40
Last seen 4 days ago
Member for 9 years, 10 months, 19 days
Difficulty Advanced
First-eugene128736871
Hi, it's kinda verbose:
1. `== None` is redundant.
2. You can use _all()_ instead of those `if`s.
3. In each if condition you have the same pattern.
return all(re.search(rgx, d) for rgx in ("^.{10,}$", "[0-9]+", "[a-z]+", "[A-Z]+"))
More
First-eugene128736871
Hi,
1. _max()_ and _dict()_ are a built-in functions.
2. Look at _sorted_, e. g.: `alpha = sorted(dict)`
3. Even if Python's exceptions are not expensive, lines 2-9 are quite ugly.
4. Look at __collections.Counter__.
I've played with your code. Result is [here](http://www.checkio.org/mission/most-
More
Hi, a few comments:
1. `else: pass` branches are redundant. They don't do anything.
2. You could write `sum += stack[-1]` instead of `sum = sum + stack[len(stack)-1]`.
3. and also `if stack:` instead of `if len(stack) > 0`.
P. S.: It's better to avoid `sum` as variable name. It's a useful built-in
More
Hi,
1. `else: continue` is redundant.
2. You could use _any()_ or _all()_ + generator expression.
More
Hi, you could use _any()_:
any(c1.endswith(c2) or c2.endswith(c1) for c1, c2 in combinations(words_set, 2))
More
First-levor2384
Hi, you could write:
if word1 != word2 and word1.endswith(word2):
return True
or even use _any()_ with _generator expression_.
More
Hi,
1. You could write alphabet as a string: "abcdefghijklmnoprstuvwxyz" as __str__ is is iterable.
2. The if is redundant as:
if condition:
return True
else:
return False
# is the same as
return condition
More
First-MichaelAzar
Hi, I have a few comments:
1. Line 2-4 for are useless. On line 8 you use x.isalpha() which returns False for white spaces, so you don't need to remove them previously. All you need to preserve is this:
text = text.lower()
# and then replace no_white with text
for x in text:
More
First-Benjamin.michelland
Hi, let me deduce from your code first:
1. You're comming from the C-world.
2. You speak French.
I have a few pieces of advices:
1. Forget about semicolons in Python :)
2. Look at [_String Methods_](https://docs.python.org/3.5/library/stdtypes.html?highlight=str#string-methods).
More
Hi, _bin()_ return __str__ so `str(bin_num)` is redundant. As `bin(number)` is a __str__ you can use _str.count()_ method.
More
Hi,
1. Look at _divmod()_.
2. Look at [this](https://docs.python.org/3.4/library/stdtypes.html#truth-value-testing).
def checkio(number):
multiple=1
while number:
number, digit = divmod(number, 10)
if digit: multiple *= digit
return multiple
More
Hi,
1. Why not?:
if len(array) > 0:
# or even:
if array:
# and:
for i in array[::2]:
# instead of
for i in array[0: len(array): 2]:
2. Look at _sum()_:
# sum of even elements:
sum(array[::2])
More
Hi, instead of lines 14-17 you could write:
return brackets == []
# or
return not brackets
More
You like typing :)
import re
VOWELS = "[AEIOUY]"
CONSONANTS = "[BCDFGHJKLMNPQRSTVWXZ]"
r = re.compile(r'\A({0}?({1}{0})*{1}?){2}\Z'.format(VOWELS, CONSONANTS, '{1}'), re.I)
def checkio(text):
return sum(bool(len(w) > 1 and r.match(w)) for w in re.split(r'
More
First-AndreyDorofeyev
1
Hi,
1. You can import string.ascii_lowercase instead of `abc`.
2. `... if ... else ...` is redundant.
return all(w.lower() in text.lower() for w in abc)
More
