9
Troy Walters
Last seen 4 years ago
Member for 10 years, 4 months, 14 days
Difficulty Normal
Nice work, but you can also do this in one line without importing the re library.
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Good job. You can clean this up a bit.
for word in words:
if text.lower().find(word) != -1:
words_found += 1
can be written more succinctly as:
for word in words:
if word in text.lower():
words_found += 1
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First-RomanZhyvitski
Although your code works, it could be A LOT simpler. I would check out some of the other answers for some ideas as to how you can accomplish this more elegantly.
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Nice work with the one-liner, but you can get rid of that ugly alphabet string using the .isupper() method.
find_message=lambda t:''.join(l for l in t if l.isupper())
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Nice. You can simplify this a bit. You don't need to make str(number) a list. A string is already an iterable.
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Too confusing. You can implement this in a much cleaner and shorter way without a while loop.
Also:
range(0,len(seq)-1) == range(len(seq)-1)
True
You dont need the 0
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Cool. Very similar to mine. Good idea to use endswith(), I forgot about that one.
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nice, but I don't think a Class is appropriate for this problem. What purpose would an instance of BooleanAlgebra class serve in OOP?
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This works, but its super complicated. You can do this without any looping or branching.
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Check out Python list comprehensions to do away with that nested for, if loop.
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Well done with relatively few lines of code. Line 5 is is very long and I can't see all of it. Might want to break that up into several lines for readability.
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