Python3.6基于正则实现的计算器示例【无优化简单注释版】

时间:2022-11-07 23:29:08

本文实例讲述了Python3.6基于正则实现的计算器。分享给大家供大家参考,具体如下:

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# -*- coding:utf-8 -*-
#!python3
import re
import copy
def my_calc(inside):
  """
  计算括号内的算术式
  :param inside:算术式
  :return:结果
  """
  while True:
    # 1、首先需要把含有优先级最高的*和/找出来
    # 这里有几种情况,(1*1) (1*-1) (-1*1)除法类似(暂时不考虑分母为0的情况)
    # 但是有了正则就方便多了
    search_ret = re.search('(\(-)?\d+(\.\d+)?[*/]-?\d+(\.\d+)?', inside)
    if search_ret is None:
      break
    ret_str = search_ret.group()
    if '(' in ret_str:
      ret_str = ret_str[1:]
    num_list = re.split('[*/]', ret_str)
    operator = re.search('[*/]', ret_str).group()
    calc_ret = 0
    if operator == '*':
      calc_ret = float(num_list[0]) * float(num_list[1])
    elif operator == '/':
      calc_ret = float(num_list[0]) / float(num_list[1])
    inside = inside.replace(ret_str, str(calc_ret))
  # */都运算完以后就可以做+-了
  while True:
    # 2、把含有+-的算术式找出来
    search_ret = re.search('(\(-)?\d+(\.\d+)?[+\-]-?\d+(\.\d+)?', inside)
    if search_ret is None:
      break
    ret_str = search_ret.group()
    if '(' in ret_str:
      ret_str = ret_str[1:]
    tmp_str = copy.copy(ret_str)
    num_1 = re.match('-?\d+(\.\d+)?', tmp_str).group()
    tmp_str = tmp_str.replace(num_1, '')
    operator = tmp_str[0]
    num_2 = tmp_str[1:]
    calc_ret = 0
    if operator == '+':
      calc_ret = float(num_1) + float(num_2)
    elif operator == '-':
      calc_ret = float(num_1) - float(num_2)
    inside = inside.replace(ret_str, str(calc_ret))
  return re.sub('[()]', '', inside)
def format_str(s):
  s = s.replace('--', '+')
  s = s.replace('-+', '-')
  s = s.replace('++', '+')
  s = s.replace('+-', '-')
  if s[0] == '+':
    s = s[1:]
  s = '('+s+')'
  return s
def un_bracket_calc(final_str): # -1*2+3-4/-5
  final_str = format_str(final_str)
  final_str = my_calc(final_str)
  return final_str
def my_math(s): # "((-1-2*-3)/(3-2)+(9*5-89)*(2*3*(3-0)))"
  while True:
    inside_bracket = re.search('[()]+[()]+', s)
    if inside_bracket is None:
      # 括号都算完了,如果还有算术式继续运算
      s = un_bracket_calc(s)
      break
    src_str = inside_bracket.group()
    ret = my_calc(src_str)
    s = s.replace(src_str, ret)
  return s
s_src = "((-1 - 2 * -3) / (3 - 2) + (9 * 5 - 9) * (2 * 3 * (3 - 0))) * -100 + 99-100 * -1-1"
s_src = s_src.replace(' ', '')
print(my_math(s_src))
s_ret = ((-1 - 2 * -3) / (3 - 2) + (9 * 5 - 9) * (2 * 3 * (3 - 0))) * -100 + 99 - 100 * -1 - 1
print(s_ret)

运行结果:

Python3.6基于正则实现的计算器示例【无优化简单注释版】

希望本文所述对大家Python程序设计有所帮助。

原文链接:https://blog.csdn.net/tyrantu1989/article/details/78473637