python计算器
作业需求：
开发一个简单的python计算器：
1、实现加减乘除及拓号优先级解析
2、用户输入 1 - 2 * ( (60-30 +(-40/5) * (9-2*5/3 + 7 /3*99/4*2998 +10 * 568/14 )) - (-4*3)/ (16-3*2) ) 等类似公式后，必须自己解析里面的(),+,-,*,/符号和公式(不能调用eval等类似功能偷懒实现)，运算后得出结果，结果必须与真实的计算器所得出的结果一致作业总结
本次作业做了好几个版本，
一开始脑子是空白的，越看题目越傻了；
后来就想的整体写不出来，我留先写乘除加减的函数。结果第一版出来不支持负数，第二版支持负数；
最后又单独写去括号；
再将去括号和计算合并，可以实现作业的要求。后来发现代码只是仅仅可以实现作业的需求，和简单的正数类的运算；输入其他中间包含负数之类的就会产生计算错误，越写越没法收场了。。。
思路重新整理，通过网上查阅各版本思路，回过头来重新整理自己的套路及方法，历尽沧桑呀。
正确的答案请阅读片段：“作业思路重新整理”作业思路
主要思路
1、找到最里边的括号
2、把加减和乘除放到函数里，不断的去调取。
具体思路
1、先写出乘除计算的函数，可以进行不包含括号的计算
2、再写出去括号的函数
3、合并代码
计算器流程图
乘除函数第一版：
#第一次完成，可以随便 乘 除 加 减 计算，但是不支持负数；
import re
def convert_list(a,arg):
list1 = [] #转换后的列表
num = ""
for i in a:
if num.isnumeric() and i.isdigit() :
list1.pop()
num = num + i
elif i == "-":
num = i
elif i == "." :
list1.pop()
num = num +i
elif num[-1:] == "." and i.isdigit():
print(num[-1:])
list1.pop()
num = num+i
else:
num = i
list1.append(num)
for i2 in list1:
arg.append(i2)
# print(a,"==",arg)
# return a
return arg
if __name__ == "__main__":
# a = "1+10.0*3.0/2-2"
a = input("please input Calculation formula:<1+3*2-1>:")
# 转换成列表
listlist = []
convert_list(a,listlist)
print(a,"===",listlist)
#计算乘除
while True:
print(a)
if "*" in listlist or "/" in listlist:
quzhi = re.search("\d+.\d+\*\d+\.\d+|\d+\.\d+\*\d+|\d+\*\d+\.\d+|\d+\*\d+|\d+.\d+\/\d+\.\d+|\d+\.\d+\/\d+|\d+\/\d+\.\d+|\d+\/\d+",a).group()
print(type(quzhi))
print(quzhi)
quzhi2 = []
quzhi2 = convert_list(quzhi,quzhi2)
print(type(quzhi),type(quzhi2))
print(quzhi2)
print(quzhi2[2])
print("=================")
result = float()
if "*" in quzhi2:
# print(True)
print(quzhi2.index("*"))
num3 = quzhi2.index("*")
list_n = quzhi2[0:num3]
print(list_n)
x = "".join(list_n)
list2_n = quzhi2[num3+1:]
y = "".join(list2_n)
print(x,y)
result = float(x) * float(y)
print(result)
re1 = listlist.index("*")
else:
result = float(quzhi2[0]) / float(quzhi2[2])
# print(listlist.index(quzhi2[1]))
# print(quzhi2)
re1 = listlist.index("/")
listlist.pop(re1 - 1)
listlist.pop(re1 - 1)
listlist.pop(re1 - 1)
listlist.insert(re1-1,str(result))
print(listlist)
#给变量a 重新赋值
a = "".join(listlist)
print(a)
else:
break
while True:
print(a)
if "+" in listlist or "-" in listlist:
quzhi = re.search("\-\d+\*\d+|\-\d+\/\d+|\-\d+\+\d+|\-\d+\-\d+|\d+.\d+\+\d+\.\d+|\d+\.\d+\+\d+|\d+\+\d+\.\d+|\d+\+\d+|\d+.\d+\-\d+\.\d+|\d+\.\d+\-\d+|\d+\-\d+\.\d+|\d+\-\d+",a).group()
