Is This An Efficient Calculator In Python?

May 27, 2024 Post a Comment

Is this an efficient calculator in Python? def calculator(): print('\nBasic Calculator.\n') num_1 = input('Enter your first number: ') operation = input('Enter your

Solution 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)

Solution 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':
        breakelse:
        print(operations[sign](int(x), int(y)))
except (ValueError, KeyError, ZeroDivisionError):
    print("Something went wrong, check your input")

Solution 3:

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

Baca Juga

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.

Solution 4:

Try this:

defcalculate(num1, num2, operator):
    operator = operator.strip()
    if operator.strip() in ['+', '-', '*', '/']:
        if operator == '/'andeval(num2) == 0:
            returnNonetry:
        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} !')

Solution 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))

Python Guru