27
Last seen 4 years ago
Member for 8 years, 7 months, 6 days
Difficulty Normal
You should confuse your code more thoroughly like this:
key=lambda x:(lambda x:(abs(x)))(x)
sorry 4 that joke )))
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`checkio = lambda _: Solution()` it`s COOLLL!
than also:
class Solution:
__lt__ = __ge__ = __le__ = __eq__ = __ne__ = __gt__ = lambda *_: True
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I like your 1-liner very much, but I can`t give 4 stars, because your code is not dry for a little :)
combine = lambda x: ''.join(sorted(x))
and
[combine(c[i] + c[j])) for i, j in ((0, 3), (1, 2), (4, 5))]
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1. No need erite 0 in
item[0:1] ===> item[:1]
2. String is a sequence so it will suffice to write
alphabet, number = item
3. Following "dry"
guard_1, guard_2 = [( chr ( ord (alphabet) + skew) + str ( int (number) -1 )) for skew in ( -1 , 1 )]
or
guard_1, guard_2 = [f'{ chr(o
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Falls on:
assert safe_code("19--3=##") == 2
but I think it`s just mark
('--', '-*-') --->replace to ---> ('--', '+'), ('*-', '*(-1)*')
right?
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grid = [[chr(ord(c) & 0xFFDF) for c in r] for r in grid]
for clear category? Why not just `c.upper()` if it has same result?
from string import ascii_letters
assert all(chr(ord(x) & 0xFFDF) == x.upper() for x in ascii_letters) == True
___
You can use `itertools.product()` for `for in
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It`s more effective to allocate resources (like "not_repeated = []") after if-statemaent with return.
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Would be more effective to use list instead str for current
self.current=[] if None else [current]
in __init__() of Text
for write() method use just
self.current.append(text)
and than
''.join(self.current)
in show().
And
self.current, self.font = previous
in restore() instead 2 lines
Othe
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Due to pep8 we should not use upper-letters for naming variables, except constants. Also would be better to give meaningful names, like "row_count, "column_count", or just 'row' and 'col'.
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There would not have been this solution without [research](https://py.checkio.org/mission/the-secret-room/publications/Merzix/python-3/caution-pure-magic/share/74103cdf0c2429d922792e9e7179f8ae/#comment-64186) conducted by [Phil15](https://py.checkio.org/user/Phil15/)
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Secret Message-Michal_Szajer
1
1
1.
char.lower() != char.upper()
that will be always True
2) instead
char.upper() == char
just use
char.isupper()
3) more efficiency would be using list, than ''.join(it)
4) list comprehension is mush shotter then simple for consruction
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First-Oleg_Domokeev
1
The more you compress the code, the more difficult to read it. But it`s very good 2-liners solution.
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You could use
import string
letters = string.ascii_lowercase() instead of calculating the sequence using magic numbers
letters = [chr(x) for x in range(97, 123)]
Also, use just
retun answer == 26 instead
return True if answer == 26 else False
or even
if answer == 26:
return True
return False
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I like to read such clear code.
It`s not need to use '.keys()' just "for i in sailors:" or even better " for name, age in sailors.items():
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May be .split() move to trim() like
def trim(key: list):
key = key.split()
for ... and
lock, key = trim(lock), trim(key) Also if you do not use variable in for use _ for name it
for _ in range(4): Other is good!
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There is a function for finding (max-min) called [numpy.ptp](http://docs.scipy.org/doc/numpy/reference/generated/numpy.ptp.html) if that's useful for you
max(indices) - min(indices) == ptp(indices)
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