# quzhi = re.search(
# "\d+.\d+\+\d+\.\d+|\d+\.\d+\+\d+|\w+\d+\+\d\.\d+|\w+\d\+\d|\d+.\d+\-\d+\.\d+|\d+\.\d+\-\d+|\w+\d+\-\d\.\d+|\w+\d\-\d",
# a).group()
print(type(quzhi))
print(quzhi)
quzhi2 = []
quzhi2 = convert_list(quzhi, quzhi2)
print(type(quzhi), type(quzhi2))
print(quzhi2)
print(quzhi2[2])
print("=================")
result = float()
if "+" in quzhi2:
# print(True)
print(quzhi2.index("+"))
num3 = quzhi2.index("+")
list_n = quzhi2[0:num3]
x = "".join(list_n)
list2_n = quzhi2[num3 + 1:]
y = "".join(list2_n)
print(x, y)
result = float(x) + float(y)
print(result)
re1 = listlist.index("+")
else:
result = float(quzhi2[0]) - float(quzhi2[2])
# print(listlist.index(quzhi2[1]))
# print(quzhi2)
re1 = listlist.index("-")
listlist.pop(re1 - 1)
listlist.pop(re1 - 1)
listlist.pop(re1 - 1)
listlist.insert(re1 - 1, str(result))
print(listlist)
# 给变量a 重新赋值
a = "".join(listlist)
print(a)
else:
break乘除函数第二版：
# 乘 除 加 减 后 精简的,优化版本,支持负数；
#修复后支持负数，其实就是： 步骤一、字符串转列表 步骤二、正则匹配
import re
def convert_list(a,arg):
list1 = [] #转换后的列表
num = ""
for i in a:
if num.isnumeric() and i.isdigit() :
list1.pop()
num = num + i
elif i == "." :
list1.pop()
num = num +i
elif num[-1:] == "." and i.isdigit():
print(num[-1:])
list1.pop()
num = num+i
else:
num = i
list1.append(num)
for i2 in list1:
arg.append(i2)
if arg[0] == "-":
arg.insert(1,arg[0]+arg[1])
arg.pop(0)
arg.pop(1)
print(arg)
return arg
def ccjj(a,arg):
# 计算乘除
while True:
if "*" in listlist or "/" in listlist:
quzhi = re.search(
"\-\d+.\d+\*\d+.\d+|\-\d+.\d+\/\d+.\d+|\-\d+\*\d+|\-\d+\/\d+|\d+.\d+\*\d+\.\d+|\d+\.\d+\*\d+|\d+\*\d+\.\d+|\d+\*\d+|\d+.\d+\/\d+\.\d+|\d+\.\d+\/\d+|\d+\/\d+\.\d+|\d+\/\d+",
a).group()
quzhi2 = []
quzhi2 = convert_list(quzhi, quzhi2)
result = float()
if "*" in quzhi2:
num3 = quzhi2.index("*")
list_n = quzhi2[0:num3]
x = "".join(list_n)
list2_n = quzhi2[num3 + 1:]
y = "".join(list2_n)
result = float(x) * float(y)
re1 = listlist.index("*")
else:
result = float(quzhi2[0]) / float(quzhi2[2])
re1 = listlist.index("/")
listlist.pop(re1 - 1)
listlist.pop(re1 - 1)
listlist.pop(re1 - 1)
listlist.insert(re1 - 1, str(result))
# 给变量a 重新赋值
a = "".join(listlist)
else:
break
while True:
if "+" in listlist or "-" in listlist:
quzhi = re.search(
"\-\d+.\d+\+\d+.\d+|\-\d+.\d+\-\d+.\d+|\-\d+\+\d+|\-\d+\-\d+|\d+.\d+\+\d+\.\d+|\d+\.\d+\+\d+|\d+\+\d+\.\d+|\d+\+\d+|\d+.\d+\-\d+\.\d+|\d+\.\d+\-\d+|\d+\-\d+\.\d+|\d+\-\d+",a).group()
quzhi2 = []
