python写计算器
#!/usr/bin/env python# -*- coding:utf-8 -*-import redef chu(arg1): #定义加减 arg = arg1[0] #beacuse price is a list ,so index 0 arg = arg.replace('--', '+').replace('++', '+').replace('
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#!/usr/bin/env python
# -*- coding:utf-8 -*-
import re
def chu(arg1): #定义加减
arg = arg1[0] #beacuse price is a list ,so index 0
arg = arg.replace('--', '+').replace('++', '+').replace('-+', '-').replace('+-', '-') #重点重点重点:就是对负数的一个替换
# r = '-9-2-5-75884-1.6666666666666667-80.0-0.6+18'
li = re.findall("[\-\+]?\d+\.?\d*", arg) #意思是提取每个数,用findall变成列表形式,然后循环相加,张宇我佩服你
# 意思是用findall提取每个值包括值前面的运算符
num = 0 #定义空值,然后循环列表让每个值相加
for i in li:
num += float(i)
return num
def multiply(arg): #definition multip
while True:
nu = re.split("(\d+\.?\d*[*\/][\-]?\d+\.?\d*)",arg,1) #['-9-2-5-', '2*5', '/3+7/3*99/4*2998+10*568/14']
print(nu)
if len(nu) == 3:
bef, cen, aft = nu #split get there price
nu_cen = re.split("[*/]",cen) #['2', '5']
nu_bef, nu_af = nu_cen
# print(nu_bef,type(nu_af))
if "*" in cen: #如果*在中间那个
nu_cen = re.split("\*", cen)
# print(nu_cen)
nu_bef, nu_af = nu_cen
sum = float(nu_bef)*float(nu_af)
# print(sum)
nu = bef + str(sum) + aft #重新组合定义arg形参
arg = nu
return multiply(arg) #返回重新定义函数
elif "/" in cen:
nu_cen = re.split("[/]", cen)
nu_bef, nu_af = nu_cen
sum = float(nu_bef)/float(nu_af)
nu = bef + str(sum) + aft
arg = nu
return multiply(arg)# ['-9-2-5-3.3333333333333335+173134.50000000003+405.7142857142857']
else: #这个时候如果不等于3,那就是只剩下加减了执行加减
return chu(nu)
#return arg
#acc = "-9-2-5-244*311-5/3-40*4/2-3/5+6*3"
#acc = "-9-2-5-75884-5/3-160/2-3/5+18"
# acc = "-9-2-5-2*5/3+7/3*99/4*2998+10*568/14"
# ac = acc.strip(" ")
# a = multiply(ac)
# print(a)
# def multiply(arg):
# return 1
origin = "1 - 2 * ( (60-30 +(-9-2-5-24*11-5/3-40*4/2-3/5+6*3) * (-9-2-5-2*5/3 + 7 /3*9/4*98 +10 * 568/14 )) - (-4*3)/ (16-3*2) )"
#寻找括号最里面的括号
while True: # 只要里面还有最里面的括号,就循环
origin = re.sub(r"\s*","",origin) #no1 strinp *****space**
print(origin)
res = re.split("\(([^()]+)\)",origin,1) #分割括号str得到第一个最里面括号内的值 no2
if len(res) == 3: #equal(等于)3,证明have bracket
before,centor,after = res #no3
print(centor)#centor是最里面的字符串,也是计算新函数乘法或除的实参
r = multiply(centor) #定义definition function multiply or ride (no4)
new_res = before + str(r) + after #重新组合
origin = new_res #重新定义origin
print(origin)
else:
final = multiply(origin)
print(final)
break
#this is comments
"""
acc = "-9-25-75884-5/3-160/2-3/5+18"
nu = re.split("(\d+\.?\d*[*][\-]?\d+\.?\d*)",acc,1)
print(type(nu[0]))
"""
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