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K-Nearest Neighbors Classifier

Predict a label by majority vote of the k closest training points.

Ce que vous allez apprendre

  • Find the closest training points
  • Predict by majority vote
  • Break ties deterministically

Problème

Afficher la traduction

K-nearest neighbors (KNN) classifies a new point by looking at the k training points closest to it and taking the most common label.

  1. Compute the Euclidean distance from query to every point.
  2. Sort the points by distance; if two distances are equal, the point with the smaller index comes first.
  3. Take the first k points and count their labels.
  4. Return the label with the most votes. If labels tie, return the tied label whose closest point comes first in the sorted order.

Implement knn_predict(points, labels, query, k).

Input

points: The training points, each a list of coordinates.

labels: The label of each point.

query: The point to classify.

k: How many neighbors vote.

Output

Return the predicted label.

Contraintes

1 ≤ k ≤ number of points ≤ 1,000

1 ≤ dimensions ≤ 20

-1,000 ≤ each coordinate ≤ 1,000

Exemples

Exemple 1
Entrée :
k = 3
query = [3.0, 3.0]
labels = ["A", "A", "B", "B"]
points = [
    [1.0, 1.0],
    [2.0, 2.0],
    [8.0, 8.0],
    [9.0, 9.0]
]

Sortie attendue :
expected = "A"
Entrée :
long long k = 3;
std::vector<double> query = {3.0, 3.0};
std::vector<std::string> labels = {"A", "A", "B", "B"};
std::vector<std::vector<double>> points = {
    {1.0, 1.0},
    {2.0, 2.0},
    {8.0, 8.0},
    {9.0, 9.0}
};

Sortie attendue :
std::string expected = "A";
Entrée :
long k = 3;
double[] query = new double[] {3.0, 3.0};
string[] labels = new string[] {"A", "A", "B", "B"};
double[][] points = new double[][] {
    new double[] {1.0, 1.0},
    new double[] {2.0, 2.0},
    new double[] {8.0, 8.0},
    new double[] {9.0, 9.0}
};

Sortie attendue :
string expected = "A";

Explication

Distances from (3, 3): (2, 2) ≈ 1.41, (1, 1) ≈ 2.83, (8, 8) ≈ 7.07. The 3 nearest labels are A, A and B, so the answer is A.

Exemple 2
Entrée :
k = 2
query = [1.0, 1.0]
labels = ["red", "blue", "blue", "red"]
points = [
    [0.0, 0.0],
    [4.0, 0.0],
    [0.0, 4.0],
    [4.0, 4.0]
]

Sortie attendue :
expected = "red"
Entrée :
long long k = 2;
std::vector<double> query = {1.0, 1.0};
std::vector<std::string> labels = {"red", "blue", "blue", "red"};
std::vector<std::vector<double>> points = {
    {0.0, 0.0},
    {4.0, 0.0},
    {0.0, 4.0},
    {4.0, 4.0}
};

Sortie attendue :
std::string expected = "red";
Entrée :
long k = 2;
double[] query = new double[] {1.0, 1.0};
string[] labels = new string[] {"red", "blue", "blue", "red"};
double[][] points = new double[][] {
    new double[] {0.0, 0.0},
    new double[] {4.0, 0.0},
    new double[] {0.0, 4.0},
    new double[] {4.0, 4.0}
};

Sortie attendue :
string expected = "red";

Explication

The nearest point is (0, 0) (red); the next two, (4, 0) and (0, 4), are at the same distance, so the smaller index (4, 0) (blue) wins the tie. Red and blue have one vote each, and red's point came first, so the answer is red.

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