KNN算法是機器學習領域中一個最基本的經典算法。它屬於無監督學習領域的算法並且在模式識別，數據挖掘和特征提取領域有着廣泛的應用。
給定一些預處理數據，通過一個屬性把這些分類坐標分成不同的組。這就是KNN的思路。
下面，舉個例子來說明一下。圖中的數據點包含兩個特征：
現在，給出數據點的另外一個節點，通過分析訓練節點來把這些節點分類。沒有分來的及誒但我們標記為白色，如下所示：
直觀來講，如果我們把那些節點花道一個圖片上，我們可能就能確定一些特征，或組。現在，給一個沒有分類的點，我們可以通過觀察它距離那個組位置最近來確定它屬於哪個組。意思就是，假如一個點距離紅色的組最近，我們就可以把這個點歸為紅色的組。簡而言之，我們可以把第一個點（2.5,7）歸類為綠色，把第二個點（5.5,4.5）歸類為紅色。
算法流程：
假設m是訓練樣本的數量，p是一個未知的節點。
1 把所有訓練的樣本放到也數組arr[]中。這個意思就是這個數組中每個元素就可以使用元組（x,y）表示。
2 偽碼
for i=0 to m: Calculate Euclidean distance d(arr[i], p).
3 標記設置S為K的最小距離。這里每個距離都和一個已經分類的數據點相關。
4 返回在S之間的大多數標簽。
實際程序C代碼：
// C++ program to find groups of unknown // Points using K nearest neighbour algorithm. #include <bits/stdc++.h> using namespace std; struct Point { int val; // Group of point double x, y; // Co-ordinate of point double distance; // Distance from test point }; // Used to sort an array of points by increasing // order of distance bool comparison(Point a, Point b) { return (a.distance < b.distance); } // This function finds classification of point p using // k nearest neighbour algorithm. It assumes only two // groups and returns 0 if p belongs to group 0, else // 1 (belongs to group 1). int classifyAPoint(Point arr[], int n, int k, Point p) { // Fill distances of all points from p for (int i = 0; i < n; i++) arr[i].distance = sqrt((arr[i].x - p.x) * (arr[i].x - p.x) + (arr[i].y - p.y) * (arr[i].y - p.y)); // Sort the Points by distance from p sort(arr, arr+n, comparison); // Now consider the first k elements and only // two groups int freq1 = 0; // Frequency of group 0 int freq2 = 0; // Frequency of group 1 for (int i = 0; i < k; i++) { if (arr[i].val == 0) freq1++; else if (arr[i].val == 1) freq2++; } return (freq1 > freq2 ? 0 : 1); } // Driver code int main() { int n = 17; // Number of data points Point arr[n]; arr[0].x = 1; arr[0].y = 12; arr[0].val = 0; arr[1].x = 2; arr[1].y = 5; arr[1].val = 0; arr[2].x = 5; arr[2].y = 3; arr[2].val = 1; arr[3].x = 3; arr[3].y = 2; arr[3].val = 1; arr[4].x = 3; arr[4].y = 6; arr[4].val = 0; arr[5].x = 1.5; arr[5].y = 9; arr[5].val = 1; arr[6].x = 7; arr[6].y = 2; arr[6].val = 1; arr[7].x = 6; arr[7].y = 1; arr[7].val = 1; arr[8].x = 3.8; arr[8].y = 3; arr[8].val = 1; arr[9].x = 3; arr[9].y = 10; arr[9].val = 0; arr[10].x = 5.6; arr[10].y = 4; arr[10].val = 1; arr[11].x = 4; arr[11].y = 2; arr[11].val = 1; arr[12].x = 3.5; arr[12].y = 8; arr[12].val = 0; arr[13].x = 2; arr[13].y = 11; arr[13].val = 0; arr[14].x = 2; arr[14].y = 5; arr[14].val = 1; arr[15].x = 2; arr[15].y = 9; arr[15].val = 0; arr[16].x = 1; arr[16].y = 7; arr[16].val = 0; /*Testing Point*/ Point p; p.x = 2.5; p.y = 7; // Parameter to decide groupr of the testing point int k = 3; printf ("The value classified to unknown point" " is %d.\n", classifyAPoint(arr, n, k, p)); return 0; }
實際程序python代碼：
1 # Python3 program to find groups of unknown 2 # Points using K nearest neighbour algorithm. 3 4 import math 5 6 def classifyAPoint(points,p,k=3): 7 ''' 8 This function finds classification of p using 9 k nearest neighbour algorithm. It assumes only two 10 groups and returns 0 if p belongs to group 0, else 11 1 (belongs to group 1). 12 13 Parameters - 14 points : Dictionary of training points having two keys - 0 and 1 15 Each key have a list of training data points belong to that 16 17 p : A touple ,test data point of form (x,y) 18 19 k : number of nearest neighbour to consider, default is 3 20 ''' 21 22 distance=[] 23 for group in points: 24 for feature in points[group]: 25 26 #calculate the euclidean distance of p from training points 27 euclidean_distance = math.sqrt((feature[0]-p[0])**2 +(feature[1]-p[1])**2) 28 29 # Add a touple of form (distance,group) in the distance list 30 distance.append((euclidean_distance,group)) 31 32 # sort the distance list in ascending order 33 # and select first k distances 34 distance = sorted(distance)[:k] 35 36 freq1 = 0 #frequency of group 0 37 freq2 = 0 #frequency og group 1 38 39 for d in distance: 40 if d[1] == 0: 41 freq1 += 1 42 elif d[1] == 1: 43 freq2 += 1 44 45 return 0 if freq1>freq2 else 1 46 47 # driver function 48 def main(): 49 50 # Dictionary of training points having two keys - 0 and 1 51 # key 0 have points belong to class 0 52 # key 1 have points belong to class 1 53 54 points = {0:[(1,12),(2,5),(3,6),(3,10),(3.5,8),(2,11),(2,9),(1,7)], 55 1:[(5,3),(3,2),(1.5,9),(7,2),(6,1),(3.8,1),(5.6,4),(4,2),(2,5)]} 56 57 # testing point p(x,y) 58 p = (2.5,7) 59 60 # Number of neighbours 61 k = 3 62 63 print("The value classified to unknown point is: {}".\ 64 format(classifyAPoint(points,p,k))) 65 66 if __name__ == '__main__': 67 main() 68 69 # This code is contributed by Atul Kumar (www.fb.com/atul.kr.007)