python – 我的候选消除算法不起作用
发布时间:2020-12-20 13:11:32 所属栏目:Python 来源:网络整理
导读:我试图在 Python中实现“候选消除算法”,但我的代码不起作用. 我写了3个函数: 一致地检查假设和训练样例之间的一致性 more_general用于查找更一般的参数 more_specific用于查找更具体的参数 但我的算法没有添加或删除G和S的假设.我无法找到问题所在.你能帮
我试图在
Python中实现“候选消除算法”,但我的代码不起作用.
我写了3个函数: >一致地检查假设和训练样例之间的一致性 但我的算法没有添加或删除G和S的假设.我无法找到问题所在.你能帮助我吗? # the general hypothesis G = [ ('?','?','?') ] # the specific hypothesis S = [('0','0','0')] # attributes: AV = (['short','far'],['cheap','expensive'],['many','none'],['yes','no']) # training examples: D = [ {'sample': ('far','cheap','many','no' ),'positive': True },{'sample': ('short','expensive',{'sample': ('far','none','yes'),'positive': False},'positive': True } ] def consistent(hypothesis,sample): return all([hypothesis[i] == sample[i] or hypothesis[i] == '?' for i in range(len(hypothesis))]) def more_general(a,b): result = False if a == '0' and b != '0': result = True elif a != '?' and b == '?': result = True return result def more_specific(a,b): result = False if a == '?' and b != '?': result = True elif a != '0' and b == '0': result = True return result for d in D: if d['positive']: G = [g for g in G if consistent(g,d['sample'])] for s in S: if not consistent(s,d['sample']): S.remove(s) # Adding to S all minimal generalizations of s by h: dd = d['sample'] if s == 0: h = dd[s] else: h = '?' if consistent(h,d['sample']) and any([more_general(g,h) for g in G]): S.append(h) #Removing from S any hypothesis that is more general than another hypothesis in S for s2 in S: if any([more_general(s2,s3) and not s2 == s3 for s3 in S]): S.remove(s2) else: S = [s for s in S if not consistent(s,d['sample'])] for g in G: if consistent(g,d['sample']): G.remove(g) # Add to G all minimal specializations h of g for ai in range(len(AV)): if g[ai] == '?': h = list(g) h[ai] = AV[ai][1 - AV[ai].index(d['sample'][ai])] h = tuple(h) if not consistent(h,d['sample']) and any([more_specific(s,h) for s in S]): G.append(h) print('Sample: {} {}nG: {}nS: {}n'.format('+' if d['positive'] else '-',d['sample'],G,S)) 解决方法
这是您的代码崩溃的地方:
S.remove(s) dd = d['sample'] if s == 0: h = dd[s] else: h = '?' if consistent(h,h) for g in G]): S.append(h) s是一个元组,但在if条件下,当你比较零时,你将它视为标量值.在if之后,h包含一个标量,但是在下一个if中,对consistent()的调用期望它的第一个参数,h是一个元组,而不是一个标量. (编辑:李大同) 【声明】本站内容均来自网络,其相关言论仅代表作者个人观点,不代表本站立场。若无意侵犯到您的权利,请及时与联系站长删除相关内容! |