python 實現計算器功能 輸入字串,輸出相應結果

會飛的兩棵樹發表於2018-12-28
import re
formul=`1 - 2 *( (6 0- 30+(0-40/5) * (9-2* 5/3 +7 /3*99/4*2998 +10 *568/14)) - (-4*3) / (16-3*2))`
def zhengli(func):
    func = func.replace("+-","-").replace("--","+")
    return func
def addjian(func):
    func = func.strip("()")
    ret = re.findall("-?d*.?d+",func)
    sum=0
    for i in ret:
        sum +=float(i)
    return str(sum)
def chengchu(func):
    func = func.strip("()")
    while("/"in func or "*" in func):
        ret = re.search(`d*.?d+[*/]-?d*.?d`,func).group()
        digital = re.split(`([*/])`,ret)         #加個括號是為了優先順序的原因,保留非數字
        result = str(float(digital[0])*float(digital[2])) if digital[1] == `*` else str(float(digital[0])/float(digital[2]))
        func = func.replace(ret,result,1)
        func = zhengli(func)
    return func


def comput(formal):
    formal = re.sub(`s`,``,formal)
    while(re.search(`([^()]*)`,formal)):
        print(formal)
        ret = re.findall(`([^()]*)`,formal)
        for i in ret:
            print("wu kuohao:"+i)
            result_chengchu = chengchu(i) if `*`in i or `/` in i else i#先算乘除
            result_addjian = addjian(result_chengchu) if `+`in result_chengchu or `-` in result_chengchu else result_chengchu#再算加減
            formal = formal.replace(i,result_addjian,1)
            formal = zhengli(formal)
    else:
        result_chengchu = chengchu(formal) if `*` in formal or `/` in formal else formal  # 先算乘除
        result_addjian = addjian(result_chengchu) if `+` in result_chengchu or `-` in result_chengchu else result_chengchu  # 再算加減
        return result_addjian


print(comput(formul))

 

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