我有一个子树列表,我从一个解析历史派生的格式如下:
解析历史记录:
main()
列表中的每个元组都有一个包含规则列表的语法字典的键。元组中的第二项是该给定键的规则索引。
语法是:
parse = [('S', 0), ('NP', 1), ('Det', 0), ('N', 0), ('VP', 1), ('V', 4), ('NP', 2), ('NP', 0), ('PN', 1), ('NP', 1), ('Det', 0), ('N', 3)]
子树列表是:
grammar = {'S': [['NP', 'VP']],
'NP': [['PN'], ['Det', 'N']],
'VP': [['V'], ['V', 'NP', 'NP']],
'PN': [['John'], ['Mary'], ['Bill']],
'Det': [['the'], ['a']],
'N': [['man'], ['woman'], ['drill sergeant'], ['dog']],
'V': [['slept'], ['cried'], ['assaulted'],
['devoured'], ['showed']]}
我使用以下代码创建了子树列表:
[Tree('S', ['NP', 'VP']), Tree('NP', ['Det', 'N']), Tree('Det', ['the']), Tree('N', ['man']), Tree('VP', ['V', 'NP', NP]), Tree('V', ['showed']), Tree('NP', ['PN']), Tree('PN', ['Mary']), Tree('NP', ['Det', 'N']), Tree('Det', ['the']), Tree('N', ['dog'])]
我打印树时得到的输出(我知道这不是正确的方法,但它至少显示了子树)如下:
for item in parse:
apple = Tree(item[0], grammar[item[0]][item[1]])
trees.append(apple)
感谢您的帮助!
:: EDIT ::
正确的输出应如下所示:
(S NP VP)
(NP Det N)
(Det the)
(N man)
(VP V NP)
(V showed)
(NP NP NP)
(NP PN)
(PN Mary)
(NP Det N)
(Det the)
(N dog)
答案 0 :(得分:0)
您需要递归构建树,但您需要区分终端和非终端。顺便说一句。你的解析序列似乎错了。我把它搞砸了:
def build_tree(parse):
assert(parse)
rule_head = parse[0][0]
rule_body = grammar[rule_head][parse[0][1]]
tree_body = []
rest = parse[1:]
for r in rule_body:
if non_term(r):
(subtree,rest) = build_tree(rest)
tree_body.append(subtree)
else:
tree_body.append(r)
return (tree(rule_head,tree_body), rest)