40
Last seen 4 days ago
Member for 9 years, 10 months, 19 days
Difficulty Advanced
First-ViacheslavHaidei
Hi, _bin()_ return __str__ which is iterable, you don't need those indices to access elements:
def checkio(number):
b_str = bin(number)
count = 0
for e in b_str:
if e == '1':
count += 1
return count
# or use str.count() metho
More
oneliner-AlexanderPavlichuk
Hi, _str()_ is redundant. _bin()_ return __str__.
More
Hi, this is not pythonic:
1. `from math import *` ugh!!! Import everything from `math` for a function which you don't need... is the reason for thumbs down.
2. Look at `//`.
3. `if not data:` instead of `if len(data) == 0:`
4 `else` branch on lines 11-17 is redundant:
def checkio(data):
More
First-ViacheslavHaidei
Hi,
1. __str__ is iterable
2. `d` id redundant.
3. Line 8 is redundant.
def checkio(number):
res = 1
s = str(number)
for e in s:
if e != '0':
res * int(s[i])
return res
# or even
for e in str(number):
More
First-ViacheslavHaidei
Hi,
n in range(len(array))
is not a good way. Check this:
0 <= n < len(array)
More
Hi, you have weird sense of clear. This solution is not pythonic, e. g.:
1. Import `string` is redundant.
2. __str__ has _str.islower()_, _str.isupper()_ and _str.isdigit()_ methods.
etc.
More
First-ViacheslavHaidei
Hi,
1. this:
if gb:
flag = True
else:
flag = False
return flag
you can simply return `gb`:
return gb
2. You can replace all that `#search` code with _str.isalpha()_.
3. In fact you don't need `gb` and `flag` variables:
def checkio(words):
count
More
First-ViacheslavHaidei
Hi, Python has __set__ which has _set.intersection()_ method and there's even a set intersection operation `&`. Also look at _sorted()_.
def checkio(first, second):
set1 = set(first.split(","))
set2 = set(second.split(","))
return ",".join(sorted(set1 & set2))
More
First-GabrielHahmann
1
Hi, two comments:
1. Don't use _list_ as variable name. It's a [built-in type](https://docs.python.org/3/library/stdtypes.html?highlight=list#list). And it's also quite confusing as there's a _list.append_ in _list_ type, but you have to call it in a different way:
list.append(some_list,
More
First-ViacheslavHaidei
Hi, __str__ is not good type for accumulating values as it's immutable. Look at _filter()_ function.
More
Hi, look at _filter()_ built-in and the _str.join()_ method.
return "".join(filter(str.isupper, text))
More
First-Wulstigemoehre
Hi, `else` branch is redundant:
if len(a) > 0:
...
return 0
More
Hi, why not str.isupper instead of:
(letter.islower() == False) and (letter.isalpha() == True) #?
More
Hi, he last if does not make sense, as you return the same in both branches. You could write: return False.
More
Hi, try to look at list comprehensions and generator expressions in python. E. g.:
result = [w for w in text if w.isupper()]
More
Hi, look at str.isupper:
# You could write:
l.isupper()
# instead of:
ord(l) in range(ord("A"), ord("Z")+1)
More
