def myAtoi(self, str):
"""
:type str: str
:rtype: int
"""
if not str:
return 0
# https://docs.python.org/2/library/itertools.html
# takewhile() pred, seq seq[0], seq[1], until pred fails
# takewhile(lambda x: x<5, [1,4,6,4,1]) --> 1 4
from itertools import takewhile as tw
s = str.lstrip()
sign = list(tw(lambda x: x in '+-', s))
digits = list(tw(lambda x: x in '0123456789', s[1:] if sign else s))
try:
result = int((sign[0] if sign else "") + "".join(digits))
except Exception:
return 0
return max(min(result, 2**31 - 1), -2**31)
def common_url_prefix(urls):
def same(values):
return all(v == values[0] for v in values[1:])
from six.moves.urllib import parse
parsed = [parse.urlparse(url) for url in urls]
# We need the same protocol and netloc (host/port), otherwise there
# is not a common prefix.
if not same([u[:2] for u in parsed]):
return None
# If we get here, it means we have a shared protocol and netloc, so
# now we just compare paths.
from itertools import takewhile as tw
parts = zip(*[p.path.split('/') for p in parsed])
commonpath = '/'.join(x[0] for x in tw(same, parts))
res_parts = parsed[0][:2] + (commonpath, '', '', '')
res = parse.urlunparse(res_parts)
if res[-1] != '/':
res += '/'
return res
# itertools.starmap tests
starmap = itertools.starmap
assert list(starmap(pow, zip(range(3), range(1,7)))) == [0**1, 1**2, 2**3]
assert list(starmap(pow, [])) == []
assert list(starmap(pow, [iter([4,5])])) == [4**5]
with assert_raises(TypeError):
starmap(pow)
# itertools.takewhile tests
from itertools import takewhile as tw
t = tw(lambda n: n < 5, [1, 2, 5, 1, 3])
assert next(t) == 1
assert next(t) == 2
with assert_raises(StopIteration):
next(t)
# not iterable
with assert_raises(TypeError):
tw(lambda n: n < 1, 1)
# not callable
t = tw(5, [1, 2])
with assert_raises(TypeError):
next(t)
# non-bool predicate
"""The 5-digit number, 16807=7^5, is also a fifth power. Similarly,
the 9-digit number, 134217728=8^9, is a ninth power.
How many n-digit positive integers exist which are also an nth power?
"""
from itertools import count, chain, takewhile as tw
ans = len(list(chain.from_iterable(tw(lambda x: len(x) > 0,
(list(filter(lambda x: len(str(x)) == n,
(m**n for m in range(1,10))))
for n in count(1))))))
print(ans)
def beginning_zeros(number: str) -> int:
return sum(1 for _ in tw(lambda n: n == '0', number))
def end_zeros(num: int) -> int:
return sum(1 for _ in tw(lambda ch: ch == '0', str(num)[::-1]))
