"python中的计算器"

Question

I am trying to make calculator that can solve expressions with basic 4 operators, like 1+2*3-4\/5, however it does not work and I do not know what is wrong. Please check my code. When I run it, I am getting infinte number of errors in 8. line return ret(parts[0]) * ret(parts[2])<\/code> Here is code

def ret(s):
    s = str(s)
    if s.isdigit():
        return float(s)
    for c in ('*','/','+','-'):
        parts = s.partition(c)
        if c == '*':
            return ret(parts[0]) * ret(parts[2])
        elif c == '/':
            return ret(parts[0]) / ret(parts[2])
        elif c == '+':
            return ret(parts[0]) + ret(parts[2])
        elif c == '-':
            return ret(parts[0]) - ret(parts[2])
print(ret('1+2'))

Answer 1

The main thing you are doing wrong is that you are checking the value of c rather that the partitioned operator. You can also unpack the result from s.partition to make things a little easier by using left and right for the actual operations.

def ret(s):
    s = str(s)
    if s.isdigit():
        return float(s)
    for c in ('-','+','*','/'):
        left, op, right = s.partition(c)
        if op == '*':
            return ret(left) * ret(right)
        elif op == '/':
            return ret(left) / ret(right)
        elif op == '+':
            return ret(left) + ret(right)
        elif op == '-':
            return ret(left) - ret(right)
print(ret('1+2'))

Also, you will need to reverse the order of your operations as you want to first do addition and subtraction, followed by multiplication and division.

What I mean is, if you have an expression like 4+4*3 , you want to divide it into

ret(4) + ret(4 * 3)

Since it is a recursive call, you want the operators with the highest precedence to be the last on the call stack so they are executed first when the function returns.

As an example:

print(ret('1+2*6'))
print(ret('3*8+6/2'))

OUTPUT

13.0
27.0

Answer 2

You partition the input string regardless, never checking if the operator is even there. .partition() returns empty strings if the partition character is not present in the input:

 >>> '1+1'.partition('*')
 ('1+1', '', '')

So you'll call s.partition('*') but never check if there is any such operator present, resulting in unconditional calls to ret() . You'll always call ret(parts[0]) * ret(parts[2]) regardless of wether * is present in s or not.

The solution is to either test for the operator first or to check the return value of .partition() . The latter is probably easiest:

for c in ('+','-','*','/'):
    parts = s.partition(c)
    if parts[1] == '*':
        return ret(parts[0]) * ret(parts[2])
    elif parts[1] == '/':
        return ret(parts[0]) / ret(parts[2])
    elif parts[1] == '+':
        return ret(parts[0]) + ret(parts[2])
    elif parts[1] == '-':
        return ret(parts[0]) - ret(parts[2])

Note that I reversed the operator order; yes, multiplication and division need to be applied before addition and subtraction, but you are working in reverse here; splitting up the expression into smaller parts, and the operations are then applied when the sub-expression has been processed.

You could use assignment unpacking to assign the 3 return values of .partition() to easier names:

for c in ('+','-','*','/'):
    left, operator, right = s.partition(c)
    if operator == '*':
        return ret(left) * ret(right)
    elif operator == '/':
        return ret(left) / ret(right)
    elif operator == '+':
        return ret(left) + ret(right)
    elif operator == '-':
        return ret(left) - ret(right)

Next you can simplify all this by using the operator module , which has functions that perform the same operations as your arithmetic operations. A map should do:

import operator
ops = {'*': operator.mul, '/': operator.div, '+': operator.add, '-': operator.sub}

for c in ('+','-','*','/'):
    left, operator, right = s.partition(c)
    if operator in ops:
        return ops[operator](ret(left), ret(right))

Answer 3

Your dispatching is incorrect. The way you defined your function it will try to split by '*', which basically calls the ret function recursively with the same arguments...

You'll have to check first whether an "operator" is present in your argument string at all.

Think again!

Answer 4

In your code you have no condition on the result of s.partition(c) , so even if the partition results in ('anything', '', '') you will do a recursion on the first if.

Edited

Answer 5

Here is a simple python calculator program, feel free to use it:

#Python calculator

def menu():
    print ("Welcome to calculator.py")
    print ("your options are:")
    print (" ")
    print ("1) Addition")
    print ("2) Subtraction")
    print ("3) Multiplication")
    print ("4) Division")
    print ("5) Quit calculator.py")
    print (" ")
    return input ("Choose your option: ")

def add(a,b):
    print (a, "+", b, "=", a + b)

def sub(a,b):
    print (b, "-", a, "=", b - a)

def mul(a,b):
    print (a, "*", b, "=", a * b)

def div(a,b):
    print (a, "/", b, "=", a / b)

loop = 1
choice = 0
while loop == 1:
    choice = menu()
    if choice == 1:
        add(input("Add this: "),input("to this: "))
    elif choice == 2:
        sub(input("Subtract this: "),input("from this: "))
    elif choice == 3:
        mul(input("Multiply this: "),input("by this: "))
    elif choice == 4:
        div(input("Divide this: "),input("by this: "))
    elif choice == 5:
        loop = 0

print ("Thank you for using calculator.py!")

