Python3实现逆波兰表达式计算

来源：互联网发布：linux系统jdk下载官网编辑：程序博客网时间：2024/06/07 06:31

import  logginglogging.basicConfig(level=logging.INFO)import re##判断字符串是否为小数def isnumber(num):    regex = re.compile(r"^(-?\d+)(\.\d*)?$")    if re.match(regex,num):        return True    else:        return False#自定义栈class PyStack(object):    def __init__(self,initSize=20,incSize=10):        self.initSize=incSize        self.incSize=incSize        self.stackList=[]        self.top=self.bottom=0    def push(self,ele):        if self.top-self.bottom>=self.initSize:            self.incSize+=self.initSize        self.stackList.append(ele)        self.top+=1    def pop(self):        if self.top-self.bottom>0:            self.top-=1            ret=self.stackList.pop()            return ret        else:            return None    def len(self):        return  self.top-self.bottom##算式转化为中缀表达式列表def equation2List():    equation=input("请输入你要运算的方程:")    result=[]    buffNum=[]    for i in equation:        if i.isdigit() or i=='.':            buffNum.append(i)        else:            if len(buffNum)>0:                result.append("".join(buffNum))                buffNum.clear()            result.append(i)    if len(buffNum)>0:        result.append("".join(buffNum))        buffNum.clear()    logging.info(result)    return result##中缀表达式转后缀表达式## 1+2*(3+4)=15##1 2 3 4 + * +def mid2EndSuffix(list):    resultList=[]    stack=PyStack()    for i in list:        if isnumber(i):            resultList.append(i)        elif ')' == i:            while stack.len()>0:                item=stack.pop()                logging.debug(")pop==%s"%(item))                if '('==item:                    break                else:                    resultList.append(item)        elif '+' == i or '-' == i:            if stack.len() == 0:                stack.push(i)                logging.debug("+-None=push==%s"%i)            else:                while stack.len()>0:                    item2=stack.pop()                    logging.debug("+-=pop==%s"%item2)                    if '(' == item2:                        stack.push(item2)                        logging.debug("+-=(push==%s"%item2)                        break                    else:                        resultList.append(item2)                stack.push(i)                logging.debug("+-lastpush==%s"%i)        elif '*' == i or '/' == i or '(' == i:            stack.push(i)            logging.debug("*/(push==%s"%i)        else:            print("输入格式有误")    while stack.len()>0:        item3=stack.pop()        logging.debug("last==pop=%s"%item3)        resultList.append(item3)    return  resultList##后缀表达式计算结果def   calEndSuffixResult(list):    stack=PyStack()    sumEnd=0    if len(list)==0:        return sumEnd    for i  in list:        if isnumber(i):            stack.push(float(i))        elif '+'==i:            a=stack.pop()            b=stack.pop()            stack.push(b+a)        elif '-'==i:            a = stack.pop()            b = stack.pop()            stack.push(b - a)        elif '*'==i:            a = stack.pop()            b = stack.pop()            stack.push(b * a)        elif '/'==i:            a = stack.pop()            b = stack.pop()            if a==0:                print('%d/%d分子不能为0'%(b,a))            else:                stack.push(b/a)    return stack.pop()if __name__=='__main__':    eList=equation2List()    midEList=mid2EndSuffix(eList)    logging.info(midEList)    lastResult=calEndSuffixResult(midEList)    print("算式的结果是:",lastResult)

目前程序还有个缺陷，控制台输入的数字一般是斜体，如果是直体会无法识别。

暂时不清楚为啥会出现直体的情况，如果有知道的同学可以解答下

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