信息检索————NDCG计算

先贴代码,原理有时间补上。

 1 import numpy as np
 2 import copy
 3 from math import log
 4 
 5 # input data
 6 ideal = np.array([5,5,4,4,3,3,2,2,1,1])
 7 relevant_score_list = np.array([5.0,5.0,4.0,4.0,3.0,3.0,2.0,2.0,1.0,1.0])
 8 recall_list = np.array([8,11,2,12,3,1,5,4,13,7])
 9 
10 def dcg(rscore, m):
11     return (2.0 ** rscore - 1.0) / np.log2(2.0 + m)
12 
13 def dcg_k(rslist,rlist,k):
14     # 计算第k个dcg值
15     dcgscore_list = [0.0 for i in range(11)]##计算1,2,...,k的dcg值
16     for i in range(k):
17         relevant_score = 0
18         if rlist[i] < len(rslist):
19             relevant_score = rslist[rlist[i]-1]
20         dcgscore_list[i+1] = dcgscore_list[i] + dcg(relevant_score, i)
21     return dcgscore_list;
22 
23 def idcg_k(ideal,k):
24     # idcg是一个系统最理想情况下的返回结果排序,也就是一个案例中所给的最佳排序
25     idcgscore_list = [0.0 for i in range(11)] #计算1,2,...,k的idcg值
26     for i in range(k):
27         idcgscore_list[i+1] = idcgscore_list[i] + dcg(ideal[i], i)
28     return idcgscore_list;
29 
30 def ndcg(rslist,rlist,k):
31     dcgscore_list = dcg_k(rslist,rlist,k)
32     # 计算归一化因子z, 最完美情况应该是recall结果按照相关性分数降序排列
33     idcgscore_list = idcg_k(ideal,k)
34 
35     ndcg_list = [0.0 for i in range(k)]
36     for i in range(k):
37         ndcg_list[i] = round(dcgscore_list[i+1] / idcgscore_list[i+1], 3)
38     print(ndcg_list)
39     
40 
41 def ndcg(ideal, rlist, k):
42     recall = [0.0 for i in range(k)]
43     for i in range(k):
44         if rlist[i] <= len(ideal):
45             recall[i] = ideal[rlist[i]-1]
46     
47     dcg, dcg_max, ndcg = 0.0, 0.0, 0.0
48     for i, (true, predict) in enumerate(zip(ideal, recall)):
49         dcg += (2 ** predict - 1) / log(2 + i)
50         dcg_max += (2 ** true -1) / log(2 + i)
51         ndcg = dcg / dcg_max
52     return ndcg
53     
54 # print(ndcg(relevant_score_list,recall_list,10))
55 # ndcg(relevant_score_list,recall_list,10)
56 ndcg(ideal, recall_list, 10)

代码中两种方式实现了ndcg的计算,调用的时候需要注意一下。

原文地址:https://www.cnblogs.com/sgatbl/p/12768187.html