quzhi2 = convert_list(quzhi, quzhi2)
result = float()
if "+" in quzhi2:
num3 = quzhi2.index("+")
list_n = quzhi2[0:num3]
x = "".join(list_n)
list2_n = quzhi2[num3 + 1:]
y = "".join(list2_n)
result = float(x) + float(y)
re1 = listlist.index("+")
else:
num3 = quzhi2.index("-")
list_n = quzhi2[0:num3]
x = "".join(list_n)
list2_n = quzhi2[num3 + 1:]
y = "".join(list2_n)
result = float(x) - float(y)
re1 = listlist.index("-")
listlist.pop(re1 - 1)
listlist.pop(re1 - 1)
listlist.pop(re1 - 1)
listlist.insert(re1 - 1, str(result))
# 给变量a 重新赋值
a = "".join(listlist)
else:
break
return a
if __name__ == "__main__":
# a = "1+10.0*3.0/2-2"
while True:
a = input("please input Calculation formula:<1+3*2-1>:")
print(a)
# 转换成列表
listlist = []
convert_list(a,listlist)
print(listlist)
a = ccjj(a,listlist)
print("=================\n计算结果为:", a)
while True:
agin = input("\n您还要继续输入吗？<y/n>:")
if agin == "y":
break
elif agin == "n":
print("\n 谢谢光临！！")
exit()
else:
print("\n{输入错误，请重新输入！！}\n")作业版本一：（去括号、加减乘除）
import re,sys
#字符串转列表函数 功能：输入的字符串格式的计算公式转换为列表的形式
def convert_list(a,arg):
list1 = [] #转换后的列表
num = ""
flag = 1
for i in a:
if i.isdigit() and num.isnumeric() :
list1.pop()
num = num + i
elif num.isdigit() and i == ".":
num = num + i
flag = 2
continue
elif num in "+-*/" and i == "-":
num = i
flag = 2
elif flag == 2 and i.isdigit():
list1.pop()
num = num + i
elif flag == 2 and i == ".":
list1.pop()
num = num + i
elif flag == 2 and i in "+-*/":
num = i
flag = 1
elif flag == 2 and i is not enumerate:
list1.append(num)
flag = 1
num = i
elif i.isdigit():
num = i
else:
num = i
list1.append(num)
for i2 in list1:
arg.append(i2)
return arg
#计算乘除加减函数 功能：传入计算公式的字符串和列表格式，进行判断计算
def ccjj(a,arg):
# 计算乘除
count = 0
while True:
if "*" in arg or "/" in arg:
quzhi = re.search(
"\-\d+\.\d+\*\d+\.\d+|\-\d+\*\d+\.\d+|\d+\*\d+|\-\d+\.\d+\/\d+\.\d+|\-\d+\/\d+\.\d+|\-\d+\.\d+\*\-\d+\.\d+|\-\d+\.\d+\*\-\d+|\-\d+\.\d+\/\-\d+\.\d+|\-\d+\.\d+\/\-\d+|\-\d+\/\-\d+\.\d+|\-\d+\/\-\d+|\d+\.\d+\*\-\d+\.\d+|\d+\.\d+\*\-\d+|\d+\*\-\d+\.\d+|\d+\*\-\d+|\d+\.\d+\/\-\d+\.\d+|\d+\.\d+\/\-\d+|\d+\/\-\d+\.\d+|\d+\/\-\d+|\d+\.\d+\*\d+\.\d+|\d+\.\d+\*\d+|\d+\*\d+\.\d+|\d+\*\d+|\d+\.\d+\/\d+\.\d+|\d+\.\d+\/\d+|\d+\/\d+\.\d+|\d+\/\d+|\d+\.\d+\*\-\d+\.\d+|\d+\.\d+\*\-\d+|\d+\*\-\d+\.\d+|\d+\*\-\d+|\d+\.\d+\/\-\d+\.\d+|\d+\.\d+\/\-\d+|\d+\/\-\d+\.\d+|\d+\/\-\d+",
a).group()
quzhi2 = []
quzhi2 = convert_list(quzhi, quzhi2)
