简单化学公式的形成

Question

给定一个化学公式，如Al2( SO4 )3，我想格式化它，方法是查看括号中的位置，并将末尾的3乘以括号中的SO4，得到Al2S3O12的输出。我在第40行收到一个语法错误，我怀疑还有更多的错误。我正在寻找我可能做错了什么，以及我如何改进这段代码。如果有帮助的话，我使用repl作为ide。提前谢谢。

def format(equation):
  a = 0
  equation_list = []
  formated = ''

  for i in range (len(equation)):
    equation_list.append(equation[i])


  for i in range(a,len(equation)):
    if equation[i] == '(':
      opening = equation[i]
    elif equation[i] == ')':
      closing = equation[i]
      a = closing
      multiplier = closing+1 
      break

  


    for i in range(len(equation_list[opening:closing+1])):

      if equation_list[i].isupper():

        if equation_list[i+1].islower():

          if equation_list[i+2].islower():
            equation_list.insert(i+3, str(multiplier))
          elif equation_list[i+2].isupper():
            equation_list.insert(i+3, str(multiplier))
          elif equation_list[i+2].isdigit():
            equation_list[i+3] *=multiplier

        
        elif equation_list[i+1].isupper():

          equation_list.insert(i+1, str(multiplier)

        elif equation_list[i+1].isdigit():

          equation_list[i+1] *= multiplier

  equation_list.remove('(')
  equation_list.remove(')')
  for i in equation_list:
    formated += equation_list[i]

  return formated





print(format('Na(Cl)2'))

Answer 1

这将处理嵌套的括号和两位数的乘数。它不会像'(CH2OH)3‘那样将分离的原子组合在一起，但这可能是一种练习。

def format(equation):
    parts = []
    partial = 0
    for char in equation+' ':
        if char.isnumeric():
            partial = partial * 10 + int(char)
            continue

        if partial:
            if parts[-1] == ')':
                parts.pop()
                i = -1
                while parts[i] != '(':
                    parts[i][1] *= partial
                    i -= 1
            else:
                parts[-1][1] = partial
            partial = 0

        if char.isupper():
            parts.append( [char, 1] )
        elif char.islower():
            parts[-1][0] += char
        elif char == '(':
            parts.append('(')
        elif char == ')':
            parts.append(')')
    build = []
    for part in parts:
        print(part)
        if part == '(':
            continue
        if part[1] == 1:
            build.append( part[0] )
        else:
            build.append( part[0] + str(part[1]))
    return ''.join(build)

print(format('Al2(SO4)3'))
print(format('Al2((SO4)3)10'))

Answer 2

我真的建议使用re (regex模块)来解决这个问题。我想出的代码是：

import re
def format(equation):
    parantheses_regex = "\((?P<inner_equation>[A-Za-z\d]+)\)(?P<multipli_number>\d+)"
    get_inner_values = "(?P<symb>[A-Z]([a-z])?)(?P<number>\d+)?"
    search_parantheses = re.finditer(parantheses_regex, equation)
    for value in search_parantheses:
        multipli = value.group("multipli_number")
        inner_equation = value.group("inner_equation")
        search_equation = re.finditer(get_inner_values, inner_equation)
        new_inner = ""
        for inner_symb in search_equation:
            number  = inner_symb.group("number") 
            symb = inner_symb.group("symb")
            if number == None:
                number = 1
            number = int(number)
            new_inner += symb + str(number * int(multipli))
        equation = re.sub("\(" + value.group(1) + "\)(?P<multipli_number>\d+)", new_inner, equation)
    return equation


print(format("Al2(SO4)3"))    

输出：

Al2S3O12