22
Last seen 5 years ago
Member for 8 years, 4 months
Difficulty Normal
Would like to receive comments/critics of my published solutions.
First-milind1992
Line 15: no need to create a special function to return negation, just use Python's:
return (not x) or y
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Lines 6 and 7 are not needed. Also, line 4 can be written just like:
if znak.isupper():
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Line 30: rather than use string concatenation, using format is preferred:
def __repr__(self):
return 'Building({}, {}, {}, {}, {})'.format(
self.south, self.west, self.width_WE, self.width_NS, self.height)
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1. collections.deque is more suitable than list for popping and inserting on different sides...
2. ...and if you choose deque, you can skip popping and inserting all together, by using deque.rotate
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Line 27: rather than use string concatenation, using format is preferred:
def __repr__(self):
return 'Building({}, {}, {}, {}, {})'.format(
self.south, self.west, self.width_WE, self.width_NS, self.height)
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Changed the line 8 to:
neighbours = sum(sum(hood) for hood in neighbourhood)
since @veky said in a recent comment that sum is "clearer, faster and more consistent with that subtraction later" than count.
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Your first line could be written as:
stripped = [c.lower() for c in text if c.isalpha()]
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Lines 5-8 can be written simply as:
constraint = min(capacity, number)
Lines 12-18 could be simplified by using list comprehension:
sides_to_paint = [i for i in range(number)] * 2
(This is a random review - sorry if this has already been mentioned)
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On line 13, more pythonic way would be the thing you did in line 8:
return not stack
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In your last line, there's no need for if-else, just have it like this:
return prepared(first_word) == prepared(second_word)
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Line 23: rather than use string concatenation, using format is preferred:
def __repr__(self):
return 'Building({}, {}, {}, {}, {})'.format(
self.south, self.west, self.width_WE, self.width_NS, self.height)
More
First-kapranowroman
Lines 4-11 could be replaced with:
if i in '({[':
lastopen.append(i)
Lines 30-33 could be replaced with:
return not lastopen # the same as: return len(lastopen) == 0
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First-derfunkenfliegenauf
No need for bool in line 13, just do:
return not stack
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Lines 2-4 changed to:
points = {point for pair in connections for point in pair}
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1. You import itertools and you don't use it
2. No need for if-else in line 3, you can just return it like this:
return sorted(first_word.lower().replace(" ", "")) == sorted(second_word.lower().replace(" ", ""))
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First-SukumarBurra
Line 2: you can create list just by: l = []
Lines 8 and 10: good job knowing integer other than 0 is 'truish' (if len(l)), but you can similarly just check if the list is not empty by: "if l", no need for 'len'.
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