用java编写一段协同过滤算法推荐音乐家
时间: 2023-07-19 19:33:00 浏览: 115
以下是一个简单的基于Java的协同过滤算法推荐音乐家的示例代码:
```
import java.util.*;
public class MusicRecommendation {
private static final int TOP_N = 5; // 推荐的音乐家数量
private static final int NEIGHBORS_SIZE = 3; // 最邻近的用户数量
private static final double SIMILARITY_THRESHOLD = 0.5; // 最小相似度阈值
// 用户-音乐家矩阵
private static final double[][] RATINGS = {
{5.0, 3.0, 4.0, 4.0, Double.NaN, 1.0},
{4.0, Double.NaN, 4.0, 5.0, 3.0, 2.0},
{Double.NaN, 4.0, 5.0, 3.0, 4.0, 4.0},
{2.0, 3.0, Double.NaN, 4.0, 3.0, 4.0},
{3.0, 4.0, 5.0, Double.NaN, 5.0, Double.NaN},
{4.0, 3.0, 3.0, 5.0, 4.0, 5.0}
};
// 用户相似度矩阵
private static double[][] similarities = new double[RATINGS.length][RATINGS.length];
public static void main(String[] args) {
// 计算用户相似度
for (int i = 0; i < similarities.length; i++) {
for (int j = 0; j < similarities.length; j++) {
if (i == j) {
similarities[i][j] = 1.0;
} else {
similarities[i][j] = calcSimilarity(RATINGS[i], RATINGS[j]);
}
}
}
// 推荐音乐家
double[] userRatings = {4.0, Double.NaN, Double.NaN, 5.0, 3.0, 4.0}; // 用户对音乐家的评分
List<Integer> topN = recommend(userRatings);
System.out.println("推荐的音乐家为:" + topN);
}
// 计算用户之间的相似度
private static double calcSimilarity(double[] ratings1, double[] ratings2) {
double numerator = 0.0, denominator1 = 0.0, denominator2 = 0.0;
for (int i = 0; i < ratings1.length; i++) {
if (!Double.isNaN(ratings1[i]) && !Double.isNaN(ratings2[i])) {
numerator += ratings1[i] * ratings2[i];
denominator1 += Math.pow(ratings1[i], 2);
denominator2 += Math.pow(ratings2[i], 2);
}
}
double denominator = Math.sqrt(denominator1) * Math.sqrt(denominator2);
return denominator == 0.0 ? 0.0 : numerator / denominator;
}
// 推荐音乐家
private static List<Integer> recommend(double[] userRatings) {
// 找到最邻近的用户
List<Integer> neighbors = new ArrayList<>();
for (int i = 0; i < similarities.length; i++) {
if (i != userRatings.length - 1 && similarities[userRatings.length - 1][i] >= SIMILARITY_THRESHOLD) {
neighbors.add(i);
}
}
if (neighbors.size() == 0) {
return Collections.emptyList();
}
// 计算音乐家的推荐指数
Map<Integer, Double> scores = new HashMap<>();
for (int i = 0; i < RATINGS[userRatings.length - 1].length; i++) {
if (Double.isNaN(userRatings[i])) {
double numerator = 0.0, denominator = 0.0;
for (int neighbor : neighbors) {
if (!Double.isNaN(RATINGS[neighbor][i])) {
numerator += similarities[userRatings.length - 1][neighbor] * RATINGS[neighbor][i];
denominator += similarities[userRatings.length - 1][neighbor];
}
}
double score = denominator == 0.0 ? 0.0 : numerator / denominator;
scores.put(i, score);
}
}
// 返回得分最高的音乐家
List<Integer> topN = new ArrayList<>();
for (int i = 0; i < TOP_N; i++) {
double maxScore = Double.NEGATIVE_INFINITY;
int maxIndex = -1;
for (int index : scores.keySet()) {
if (scores.get(index) > maxScore) {
maxScore = scores.get(index);
maxIndex = index;
}
}
if (maxIndex != -1) {
topN.add(maxIndex);
scores.remove(maxIndex);
}
}
return topN;
}
}
```
该算法使用了一个用户-音乐家矩阵来存储用户对音乐家的评分,对于未评分的音乐家,使用协同过滤算法来预测用户的评分,并返回得分最高的音乐家作为推荐结果。
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