Keeping the way that you're doing this at the moment, you can:
- Change consonants to a string. This allows for easier readability, and may give better performance than a list.
- Name constants in
UPPER_SNAKE_CASE.
- Using
itertools.pairwise, you can remove all bar one if. This is as you can catigorise each letter, l in consonants. And then compair if they are the same.
- You should use a function, as it can be faster
- You should use a
if __name__ == '__main__' guard.
- You should use
print ''.join(letters), rather than for j in letters: sys.stdout.write(j).
- You can
yield from the function instead, so that if it's not fast enough, you can group prints. Or use it in other ways.
This can lead to:
import itertools
CONSONANTS = 'bcdfghjklmnpqrstvwxz'
VOWELS = 'aeiou'
def pairwise(iterable):
"s -> (s0,s1), (s1,s2), (s2, s3), ..."
a, b = itertools.tee(iterable)
next(b, None)
return izip(a, b)
def generate_answers(n):
consonants = CONSONANTS
for answer in itertools.product(consonants + VOWELS, repeat=n):
letter_categories = (l in consonants for l in answer)
if all(a != b for a, b in pairwise(letter_categories)):
yield ''.join(answer)
if __name__ == '__main__':
for answer in generate_answers(int(raw_input().strip())):
print(answer)
However, this is based around an inefficant usage of itertools.product.
You instead want to pass it the arguments so that it'll create them efficiantly.
This is easy to do when n is even:
def generate_answers(n):
return chain(product(CONSONANTS, VOWELS, repeat=n//2),
product(VOWELS, CONSONANTS, repeat=n//2))
To, do this efficiantly when n is odd, is also quite easy.
You want to do roughly the same thing, but if the first item of the products is not the same as the first character of the first argument, then stop looping.
And so you can use:
import itertools
CONSONANTS = 'bcdfghjklmnpqrstvwxz'
VOWELS = 'aeiou'
def product(a, b, repeat):
if repeat % 2 == 0:
for ret in itertools.product(a, b, repeat=repeat//2):
yield ret
else:
for ret in itertools.product(b, a, repeat=repeat//2+1):
if ret[0] != b[0]:
break
yield ret[1:]
def generate_answers(n):
return itertools.chain(product(CONSONANTS, VOWELS, repeat=n),
product(VOWELS, CONSONANTS, repeat=n))
if __name__ == '__main__':
for answer in generate_answers(int(raw_input().strip())):
print(''.join(answer))
