40
Last seen 4 days ago
Member for 9 years, 10 months, 19 days
Difficulty Advanced
Hi, look at [Counter](https://docs.python.org/3.4/library/collections.html#collections.Counter).
More
Hi as data is a list of integers you don't need _operator.add()_:
return reduce(int.__add__, data)
More
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
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
Hi,
n in range(len(array))
is not a good way. Check this:
0 <= n < len(array)
More
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
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
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
Hi. A small note: Using dictionary instead of if .. elif .. else could save you some typing.
More
Hi,
1. Line 2 is redundant: __tuple__ is iterable.
2. Line 3 and 4 are redundant: `return ','.join(x).replace('right', 'left')`
Regards,
suic
More
Hi,
1. _issufix()_ is redundant see [_str.endswith()_](https://docs.python.org/3.5/library/stdtypes.html?highlight=str.endswith#str.endswith)
2. `is not` and `!=` don't do the same thing (look at [this](https://docs.python.org/3.5/library/stdtypes.html#comparisons))
3. Instead of:
if word1 is
More