print("乘除法取的值：",quzhi,quzhi2)
result = float()
if "*" in quzhi2:
result = float(quzhi2[0]) * float(quzhi2[2])
re1 = arg.index("*")
# print("乘法完之后：",result)
else:
result = float(quzhi2[0]) / float(quzhi2[2])
re1 = arg.index("/")
arg.pop(re1 - 1)
arg.pop(re1 - 1)
arg.pop(re1 - 1)
if float(result) < 0:
pass
else:
if float(quzhi2[2]) < 0 and count !=0:
print("retrun_a:::111", result)
# print(return_a)
return_a = str(-float(result))
print("retrun_a:::222", result)
if a[0] == "-" and count ==0 :
print("retrun_a:::11", result)
# print(return_a)
result = str(-float(result))
print("retrun_a:::22", result)
arg.insert(re1 - 1, str(result))
print("乘除法结果加入到列表：",arg)
# 给变量a 重新赋值
a = "".join(arg)
count +=1
else:
break
#计算加减
while True:
count = 0
if "+" in arg or "-" in arg:
quzhi = ""
try:
quzhi = re.search(
"\-\d+\.\d+\-\d+\.\d+|\-\d+\.\d+\-\d+|\-\d+\.\d+\+\d+\.\d+|\-\d+\.\d+\+\d+|\-\d+\+\d+\.\d+|\d+\+\d+|\-\d+\.\d+\-\d+\.\d+|\-\d+\-\d+\.\d+|\-\d+\.\d+\+\-\d+\.\d+|\-\d+\.\d+\+\-\d+|\-\d+\.\d+\-\-\d+\.\d+|\-\d+\.\d+\-\-\d+|\-\d+\-\-\d+\.\d+|\-\d+\-\-\d+|\d+\.\d+\+\-\d+\.\d+|\d+\.\d+\+\-\d+|\d+\+\-\d+\.\d+|\d+\+\-\d+|\d+\.\d+\-\-\d+\.\d+|\d+\.\d+\-\-\d+|\d+\-\-\d+\.\d+|\d+\-\-\d+|\d+\.\d+\+\d+\.\d+|\d+\.\d+\+\d+|\d+\+\d+\.\d+|\d+\+\d+|\d+\.\d+\-\d+\.\d+|\d+\.\d+\-\d+|\d+\-\d+\.\d+|\d+\-\d+|\d+\.\d+\+\-\d+\.\d+|\d+\.\d+\+\-\d+|\d+\+\-\d+\.\d+|\d+\+\-\d+|\d+\.\d+\-\-\d+\.\d+|\d+\.\d+\-\-\d+|\d+\-\-\d+\.\d+|\d+\-\-\d+",
a).group()
except:
pass
print("加减法取的值：",quzhi)
quzhi2 = []
quzhi2 = convert_list(quzhi, quzhi2)
# print(quzhi,quzhi2)
result = float()
if "+" in quzhi2:
result = float(quzhi2[0]) + float(quzhi2[2])
re1 = arg.index("+")
# print(result)
else:
result = float(quzhi2[0]) - float(quzhi2[2])
re1 = arg.index("-")
print("加减法得出的结果：",result)
arg.pop(re1 - 1)
arg.pop(re1 - 1)
arg.pop(re1 - 1)
if float(result) <= 0:
pass
else:
if float(quzhi2[2]) < 0 and count != 0:
print("retrun_a:::111", result)
# print(return_a)
return_a = str(-float(result))
print("retrun_a:::222", result)
if a[0] == "-" and count ==0 :
print("retrun_a:::11", result)
# print(return_a)
result = str(-float(result))
print("retrun_a:::22", result)
arg.insert(re1 - 1, str(result))
# 给变量a 重新赋值
a = "".join(arg)
count =+1
# print("***:",arg)
if len(arg) < 3:
break
else:
break
return a
#去括号函数 功能：把输入的公式先计算括号里边的内容
def qukuohao(a,arg):
listlist = []
convert_list(a,listlist)
countcount = 1
while True:
