PEG 节点访问者

Question

想象一个像

这样的小PEG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   = ","
    ws      = ~"\s*"
    """
)

tree = grammar.parse("(5, 4, 3)")
print(tree)

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 term 中获取 number 正则表达式部分？我知道我可以使用 NodeVisitor class 并检查每个数字，但我想从 term.

中获取正则表达式部分

Answer 1

使用 NodeVisitor class 并以这种方式遍历树可能更好，但这是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">

PEG 节点访问者

PEG NodeVisitor

python

parsing

peg