Answer 6

how to condition users input of number 1 & number 2, if its in string. with if/else or try/except.

print("\n Select A Number \n")
print("Press 1 for Addition")
print("Press 2 for Subtraction")
print("Press 3 for multiplication")
print("Press 4 for Division \n")

while True:
    user = input("Enter A Choice(1/2/3/4): ")
    if user in ('1','2','3','4'):
        num1 = int(input("Enter First Number: "))
        num2 = int(input("Enter Second Number: "))

        if user == '1':
            print(f"{num1} + {num2} = {num1 + num2}")
            print("Thanks For Usnig This Calculator")
        elif user == '2':
            print(f"{num1} - {num2} = {num1 - num2}")
            print("Thanks For Usnig This Calculator")
        elif user == '3':
            print(f"{num1} * {num2} = {num1 * num2}")
            print("Thanks For Usnig This Calculator")
        elif user == '4':
            print(f"{num1} / {num2} = {num1 / num2}")
            print("Thanks For Usnig This Calculator")
        else:
            print("Please Enter Correct Choice")
    else:
        print("Invalid")

Answer 7

#1. Create a class with the name Calculator
class Calculator:
    #2. Declare two class variables num1 and num2
    num1=""
    num2=""
    result=""
    option=""
    #3. Initialize the class variables num1 and num2 in constructor __init__()
    def __init__(self):
        self.num1=0
        self.num2=0
        self.result=0
        self.option=0
    #4. Make getter and setter function for variables
    def getNum1(self):
        return self.num1
    def setNum1(self,n1):
        self.num1=n1 
    def getNum2(self):
        return self.num2
    def setNum2(self,n2):
        self.num2=n2
    def getResult(self):
        return self.result
    def setResult(self,r):
        self.result=int(r)
    def getOption(self):
        return self.option
    def setOption(self,op):
        self.option=op
    
    #5. Create Main Manu function to display choices
    def mainMenu(self):
        print("==============")
        print("==CALCULATOR==")
        print("==============")
        print("Menu:")    
        print("1. Addtion")
        print("2. Subtraction")
        print("3. Multiplication")
        print("4. Division")
        print("5. Remainer")
        print("6. Exit")
        print("")
        self.setOption(int(input("Enter Option:")))
        self.setNum1(int(input("Enter Number1:")))
        self.setNum2(int(input("Enter Number2:")))
        
    #6. Make a function Take Input
    def takeInput(self):
        if self.option==1:
            self.setResult(self.getNum1()+self.getNum2())
        elif self.option==2:
            self.setResult(self.getNum1()-self.getNum2())
        elif self.option==3:
            self.setResult(self.getNum1()*self.getNum2())
        elif self.option==4:
            self.setResult(self.getNum1()/self.getNum2())
        elif self.option==5:
            self.setResult(self.getNum1()%self.getNum2())
            
    #7. Make a display function to display the output
    def display(self):
        resultOption=""
        print("")
        if self.option==1:
            resultOption="Addition"
        elif self.option==2:
            resultOption="Subtraction"
        elif self.option==3:
            resultOption="Multiplication"
        elif self.option==4:
            resultOption="Division"
        elif self.option==5:
            resultOption="Remainder"
        if self.option==1 or self.option==2 or self.option==3 or self.option==4 or self.option==5:
            print(resultOption+":"+str(self.getResult()))
        else:
            print("Invalid Input")
        self.option=input("press x to exit, 0 to continue...")
        if self.option=="x":
            pass
        else:
            self.option=int(self.option)
            

#8. Create Object of Calculator Class  
c1=Calculator()
op=0

#9. Get option in while loop until x is pressed
while(c1.getOption()!='x'):
    #10.    Call display menu, input and display functions to display the output
    c1.mainMenu()
    c1.takeInput()
    c1.display()

Answer 8

I have an alternative to your code. The user can enter stuff like: 8*6/4-3+3 and this will still work. Very compact and if you want a command to exit out of the:

while True:

Just add:

if(y == "end" or y == "End"):
    break

And it will exit out of the "while True:"

Code (Python v3.3.0):

while True:
    x = "x="
    y = input(" >> ")
    x += y
    exec(x)
    print(x)

Answer 9

Here is simple code in python for creating a calculator same as in Python terminal.

number = input("")
print number