Question

I’m sure there is a really simple answer to this but I can’t find it after searching around for a while.

prefixes = "JKLMNOPQ"
suffix = "ack"

for letter in prefixes:
    if letter == "Q" or letter == "O":
        print letter + "u" + suffix
    else:
        print letter + suffix

The above code works perfectly, the condition on the if statement seems a bit long-winded so I tried:

if letter == "Q" or "O":

which is shorter but doesn’t work. I worked out that it doesn’t work because “O” is a boolean expression which will always be True and that it doesn’t consider anything on the left side of “or”.

I tried putting brackets around it like so:

if letter == ("Q" or "O"):

but that only matches Q and not O.

Is there any shorthand way of getting the code to work or do I have to use the long-winded way that’s working for me?

Answer 1

You can use letter in 'OQ' or re.match('[OQ]', letter)

More thoroughly, I’d probably define a string of prefixes that should get the ‘u’.

prefixes = 'JKLMNOPQ'
prefixes_taking_u = 'OQ'
suffix = 'ack'
for p in prefixes:
    u_part = 'u' if p in prefixes_taking_u else ''
    print p + u_part + suffix

In your code, letter == ("O" or "Q") first evaluates ("O" or "Q") (“O” isn’t false, so the result is “O”) and so this is the same as letter == "O".

Your other attempt, letter == "O" or "Q" first evaluates letter == "O" and if it’s True, it gives the answer True otherwise it’ll give the answer “Q”.

You can find the documentation for how or works in the language reference.