import datasets
from sklearn.cross_validation import train_test_split

X_train, X_temp, y_train, y_temp = \
train_test_split(iris.data, iris.target, test_size=.4)
X_validation, X_test, y_validation, y_test = \
train_test_split(X_temp, y_temp, test_size=.5)

neighbor = KNeighborsClassifier().fit(X_train, y_train)

print(neighbor.score(X_train, y_train))
print(neighbor.score(X_validation, y_validation))


Using train_test_split function from cross_validation module, it first splits the data in the ratio 60:40, then splits the latter in half. Now we have 60% for the training set, 20% for the validation and test set.

Next it initializes k-NN with no special configuration and fits it. Finally, it prints the training set accuracy and the validation set accuracy to see if the model is overfitting. Since the validation set contains only 30 samples, the result varies a lot. Sometimes I got 100% and sometimes 90%. The most common one seems to be .967, meaning it got only one sample wrong. I've tried the various combinations of the 3 parameters, but the default setting is hard to beat in this dataset.

Once you find the good parameters, you can get the true score on the test set.

print(neighbor.score(X_test, y_test))


I encourage you to play with the code and see how changing each parameter affects the accuracy.

### Conclusion

If you have questions or comments, tweet @kenzotakahashi and I'll be happy to help.