40
Last seen 4 days ago
Member for 9 years, 10 months, 19 days
Difficulty Advanced
First-artemoniux
Hi,
1. Why not `str(number)` instead of `string='{}'.format(number)`.
2. I'm sorry but, lines 5-7 are so funny :)
for i in string:
if not i == '0':
result *= int(i)
# or even:
for i in str(number):
if not i == '0':
result *= int(i)
3. Look at
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First-pavelnenov
Hi, you don't need the ...if..else:
return re.match("^(?=.*[a-z])(?=.*[A-Z])(?=.+[0-9]).{10,}$",data)
# or if you want:
return bool(re.match("^(?=.*[a-z])(?=.*[A-Z])(?=.+[0-9]).{10,}$",data))
More
Hi, the last if is redundant:
return length >= 10 and contains_lower and contains_upper and contains_digit
If data is to short you don't need to continue:
if len(data) < 10:
return False
etc.
Shortened version:
def checkio(data):
if len(data) < 10: return False
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First-talha.zaman
1
One-liner:
checkio = lambda data: len(data) >= 10 and all(map(set(data).__and__, map(set, (string.ascii_lowercase, string.ascii_uppercase, string.digits))))
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First-artemoniux
Hi, `string` and `result` are redundant. You can chain methods in Python:
return ','.join(phrases).replace('right','left')
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First-artemoniux
Hi, __bool__ is subtype of __int__ in Python, Therefore all the nested `if..else` statements are redundant e. g.:
if O == c:
R = x and y
elif O == d:
R = x or y
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First-artemoniux
Hi, you could unpack `date1` and `date1` with `*` e. g.:
date_1 = datetime.date(*date1)
and `d` is redundant:
return abs(date_2 - date_1)
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Perhaps a bit overcomplicated?-NerdAlert1101
Hi,
1. Creative?
2. That elif...else is redundant. More straightforward approach would be better here.
3. First if is redundant. See the preconditions.
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First-LukeMurray
Hi,
1. the last if is redundant.
2. Once an if condition is met, you can continue. E. g.:
if char.isdigit():
digitCount += 1
continue
3. You can move the last if inside the loop:
for ...:
...
if digitCount > 0 and upperCount > 0 and lowerCount > 0:
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First-LukeMurray
1
Hi,
1. Don't use _max_ as variable name. _max()_ is a useful built-in function.
2. Look at collections.Counter.
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First-LukeMurray
1
Ugh... this is so repetitive.
Shortened:
def checkio(n):
nums = (('M',1000), ('CM',900), ('D',500),
('CD',400), ('C',100), ('XC',90),
('L',50), ('XL',40), ('X',10),
('IX',9), ('V',5), ('IV',4),
('I',1))
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Hi,
1. All the `== True` are redundant.
2. The last `if` is redundant.
3. When `data` is too short, you don't need the for loop, you can immediately return.
if len(data) < 10:
return False
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Two-liner, short, universal-AntoniGoldstein
Hi,
1. Why two-liner and why c?
return ' '.join(sum(((v,) for k,v in {3: 'Fizz', 5: 'Buzz'}.items() if number % k == 0),())) or str(number)
2. Why _sum()_?
return ' '.join(v for k, v in {3: 'Fizz', 5: 'Buzz'}.items() if number % k == 0) or str(number)
3. Why not one-liner?
check
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