print("\033[32m[[第%s次计算前：]]\033[0m"%(countcount),a)
try:
index1 = listlist.index(")")
# print("查看)的第几个位置：",index1)
# print(len(listlist))
except:
pass
for i in range(index1,-1,-1):
try:
if listlist[i] == "(":
new_list = listlist[i+1:index1] #获取括号里的内容
# print("new_list括号里的内容-列表形式---：",new_list) #打印获取括号里的内容，不包括括号
# print(listlist[i:index1+1])
del_count = len(listlist[i:index1+1]) #获取括号以及括号里的内容的长度，用于后边listlist.pop()
# print("del_count",del_count) #打印上边的长度
#以下if判断用于 判断括号里是否有计算公式，没有就删除listlist列表里的括号及里边内容并插入计算结果，再进行下次循环；
if len(new_list) < 3: #如果new_list的长度小于3，代表括号里边没有计算公式了；
a = "".join(new_list) #接下来可以直接去掉括号
listlist.pop(index1 - 2)
listlist.pop(index1 - 2)
listlist.pop(index1 - 2)
listlist.insert(index1 - 2,a) #去掉括号后，把括号里边的内容插入到listlist列表
# print("重新整理后的列表", listlist)
break
new_a = "".join(new_list) #括号里边的内容转化成字符串，为了运行 convert_list函数做准备
nn_list = [] #定义括号里的字符串内容转换后 用作的列表
#运行 字符串转成列表函数
convert_list(new_a,nn_list) #转换括号里的字符串new_a成新的列表nn_list
#运行 乘除加减函数
return_a = ccjj(new_a,nn_list) #将括号里的字符串形式 和 括号里字符串列表形式 传进去计算；结果return给a
#判断 删除括号以及括号里边的内容的次数，并删除括号以及括号里边的内容
count = 0
for i in range(0,del_count):
# print("----listlist前:", listlist)
if count >= del_count:
break
else:
listlist.pop(index1 - (del_count-1))
count += 1
#将括号里边计算出来的结果 加入到listlist列表中
listlist.insert(index1-(del_count-1),return_a)
a = "".join(listlist)
print("\033[31m[[第%s次计算后：]]\033[0m" % (countcount), a)
countcount += 1 #用于打印去括号计算的次数
break
except:
pass
else:
break
#如果计算到最后列表里边没有括号了，直接进行 乘除加减 计算
a = ccjj(a,listlist)
# print("\033[34m最后的计算结果是：%s\033[0m"%a)
print("\033[31m[[第%s次计算后：]]\033[0m" % (countcount), a)
return a
if __name__ == "__main__":
# a = "1-(5*2-5*3/2)+10"
count = 0
while True:
msg = "------------------------------------------------------------------------------\n" \
"请输入计算公式:\n" \
"如①：<1+3*2-1>\n" \
"如②：1-2*((60-30+(-40/5)*(9-2*5/3+7/3*99/4*2998+10*568/14))-(-4*3)/(16-3*2))\n" \
"------------------------------------------------------------------------------"
print(msg)
a = input("\033[42m please input Calculation formula:\033[0m")
a = a.replace(" ", "") # 去空格
if count < 3:
# 转换成列表
listlist = []
if "(" in a:
#去括号
a = qukuohao(a,listlist)
else:
convert_list(a, listlist)
# print(listlist)
# print(a)
a = ccjj(a, listlist)
p = len(a) + 20
for i in range(p):
sys.stdout.write("#")
sys.stdout.flush()
# time.sleep(0.1)
sys.stdout.write("\n")
print("\033[34m最后的计算结果是：%s\033[0m"%a)
for i in range(p):
sys.stdout.write("#")
