Is this an efficient calculator in Python?

问题

def calculator():

    print("\nBasic Calculator.\n")

    num_1 = input("Enter your first number: ")
    operation = input("Enter your operation: ")
    num_2 = input("Enter your second number: ")

    if operation == ("+"):
        sum = float(num_1) + float(num_2)
        print ("The answer is:",(sum))

    elif operation == ("-"):
        sum = float(num_1) - float(num_2)
        print ("The answer is:",(sum))

    elif operation == ("*"):
        sum = float(num_1) * float(num_2)
        print ("The answer is:",(sum))

    elif operation == ("/") and num_2 == ("0"):
        print ("\nYou cannot divide by zero.")

    elif operation == ("/"):
        sum = float(num_1) / float(num_2)
        print ("The answer is:",(sum))

    else:
        print("Invalid Operation.")

    restart = input("\nDo you want to enter another equation? Yes or No?").lower()

    if restart == ("yes"):
        calculator()

    else:
        print ("\nEnding Program.")
        quit()

calculator()

回答1:

You can use eval()

a = 1
b = 2
operation = '/'
print(eval(f'{a} {operation} {b}'))

0.5

Handle shenanigans by users:

a = 1
b = 0
operation = '/'
try:
    print(eval(f'{a} {operation} {b}'))
except Exception as exp:
    print(exp)

回答2:

Here is another basic example:

operations = {
    '+': lambda x, y: x + y,
    '-': lambda x, y: x - y,
    '*': lambda x, y: x * y,
    '/': lambda x, y: x / y,
}

try:
    x, sign, y = input("Enter expression separated by whitespace(ex: 2 + 3): ").split()
    if sign == '0':
        break
    else:
        print(operations[sign](int(x), int(y)))
except (ValueError, KeyError, ZeroDivisionError):
    print("Something went wrong, check your input")

回答3:

It's decent, but reviews of working code belong on CodeReview.SE, not here.

Call the result variable result instead of sum, that obviously only is meaningful for addition.
As per AlexanderLekontsev's answer, you don't need a huge big if...else ladder that always computes result and prints the output. A dispatch dictionary to (binary or unary) lambda function is better. You could have all the functions be binary, and default arg2=None, that way you can handle unary functions.
You're assuming the user types in valid floats in response to num_1, num_2. But what if they press return? or type pi or e? or 'help' or :-D etc. You should catch the exception ValueError: could not convert string to floatand display the user's invalid input back to them, "expected a number"(/"operator").
- You only need num_2 if operation is a binary not a unary operation, but future stuff like sqrt, log, log10, trigonometrics (sin, cos, tan), hyperbolics and their inverses (arc-fns) are all unary operations. Just something to keep in mind for the future. Don't hardwire your parser to one expected input sequence.
- Inputting numbers could get more complicated in future. What if you wanted to support both hexadecimal 7e8 and float/general/exponential notation 7e8? You might need multiple try...except clauses. You might add a HEX mode in future. But then you'll need to generalize from num1 to say arg1, and if arg1 == HEX then enable(/toggle) hex mode, and recurse/loop.
Suggest printing print("Invalid Operation: must be +,-,*,/,..."), this actually tells the user which operations are legal. So: % isn't, neither is ^, neither is log, cos, sqrt etc.
So if you implement the above, you can support things like e^x
Supporting parentheses would require recursion.

回答4:

Try this:

def calculate(num1, num2, operator):
    operator = operator.strip()
    if operator.strip() in ['+', '-', '*', '/']:
        if operator == '/' and eval(num2) == 0:
            return None
    try:
        result = eval(f'float({num1.strip()}) {operator} float({num2.strip()})')
    except:
        return ""
    return result

num1 = '3'
num2 = '5'
operator =  '+'

result = calculate(num1, num2, operator)

if result == '':
    print('Wrong expression !')
elif result == None:
    print('Dive bye zero !')
else:
    print(f'The answe is {result} !')

回答5:

Your code is alright but we can improvise by using eval()

print(" Basic Calculator ")
i = ""
while i != 'exit':
    i = input(" Enter the expression to evaluate or type 'exit' to exit : ")
    print(eval(i))

回答6:

Here is a very clean and short calculator script:

num1 = float(input("Enter a number: "))
op = (input("Enter an operation: "))
num2 = float(input("Enter another number: "))
if op == "*":   
    print(num1 * num2)
elif op == "/":
    print(num1 / num2)
elif op == "+":
    print(num1 + num2)
elif op == "-":
    print(num1 - num2)
elif op == "^":
    print(num1 ** num2)
else:
    print("error, you did not enter a supported operation")

Also if you were wondering ** means ^ or to the power of.

来源：https://stackoverflow.com/questions/62381867/is-this-an-efficient-calculator-in-python

标签

python

performance

calculator