🌈 第16课：多样性优化

🎯 学习目标

🗂️ 课程定位

📖 核心理论

多样性是推荐系统的重要质量维度。单一追求相关性会导致信息茧房——用户总看到相似内容。多样性优化在保持相关性的同时，确保推荐列表内容丰富。

📐 数学公式

💻 代码实现

import numpy as np
np.random.seed(42)
print("="*60+"\n多样性优化\n"+"="*60)
n_items=20;n_features=8
item_embs=np.random.randn(n_items,n_features)
categories=np.random.choice(["A","B","C","D"],n_items)
scores=np.random.random(n_items)*5+1
print(f"物品:{n_items} 类别:{dict(zip(*np.unique(categories,return_counts=True)))}")
# 多样性度量
def intra_list_diversity(items,embs):
    if len(items)<2:return 0
    div=0;n=0
    for i in range(len(items)):
        for j in range(i+1,len(items)):
            a,b=embs[items[i]],embs[items[j]]
            div+=1-np.dot(a,b)/(np.linalg.norm(a)*np.linalg.norm(b)+1e-10)
            n+=1
    return div/n
def category_coverage(items,cats):
    return len(set(cats[i]for i in items))/len(set(cats))
def entropy_diversity(items,cats):
    from collections import Counter
    c=Counter(cats[i]for i in items)
    total=len(items)
    return -sum((cnt/total)*np.log2(cnt/total)for cnt in c.values())
# 基线(纯相关性排序)
baseline=np.argsort(-scores)[:10]
print(f"\n基线(ILD={intra_list_diversity(baseline,item_embs):.4f} 覆盖={category_coverage(baseline,categories):.2f} 熵={entropy_diversity(baseline,categories):.4f})")
# MMR
def mmr(scores,embs,lam=0.5,k=10):
    sel=[];cand=list(range(len(scores)))
    for _ in range(min(k,len(scores))):
        best=None;best_val=-999
        for i in cand:
            rel=scores[i]
            if sel:
                max_sim=max(abs(np.dot(embs[i],embs[j])/(np.linalg.norm(embs[i])*np.linalg.norm(embs[j])+1e-10))for j in sel)
            else:max_sim=0
            val=lam*rel-(1-lam)*max_sim*5
            if val>best_val:best_val=val;best=i
        sel.append(best);cand.remove(best)
    return sel
mmr_result=mmr(scores,item_embs,lam=0.5)
print(f"MMR(ILD={intra_list_diversity(mmr_result,item_embs):.4f} 覆盖={category_coverage(mmr_result,categories):.2f} 熵={entropy_diversity(mmr_result,categories):.4f})")
# DPP
print("\n--- DPP多样性 ---")
def dpp_select(scores,embs,k=10,gamma=0.5):
    n=len(scores)
    L=np.zeros((n,n))
    for i in range(n):
        for j in range(n):
            sim=np.dot(embs[i],embs[j])/(np.linalg.norm(embs[i])*np.linalg.norm(embs[j])+1e-10)
            L[i,j]=gamma*np.sqrt(scores[i])*sim*np.sqrt(scores[j])
    # 贪心选择
    sel=[]
    for _ in range(min(k,n)):
        best=None;best_det=0
        for i in range(n):
            if i in sel:continue
            trial=sel+[i]
            sub=L[np.ix_(trial,trial)]
            det=np.linalg.det(sub+1e-10*np.eye(len(trial)))
            if det>best_det:best_det=det;best=i
        if best is not None:sel.append(best)
    return sel
dpp_result=dpp_select(scores,item_embs,k=10,gamma=0.3)
print(f"DPP(ILD={intra_list_diversity(dpp_result,item_embs):.4f} 覆盖={category_coverage(dpp_result,categories):.2f} 熵={entropy_diversity(dpp_result,categories):.4f})")
# λ影响
print("\nMMR λ参数影响:")
for lam in [0.1,0.3,0.5,0.7,0.9]:
    res=mmr(scores,item_embs,lam=lam)
    avg_score=np.mean([scores[i]for i in res])
    ild=intra_list_diversity(res,item_embs)
    print(f"  λ={lam}: 平均分={avg_score:.2f} ILD={ild:.4f}")

🔴 运行结果

✅ 验证通过

🔍 代码解读

💡 关键洞察

📝 本章小结

🧪 练习

📚 延伸阅读

💼 面试考点

🏭 工业案例