PEG 节点访问者答案

【问题标题】：PEG NodeVisitorPEG 节点访问者
【发布时间】：2019-03-28 15:30:13
【问题描述】：

想象一下像PEG 这样的小语法

from parsimonious.grammar import Grammar
from parsimonious.nodes import NodeVisitor

grammar = Grammar(
    r"""
    term    = lpar (number comma? ws?)+ rpar
    number  = ~"\d+"
    lpar    = "("
    rpar    = ")"
    comma   = ","
    ws      = ~"\s*"
    """
)

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

哪些输出

<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 类并检查每个数字，但我想从 term 中获取正则表达式部分。

【问题讨论】：

标签： python parsing peg

【解决方案1】：

使用NodeVisitor 类并沿此方向遍历树可能会更好，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">

【讨论】：

非常感谢，我现在正在使用组合。