I have a menu that will return 'e' unless the input is d or D. I would like to do it without making another variable and doing it on one line

encrypt = 'd' if (raw_input("Encrypt or Decrypt a file(E/d):") == ('d' or 'D')) else 'e'

[Edit] Ok here is a harder one

How can I do the same for this

file_text = 'a.txt' if (raw_input("File name(a.txt):")=='a.txt' else [What I typed in]

• python
• input
• validation

Use the in operator:

encrypt = 'd' if raw_input("Encrypt or decrypt a file (E/d):") in ('d', 'D') else 'e'

Alternatively, you can just convert the input to lowercase and compare it to 'd':

encrypt = 'd' if raw_input("Encrypt or decrypt a file (E/d):").lower() == 'd' else 'e'

Finally, if you want to ensure that they enter e or d, you can wrap it up in a while loop:

while True:
    encrypt = raw_input("Encrypt or decrypt a file (E/d):")

    # Convert to lowercase
    encrypt = encrypt.lower()

    # If it's e or d then break out of the loop
    if encrypt in ('e', 'd'):
        break

    # Otherwise, it'll loop back and ask them to input again

Edit: To answer your second question, you can use a lambda for it I guess?

file_text = (lambda default, inp: default if inp.lower() == default else inp)("a.txt", raw_input("File name(a.txt):"))

Although, this is clearly a bit obtuse and too "clever" by half.

Not really meant seriously but another 1-line solution (I don't think it's readable):

encrypt = {'d':'d','D':'d'}.get(raw_input("Encrypt or decrypt a file (E/d):"), 'e')

At least it's short. Sometimes a dictionary is actually useful for similar situations (if there are more choices).