python - 使用 parsec.py 解析字符串

Question

我想解析这样的字符串：

Object A -> Object B [AB_name] Object B -> Object C [BC_name] ...

我的目标是获得三个列表：

父母= ['对象A'，'对象B'，...]

孩子 = ['对象 B', '对象 C', ...]

PC_names = ['AB_name','BC_name', ...]

我已经有了一个可行的解决方案，但它难以理解，因此难以维护并且不是很健壮。基本上，我的代码在字符串上迭代两个循环，从而将子字符串添加到几个列表中。

为了解决这个问题，我阅读了有关 parsec.py 库的信息，但到目前为止，我还没有为像我这样的新手找到一个很好的例子。我已经尝试在其他文章和文档的帮助下弄清楚它是如何工作的，但现在收效甚微。

我很高兴每一个提示。

test_input.txt：

Society {
    A man -> a child [was once]
    A man -> an automobile [has]
    A woman -> a person [is]
    A man -> a person [is]
    A man -> a child [was once] 
}

我当前的代码：

from typing import List
from parsec import *

class Type(object):
    label: str

class Aspect(object):
    domain: Type
    codomain: Type
    label: str
    def __init__(self) -> None:
        self.domain = Type()
        self.codomain = Type()
    
class Olog(object):
    name: str
    aspects: List[Aspect]
    def __init__(self):
        self.aspects = []

with open ('testinput.txt', 'r') as f:
        f_content = f.read()
        olog_name = f_content.split("{")[0]
        first = Olog()
        first.aspects = []
        first.name = olog_name
        olog_data = f_content.split("{")[1]
        olog_data_lines = olog_data.split(']')

        orientation = str

        counter1 = 0 
        counter2 = 0
        domain_str = ''
        codomain_str = ''
        type_comma = Type()
        type_comma.label = ","
        string_store = str
        string_store = ''
        type_store = Type()
        type_store_split = [Type]
        

        for lines in olog_data_lines:
            first_type = ''
            second_type = ''
            aspect_label = str
            first_T = Type()
            second_T = Type()
            lines += ']'
            lines_split = lines.split()
            type_in_list = False

            for word in lines_split:
                if word == "}" and counter1 == 0:
                        print("Olog is empty")
                if word == "}":
                        print(">>>Olog was saved")
                        break

                if word == "->":

                    counter1 +=1
                if counter1 == counter2 and lines_split.index(word) == 0:

                    first_type = word
                if counter1 == counter2 and not lines_split.index(word) == 0:
                    first_type = first_type + (" " + word)
                if  word == "->": 
                        orientation = "->"
                        string_store = string_store + first_type + ", "

                        type_store.label = string_store
                        type_store_split = type_store.label.split(",")
                        
                        
                        for types in type_store_split: 
                                                
                            if types == first_type:

                                domain_str = int(type_store_split.index(types))
                                type_in_list = True
                                break
                        if not type_in_list:

                            domain_str = int(len(type_store_split)-2)

                if not counter1 == counter2:
                    if word[0] == "[":
                        aspect_label = (lines.split('[', 1)[1].split(']')[0])
                    else: second_type = second_type.replace('->','', 1) + " " + word
                    
                    if (word[len(word)-1]=="]"):
                        second_T.label = second_type
                        string_store = string_store + second_type + ", "
                        type_store.label = string_store
                        type_store_split = type_store.label.split(",")

                        for types in type_store_split:               
                            if types == second_type:
                                codomain_str = int(type_store_split.index(types))
                                second_T.label = codomain_str
                                break
                            elif types == type_store_split[len(type_store_split)-1]:
                                codomain_str = int(len(type_store_split)-2)
                                second_T.label = codomain_str

                        aspect_A = Aspect()
                        aspect_A.label = aspect_label
                        aspect_A.domain = Type()
                        aspect_A.codomain = Type()
                        aspect_A.domain.label = domain_str
                        aspect_A.codomain.label = codomain_str
                        first.aspects.append(aspect_A)
                        counter2 += 1
                    
``

标签： pythonstringparsingmonadsparsec

Answer 1

该解决方案使用re递归解析输入行并遍历结果，yield返回父母、孩子和pc_names：

import re, collections
def parse_line(l):
   return [re.findall('\[.*?\]|[\w\s]+', i.strip()) for i in re.split('\s*\-\>\s*', l)]

lines = [parse_line(i) for i in open('test_input.txt') if not re.findall('[\{\}]', i)]
def get_vals(d, s = []):
  if len(d) > 1:
     yield ('pc_names', d[-1][1:-1])
  if not (c:=[b for a, b in lines if d[0] == a[0]]):
     yield ('children', d[0])
     if (k:=[a for a, _ in lines if a[0] not in s]):
        yield from get_vals(k[0], s+[d[0]])
  else:
     yield ('parents', d[0])
     for i in c:
        yield from get_vals(i, s+[d[0]])

result = collections.defaultdict(set)
for a, b in get_vals(lines[0][0]):
   result[a].add(b)

print({a:list(b) for a, b in result.items()})

输出：

{'parents': ['A woman', 'A man'], 
 'pc_name': ['was once', 'is', 'has'], 
 'children': ['a person ', 'an automobile ', 'a child ']}

第二个test_input.txt内容：

Object A -> Object B [AB_name] 
Object B -> Object C [BC_name]

结果：

{'parents': ['Object B', 'Object A'], 
 'pc_names': ['AB_name', 'BC_name'], 
 'children': ['Object B ', 'Object C ']}