PEG NodeVisitor
PEG NodeVisitor
提问于 2019-03-06 14:34:16
想象一下一个小的PEG语法，比如
from parsimonious.grammar import Grammar
from parsimonious.nodes import NodeVisitor
grammar = Grammar(
r"""
term = lpar (number comma? ws?)+ rpar
number = ~"\d+"
lpar = "("
rpar = ")"
comma = ","
ws = ~"\s*"
"""
)
tree = grammar.parse("(5, 4, 3)")
print(tree)哪种输出
<Node called "term" matching "(5, 4, 3)">
<Node called "lpar" matching "(">
<Node matching "5, 4, 3">
<Node matching "5, ">
<RegexNode called "number" matching "5">
<Node matching ",">
<Node called "comma" matching ",">
<Node matching " ">
<RegexNode called "ws" matching " ">
<Node matching "4, ">
<RegexNode called "number" matching "4">
<Node matching ",">
<Node called "comma" matching ",">
<Node matching " ">
<RegexNode called "ws" matching " ">
<Node matching "3">
<RegexNode called "number" matching "3">
<Node matching "">
<Node matching "">
<RegexNode called "ws" matching "">
<Node called "rpar" matching ")">在本例中，如何从term获取number正则表达式部分？我知道我可以使用NodeVisitor类来检查每个数字，但我想从term中获取正则表达式部分。
回答 1
Stack Overflow用户
回答已采纳
发布于 2019-03-06 23:03:27
使用NodeVisitor类并以这种方式遍历树可能会更好，但这里有另一个简单的解决方案：
from parsimonious.grammar import Grammar
from parsimonious.nodes import NodeVisitor
grammar = Grammar(
r"""
term = lpar (number comma? ws?)+ rpar
number = ~"\d+"
lpar = "("
rpar = ")"
comma = ","
ws = ~"\s*"
"""
)
tree = grammar.parse("(5, 4, 3)")
def walk(node):
if node.expr_name == 'number':
print(node)
for child in node.children:
walk(child)
walk(tree)
# <RegexNode called "number" matching "5">
# <RegexNode called "number" matching "4">
# <RegexNode called "number" matching "3">页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/55016966
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