re实现计算器

Posted 2020-12-16 rxybk

import re
def mul_div(exp):
    while re.search("[+-]{2,}",exp):
        exp = exp.replace("--", "+")
        exp = exp.replace("-+", "-")
        exp = exp.replace("++", "+")
        exp = exp.replace("+-", "-")
    if ‘*‘ in exp:
        a,b=exp.split(‘*‘)
        return str(float(a)*float(b))
    if ‘/‘ in exp:
        a,b=exp.split(‘/‘)
        return str(float(a)/float(b))
    if "+" in exp:
        a,b=exp.split("+")
        return str(float(a)+float(b))
    if "-" in exp:
        a , b = exp.split("-")
        return str(float(a) - float(b))
def remove_muldiv(srt):
    while True:
        ret=re.search("d+(.d+)?[/*]-?d+(.d+)?",srt) #3*4
        if ret: #3*4
            son_srt=ret.group()  #3*4  2*-1388335.8476190479
            res=mul_div(son_srt)   #12
            srt=srt.replace(son_srt,res)   #替换
        else:
            return add_div(srt)
def add_div(s):
    while True:
        sre=re.search("[+-]?d+(.d+)?[+-]-?d+(.d+)?",s) #1+6
        if sre:
            s1=sre.group() # 1+6
            s2=mul_div(s1) #7
            s=s.replace(s1,s2)
        else:
            return s
def main(s0):
    s0=s0.replace(" ","")
    shj=re.findall("[a-z]|[A-Z]",s0)
    if s0.count("(") == s0.count(")") and s0.count("（") == s0.count("）") and not shj:
        while True:
            s=re.findall("(([^()]*))",s0)  #s=[‘16-8‘]
            ss=re.findall("([^()]*)",s0)  #s=[‘(16-8)‘]
            if s :
                srd=remove_muldiv(s[0])  # 8.0
                s0=s0.replace(ss[0],srd)
            elif s0:
                print(remove_muldiv(s0))
                break
    else:
        print("输入错误")
main(‘1 - 2 * ( ( (6 0) -3 0  +(-40/5) * (9-2*5/3 + 7 /3*99/4*2998 +10 * 568/14 )) - (-4*3)/ (16-3*2) )‘)
main("56464+4864564+874*(8745+123-5564+1243+(7+665+-448))")

这个的加减法可能比较low，更高级的加减法看另外一篇文章