Python 代码无法正常工作

Question

Hi this is my second post on this code I dont get errors but the if else statement is not working how I want the if(calculate[2] == opperations: code is not working, it wont register even if i have a - or + there. Any ideas why? It gives me the response Sorry invalid text3.

calculate = input("Enter the problem in the format x + y = z ")

opperations = ["+", "-", "*", "/"]
numbers = ["0", "1", "2", "3", "4", "5", "6", "7", "8", "9"]
space = " "
a = calculate[0]
a = int(a)
b = calculate[4]
b = int(b)

def opperation():
if opperations == "+":
    A = 1
elif opperations == "-":
    A = 2
elif opperations == "*":
    A = 3
elif opperations == "/":
    A = 4

if calculate[0] in numbers:
    if len(calculate) > 1:
        if calculate[2] == opperations:
            if calculate[1] == space:
                if A == 1:
                    c = a + b
                    print (c)
                elif A == 2:
                    c = a - b
                    print (c)
                elif A == 3:
                    c = a * b
                    print(c)
                elif A == 4:
                    c = a / b
                    print(c)
                else:
                    print("Sorry invalid text5")
            else:
                print("Sorry invalid text4")
        else:
            print("Sorry invalid text3")
    else:
        print("Sorry invalid text2")
else:
    print("Sorry invalid text1") 

Answer 1

The condition

if calculate[2] == opperations:

will compare calculate[2] with the whole list ["+", "-", "*", "/"] . You can instead use the in keyword, to check if a element is inside a sequence:

if calculate[2] in opperations:
    # ...

Also , in the part where you compare opperations with "+" , "- ... You should be comparing them with calculate[2] :

if calculate[2] == "+":
    A = 1
elif calculate[2] == "-":
    A = 2
elif calculate[2] == "*":
    A = 3
elif calculate[2] == "/":
    A = 4

Notes: Your approach will just handle operations with 1 digit numbers. I would recommend you to use the split(" ") method so you can use it like:

s = "100 + 20"
parts = s.split(" ")   # split by space
parts[0] # 100
parts[1] # +
parts[2] # 20

Answer 2

In addition to Christian's suggestions, the if elif chain could be replaced by a dictionary lookup. In this example lambda a, b: a+b is a function that takes two values a and b and return there sum. That means opperations["+"] returns a function and opperations["+"](a,b) returns the sum of a and b .

opperations = { "+": lambda a, b: a+b, 
               "-": lambda a, b: a-b,
               "*": lambda a, b: a*b,
               "/": lambda a, b: a/b }

calculate = raw_input("Enter the problem in the format x + y = z ")

numbers = ["0", "1", "2", "3", "4", "5", "6", "7", "8", "9"]
space = " "
a = calculate[0]
a = int(a)
b = calculate[4]
b = int(b)

if calculate[0] in numbers:
    if len(calculate) > 1:
        if calculate[2] in opperations:
            if calculate[1] == space:
                c = opperations[calculate[2]](a,b)
                print (c)
            else:
                print("Sorry invalid text4")
        else:
            print("Sorry invalid text3")
    else:
        print("Sorry invalid text2")
else:
    print("Sorry invalid text1") 

Answer 3

I think it would be easier to make a new python file and then create a function in it and then import it:

def calculate(calculation):
    with open('calculation_file', 'w') as FILE:
        FILE.write(
        '''
        def calculate():
            return ''' + calculation + '''
        ''')
    import calculation_file as c
    return c.calculate

it creates a file and then imports it and returns what the calculation_file There are probably some bugs though because I have not run it yet also you probably need to make the variables a , b an intager with int(a)