sys.stdout.flush()
# time.sleep(0.1)
sys.stdout.write("\n")
elif count > 2 :
agin = input("\n您还要继续输入吗？<y/n>:")
if agin == "y":
break
elif agin == "n":
print("\n 谢谢光临！！")
exit()
else:
print("\n{输入错误，请重新输入！！}\n")存在问题
只支持作业上的计算和简单的加减计算；
1-2*((60-30+(-40/5)*(9-2*5/3+7/3*99/4*2998+10*568/14))-(-4*3)/(16-3*2))
但是计算公式中间有负数之类的就会计算错误：
例如：1-(5*2-5*3/2)+10 计算结果正确是8.5作业思路重新整理
1、加减乘除正则匹配重新整理；
2、计算结果替换 用上x.replace(source,dest) ；
3、字符串转列表循环修复；
4、去括号
获取括号里边的内容用正则匹配，而不是for循环；
计算结果替换 用上x.replace(source,dest) ；
5、用户输入公式后，先判断输入的是否包含非公式的不规则字符，有则提示“输入错误！”；
6、用户输入公式后，判断输入的字符串是否包含计算公式，不包含则不进入计算函数；
7、关键问题：将所有的加和减，都以加的形式计算；
例如：9-2*5 再匹配乘除计算的时候先是：-2*5；然后再9+-10
思路重新整理之前的问题：
1、正则匹配没有精简，也没有最佳匹配；
2、字符串转列表存在缺陷；
3、可能出现的公式形式没有想的周全，有些偏面；
4、知识点没能够合理运用，字符串替换；
5、最大问题是没有想清楚有负数的各种形式时候如何计算；最终作业版本
#!/bin/bash/env python
#_*_ coding:utf-8 _*_
#python version: 3.x
#author: 506554897
import re,sys
#字符串转列表函数
def convert_list(a,arg):
list1 = [] #转换后的列表
num = ""
flag = 1
count = 0
for i in a:
if i.isdigit() and num.isnumeric() :
list1.pop()
num = num + i
elif num.isdigit() and i == ".":
num = num + i
flag = 2
continue
elif num in "+-*/" and i == "-":
num = i
flag = 2
elif flag == 2 and i.isdigit():
list1.pop()
num = num + i
elif flag == 2 and i == ".":
list1.pop()
num = num + i
elif flag == 2 and i in "+-*/":
num = i
flag = 1
elif flag == 2 and i is not enumerate:
list1.append(num)
flag = 1
num = i
elif count == 0 and i == "+":
continue
elif i.isdigit():
num = i
else:
num = i
list1.append(num)
count +=1
for i2 in list1:
arg.append(i2)
# print("转成列表后的字符串和列表：",a,arg)
return arg
#加减乘除函数
def jjcc(a):
#乘除循环
while True:
arg = []
if "*" in a or "/" in a:
quzhi = re.search("\+?\-?\d+\.?\d*[*|/]\-?\d+\.?\d*", a).group()
quzhi2 = [] #用于保存正则取值后的列表
quzhi2 = convert_list(quzhi,arg)
# print("乘除取到的值：",quzhi,type(quzhi),type(quzhi2),quzhi2)
if quzhi2[1] == "*":
c = float(quzhi2[0]) * float(quzhi2[2])
elif quzhi2[1] == "/":
c = float(quzhi2[0]) / float(quzhi2[2])
if c > 0 : #用于结果是 +8 类似的，给取绝对值，去掉+ ；
a = a.replace(quzhi, "+" + str(c))
else:
a = a.replace(quzhi, str(c))
# print("乘除替换后的值：",a)
else:
break
#加减循环
while True:
#判断字符串是否还有计算公式的存在
arg = []
judge = convert_list(a,arg) #判断字符串 里边是否已经没有 加减乘除 的公式 的列表
if len(judge) < 3: #如果judge列表元素的长度小于3，代表已经没有计算公式了
if c > 0:
a = str(abs(float(a))) #用于结果是 +8 类似的，给取绝对值，去掉+ ；
# print("最后的结果：",a)
break
arg = []
if "+" in a or "-" in a:
quzhi = re.search("\+?\-?\d+\.?\d*[+|-]\-?\d+\.?\d*", a).group()
