I am sorry, but I am at a loss here everyone. No real idea where to start. I am new to Python and its been many years since I last tried programming of any kind. I was never very experienced. I am following this Python course online and have come to a dead end on an exercise. It is going off of the previous example.
x = int(input())
if x>=1 and x<=26:
print('letter', x, 'in the alphabet:', chr(ord('A')+(x-1)))
else:
print('invalid input:', x)
This example program converts numbers to letters. So, 'A' = 1 and 'B' = 2. It now wants me to make a program just like this but the opposite. I am supposed to start with this line:
letter = input()
Here is what the exercise describes:
Coding Exercise: 26 Letters Write a program which does the reverse of the example above: it should take a character as input and output the corresponding number (between 1 and 26). Your program should only accept capital letters. As error-checking, print invalid if the input is not a capital letter
Now this is I can do, but not the way it wants me to do it, nor the way I should. I could do something like this:
if letter == 'A':
letter = int(1)
print(letter)
elif letter == 'B':
letter = int(2)
print(letter)
Of course that would be very long and sloppy. Does anyone have any advice for me? This one has me stuck. I'm sorry, I know this is probably very simple, this just seems much more complicated to me than it should be. I have tried looking this up but everything I find online is different enough that I cannot figure out how to implement it into this code. To be honest the chr and ord stuff confuses me greatly. I appreciate it for anyone who stuck around long enough to read this. Thank you. I have no idea what to do anymore with this one. Also, this is my very first time posting anything here, so if I did anything sloppy or wrong, I apologize and will delete/edit it as soon as I can.
The problem can be found here: https://cscircles.cemc.uwaterloo.ca/9-else-and-or-not/
Adding a little information. CScircles requires that A = 1 instead of 0. Thanks for the quick replies. I will be working on this again later today.