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GS与NT之间的DIS距离和LS与NT之间的DIS距离 #5

@0Pluto

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@0Pluto

您好,我想问一下,关于您论文第4页的DIS Distance这一小节,最后提到

Similarly, the DIS distance between GS and NT is larger than that between LS and NT (both less than 0)

可是我用第5页的示例数据计算了一下,根据您的描述,GS对应R_7,LS对应R_8。GS和NT的DIS距离应该是小于LS和NT之间的DIS距离的呀

  NT = [[1, 2, 3, 4, 12, 20, 28, 29, 30, 38, 46, 54],
        [1, 2, 3, 4, 12, 20, 28, 36, 44, 52, 53, 54]]
  GS = [1, 10, 19, 28, 36, 45, 54]
  LS = [1, 2, 11, 20, 28, 36, 44, 52, 53, 54]
  # get_both_count(a,b) 得到a,b列表中共同元素个数
  s1 = (len(GS) - len(NT[0])) / (get_both_count(GS, NT[0]))
  s11 = (len(GS) - len(NT[1])) / (get_both_count(GS, NT[1]))
  s2 = (len(LS) - len(NT[0])) / (get_both_count(LS, NT[0]))
  s21 = (len(LS) - len(NT[1])) / (get_both_count(LS, NT[1]))
  print('GS\ts1:{:.4f}\ts11:{:.4f}'.format(s1, s11))
  print('LS\ts2:{:.4f}\ts22:{:.4f}'.format(s2, s21))

输出结果:

GS	s1:-1.6667	s11:-1.2500
LS	s2:-0.4000	s22:-0.2222

包括您后面第10页还提到会将异常分数降序排列,前n_GS$个即为GS,剩下的n_LS个即为LS,这就是错误的吧?我看了半天不太理解,也有可能是我看漏了什么地方,希望能够解答一下。

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