quzhi2 = []
quzhi2 = convert_list(quzhi, arg)
# print("加减取的值：",quzhi,type(quzhi),quzhi2,type(quzhi2))
if quzhi2[1] == "+":
c = float(quzhi2[0]) + float(quzhi2[2])
elif quzhi2[1] == "-":
c = float(quzhi2[0]) - float(quzhi2[2])
a = a.replace(quzhi, str(c))
# print("加减后的值：",a)
else:
break
return a
#去括号函数
def remove_parenthesis(a):
#去括号循环
countcount = 1 # 用于计算进行了几次去括号
while True:
if ")" in a:
print("\033[32m[[第%s次括号计算前：]]\033[0m" % (countcount), a)
quzhi = re.search("\([^()]+\)",a).group()
# print(quzhi)
quzhi2 = quzhi.strip("()") #去掉字符串里边的"()"
# print("去括号取到的值：",quzhi2)
result = jjcc(quzhi2) #运行加减乘除函数，进行计算
a = a.replace(quzhi,result) #将计算结果替换前边正则匹配的字符串
# print("<<去括号替换后的结果：>>",a)
else:
break
print("\033[31m[[第%s次括号计算后：]]\033[0m" % (countcount), a)
countcount += 1 # 用于打印去括号计算的次数
return a
if __name__ == "__main__":
# a = "1+10.0*3.0/2-2"
count = 0 #计算用户输入的公式次数
while True:
msg = "------------------------------------------------------------------------------\n" \
"\033[31m输入计算公式例子:\n\033[0m" \
"如①：1+3*2-1\n" \
"如②：1-2*((60-30+(-40/5)*(9-2*5/3+7/3*99/4*2998+10*568/14))-(-4*3)/(16-3*2))\n" \
"------------------------------------------------------------------------------"
print(msg)
a = input("\033[47m please input Calculation formula:\033[0m")
a = a.replace(" ", "") # 去空格
#判断是否输入计算公式以外的字符，例如中文、字母
try:
panduan = re.search("[^+|\-|*|/|(|)|\d]+",a).group()
if panduan is not None: # 如果输入字符串包含不规则就提示输入错误！！
print("\033[41m输入错误！！！\033[0m")
continue
except:
pass
#判断输入公式的次数
if count < 3:
# 检查输入的是否是公式
arg = []
judge = convert_list(a, arg) # 判断字符串 里边有没有 加减乘除 的公式 的列表
if len(judge) < 3: #如果judge的元素小于3，代表输入的字符串没有计算公司
if float(judge[0]) > 0:
a = str(abs(float(a))) #如果输入的+8 进行绝对值把 + 去掉
print("\033[31m输入的没有计算公式！\033[0m")
# 如果输入的字符串有公式
else:
a = remove_parenthesis(a) #去括号
print("\033[32m<<去完括号后的结果:>>\033[0m", a)
a = jjcc(a)
#统计“最后的计算结果是：XX”字符串的长度；“最后的计算结果是：”的长度是20 + a的长度
p = len(a) + 20
#根据p的长度打印 n个#
for i in range(p):
sys.stdout.write("#")
sys.stdout.flush()
# time.sleep(0.1)
sys.stdout.write("\n")
#打印计算结果
print("\033[34m最后的计算结果是：%s\033[0m"%a)
# 根据p的长度打印 n个#
for i in range(p):
sys.stdout.write("#")
sys.stdout.flush()
# time.sleep(0.1)
sys.stdout.write("\n\n")
#如果输入公式次数大于3次提示是否继续输入
elif count > 2 :
agin = input("\n您还要继续输入吗？<y/n>:")
if agin == "y":
break
elif agin == "n":
print("\n 谢谢光临！！")
exit()
else:
print("\n{输入错误，请重新输入！！}\n")转载于:https://blog.51cto.com/506554897/2052662