How can I use pyparsing to parse nested expressions that have multiple opener/closer types?

…衆ロ難τιáo~ 提交于 2019-11-28 23:25:12

Here's a pyparsing solution that uses a self-modifying grammar to dynamically match the correct closing brace character.

from pyparsing import *

data = '(gimme [some {nested, nested [lists]}])'

opening = oneOf("( { [")
nonBracePrintables = ''.join(c for c in printables if c not in '(){}[]')
closingFor = dict(zip("({[",")}]"))
closing = Forward()
# initialize closing with an expression
closing << NoMatch()
closingStack = []
def pushClosing(t):
    closingStack.append(closing.expr)
    closing << Literal( closingFor[t[0]] )
def popClosing():
    closing << closingStack.pop()
opening.setParseAction(pushClosing)
closing.setParseAction(popClosing)

matchedNesting = nestedExpr( opening, closing, Word(alphas) | Word(nonBracePrintables) )

print matchedNesting.parseString(data).asList()

prints:

[['gimme', ['some', ['nested', ',', 'nested', ['lists']]]]]

Updated: I posted the above solution because I had actually written it over a year ago as an experiment. I just took a closer look at your original post, and it made me think of the recursive type definition created by the operatorPrecedence method, and so I redid this solution, using your original approach - much simpler to follow! (might have a left-recursion issue with the right input data though, not thoroughly tested):

from pyparsing import *

enclosed = Forward()
nestedParens = nestedExpr('(', ')', content=enclosed) 
nestedBrackets = nestedExpr('[', ']', content=enclosed) 
nestedCurlies = nestedExpr('{', '}', content=enclosed) 
enclosed << (Word(alphas) | ',' | nestedParens | nestedBrackets | nestedCurlies)


data = '(gimme [some {nested, nested [lists]}])' 

print enclosed.parseString(data).asList()

Gives:

[['gimme', ['some', ['nested', ',', 'nested', ['lists']]]]]

This should do the trick for you. I tested it on your example:

import re
import ast

def parse(s):
    s = re.sub("[\{\(\[]", '[', s)
    s = re.sub("[\}\)\]]", ']', s)
    answer = ''
    for i,char in enumerate(s):
        if char == '[':
            answer += char + "'"
        elif char == '[':
            answer += "'" + char + "'"
        elif char == ']':
            answer += char
        else:
            answer += char
            if s[i+1] in '[]':
                answer += "', "
    ast.literal_eval("s=%s" %answer)
    return s

Comment if you need more

标签
易学教程内所有资源均来自网络或用户发布的内容,如有违反法律规定的内容欢迎反馈
该文章没有解决你所遇到的问题?点击提问,说说你的问题,让更多的人一起探讨吧!