python - 试图制作程序来格式化化学方程式

Question

给定类似 Na(Cl2)3Al(AlCl2)4 的输入我正在尝试制作一个没有任何库的程序来执行此操作，而我制作的代码不起作用，我无法弄清楚原因。
有人可以告诉我我哪里出错了(或帮我解决)？示例输入的输出应类似于 NaCl6AlAl4Cl8其中括号也被删除。提前感谢您的帮助。

def format(equation):
  equation_list = list(equation)
  formated = ''

  
  for i in range(len(equation_list)): 
    if equation_list[i] == '(':
      opening = i
    if equation_list[i] == ')':
      closing = i
      
  
      for i in range(opening+1,closing):
        if equation_list[i].isdigit():
          equation_list[i] = str(int(equation_list[i])*mult)
        
        if equation_list[i].upper():
          if equation_list[i+1].isupper():
            equation_list[i+1] = str(mult)
          elif equation_list[i+1].isdigit():
            equation_list[i+1] = str(int(equation_list[i+1])*mult)
          else:
            if equation_list[i+2].isupper():
              equation_list[i+2] = str(mult)
            elif equation_list[i+2].isdigit():
              equation_list[i+2] = str(int(equation_list[i+2])*mult)
            else:
              if equation[i+3].isdigit():
                equation_list[i+3] = str(int(equation_list[i+3])*mult)
              else:
                equation[i+3] = str(mult)

        

  for i in equation_list:
    formated+=i


  return formated
print(format('Na(Cl2)3Al(AlCl2)4'))

Answer 1

如 pointed duplicate没有投票回答，我想根据正则表达式提出一个。我很确定它不匹配所有情况，但至少你的

import re

EQ_PATTERN = re.compile(r"([A-Z][a-z]+\d*|\((?:[A-Z][a-z]+\d*)+\)\d*)")
BLOCK_PATTERN = re.compile(r"\(((?:[A-Z][a-z]+\d*)+)\)(\d*)")
ELEMENT_PATTERN = re.compile(r"([A-Z][a-z]+)(\d*)")

def format_element(element: str, amount: str, factor: int):
    return f"{element}{int(amount or '1') * factor}"

def format_block(block: str):
    if '(' not in block:
        return block
    inside, factor = BLOCK_PATTERN.findall(block)[0]
    return "".join(format_element(element, amount, int(factor))
                   for element, amount in ELEMENT_PATTERN.findall(inside))

def format_equation(eq: str):
    return "".join(map(format_block, EQ_PATTERN.findall(eq)))

通过一些测试

for eq in ('(Al3Cl2Ag6Au)4', 'Na(Cl2)3Al(AlCl2)4', 'Na2(Cl2)3Al(Al3Cl2)4'):
    print(f"{eq:20s} ==> {format_equation(eq)}")

(Al3Cl2Ag6Au)4       ==> Al12Cl8Ag24Au4
Na(Cl2)3Al(AlCl2)4   ==> NaCl6AlAl4Cl8
Na2(Cl2)3Al(Al3Cl2)4 ==> Na2Cl6AlAl12Cl8