Python PEG节点检测器
想象一下像这样的一个小语法Python PEG节点检测器,python,parsing,peg,Python,Parsing,Peg,想象一下像这样的一个小语法 from parsimonious.grammar import Grammar from parsimonious.nodes import NodeVisitor grammar = Grammar( r""" term = lpar (number comma? ws?)+ rpar number = ~"\d+" lpar = "(" rpar = ")" comma = ","
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)
哪个输出
<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 ")">
如何从本例中的术语中获取数字regex部分?我知道我可以使用NodeVisitor类并检查每个数字,但我希望在学期内获得正则表达式部分。使用NodeVisitor类并以这种方式遍历树可能更好,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">
非常感谢,我现在正在使用一个组合。