> I've got a small function that I'm using to check whether a password is of a
> certain length and contains mixed case, numbers and punctuation.
>
> Originally I was using multiple "if re.search" for the patterns but it looked
> terrible so I've read up on list comprehensions and it's slightly improved. I
> just can't escape the feeling that there's a more elegant way to do this that
> I'm missing.
>
> I've been looking through all the python stuff that I thought might be
> relevant (lambda, map, filter, set, frozenset, etc) but nothing has come
> together. Just wondering if anyone has suggested reading material for
> alternate ways they'd handle this code.
set does seem to have what you want: isdisjoint() could do the trick.
Eg:
if set(punctuation).isdisjoint(password) or
set(digits).isdisjoint(password) or set(ascii_uppercase).isdisjoint(password)
or set(ascii_lowercase).isdisjoint(password):
return False
return True
You could even confuse yourself and put it all on one line:
return not any(set(chars).isdisjoint(password) for chars in [punctuation,
digits, ascii_uppercase, ascii_lowercase])
but I wouldn't recomended it. I guess this can even shortened further.
(note: the 'not' operator is needed, because isdisjoint returns True for what
you probably prefer as False.)
Evert
> CODE:
>
> from string import ascii_lowercase, ascii_uppercase, digits, punctuation
>
>
> def complex_password(password):
> """Checks password for sufficient complexity."""
> if len(password) < 12:
> return False
> if len([c for c in password if c in punctuation]) == 0:
> return False
> if len([c for c in password if c in digits]) == 0:
> return False
> if len([c for c in password if c in ascii_uppercase]) == 0:
> return False
> if len([c for c in password if c in ascii_lowercase]) == 0:
> return False
> return True
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