# !/usr/bin/env python3
# _*_coding:utf-8_*_
'''
实现模拟计算器的功能:
公式: - * ( (- +(-/) * (-*/ + /*/* + * / )) - (-*)/ (-*) )
思路:
括号的计算优先级最高,应该优先计算括号内的式子,然后将值代替会原式子,循环,接着做惩处运算,最后是加减运算,然后弹出结果,
# print(eval('1-2*((60-30+(-40/5)*(9-2*5/3+7/3*99/4*2998+10*568/14))-(-4*3)/(16-3*2))'))
'''
import re
symbol = ['', '', '', '', '', '', '', '', '', '', '-', '+', '*', '/', '(', ')']
def check(formula):
'''
检查输入的公式的正确性:
除了十个数字和六个符号,其他的一律判错,但是这样会限制运算规则,目前先是这样规划
:param formula:
:return:
'''
symbol_flag = True
formula = re.sub(r'\s*', '', formula)
cal_list = list(formula)
for item in cal_list:
if item not in symbol:
symbol_flag = False
if not symbol_flag:
return
else:
return formula
def change(formula):
'''
替换出现两个运算符在一起的情况:
:param formula:
:return:
'''
formula = formula.replace('--', '+')
formula = formula.replace('++', '+')
formula = formula.replace('**', '*')
formula = formula.replace('//', '=')
formula = formula.replace('*+', '*')
formula = formula.replace('/+', '/')
return formula
def chengchu(arg):
val = arg[]
match = re.search(r'\d+\.*\d*[\*\/]+[\+\-]?\d+\.*\d*', val)
if not match:
return
content = re.search(r'\d+\.*\d*[\*\/]+[\+\-]?\d+\.*\d*', val).group()
if len(content.split('*')) > :
n1, n2 = content.split('*')
value = float(n1) * float(n2)
else:
n1, n2 = content.split('/')
value = float(n1) / float(n2)
before, after = re.split(r'\d+\.*\d*[\*\/]+[\+\-]?\d+\.*\d*', val, )
new_str = '%s%s%s' % (before, value, after)
arg[] = new_str
print(arg)
chengchu(arg)
def jiajian(arg):
while True:
if arg[].__contains__('+-') or arg[].__contains__("++") or arg[].__contains__('-+') or arg[].__contains__(
"--"):
arg[] = arg[].replace('++', '+')
arg[] = arg[].replace('+-', '-')
arg[] = arg[].replace('--', '+')
arg[] = arg[].replace('-+', '-')
else:
break
if arg[].startswith('-'):
arg[] +=
arg[] = arg[].replace('-', '&')
arg[] = arg[].replace('+', '-')
arg[] = arg[].replace('&', '+')
arg[] = arg[][:]
val = arg[]
match = re.search(r'\d+\.?\d*[-+]{1}\d+\.?\d*', val)
if not match:
return
content = re.search(r'\d+\.?\d*[-+]{1}\d+\.?\d*', val).group()
if len(content.split('+')) > :
n1, n2 = content.split('+')
value = float(n1) + float(n2)
else:
n1, n2 = content.split('-')
value = float(n1) - float(n2)
before, after = re.split(r'\d+\.?\d*[-+]{1}\d+\.?\d*', val, )
new_str = '%s%s%s' % (before, value, after)
arg[] = new_str
jiajian(arg)
def compute(formula):
'''
操作加减乘除:根据操作
:param formula:
:return:
'''
inp = [formula, ] #
chengchu(inp)
jiajian(inp)
if divmod(inp[], )[] == :
result = float(inp[])
result = float(result) * -
else:
result = float(inp[])
return result
def calcultor(formula):
'''
递归处理括号,并计算括号内的式子,将结果返回做替换,
:param formula:
:return:
'''
# 没有括号了
if not re.search(r'\(([-+*/]*\d+\.*\d*){2,}\)', formula):
final = compute(formula)
return final
# 取出其中的一个括号,然后计算
content = re.search(r'\(([-+*/]*\d\.*\d*){2,}\)', formula).group()
# 分裂表达式
before, nothing, after = re.split(r'\(([-+*/]*\d+\.*\d*){2,}\)', formula, )
# 计算括号里的内容
content = content[:len(content) - ]
ret = compute(content)
# 计算完后然后拿到结果,然后重新拼接成一个新的式子
formula = '%s%s%s' % (before, ret, after)
return calcultor(formula)
# s='1-2*((60-30+(-40/5)*(9-2*5/3+7/3*99/4*2998+10*568/14))-(-4*3)/(16-3*2))'
# calcultor(s)
def main():
print('欢迎使用计算器'.center(, '#'))
while True:
formula = input('计算式>>:').strip()
if formula == 'q':
exit()
elif len(formula) == :
continue
else:
formula=check(formula)
if formula == :
continue
else:
result = calcultor(formula)
print('计算结果为:\033[1;31;1m %s\033[0m' % (result))
try:
print('正确答案是:\033[1;31;1m %s\033[0m' %(eval(formula)))
except:
print('无法计算')
if __name__ == '__main__':
main()