15
Last seen 5 years ago
Member for 9 years, 3 months, 23 days
Difficulty Normal
regexp-lv_shadowoflv
2
The quantifiers (plus signs) in your regex are superfluous. There is no need to match more than one character of each of the character classes.
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if foo == bar:
return True
else:
return False
can be written more concisely as:
return foo == bar
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I expect there to be performance penalties at least for:
* lowercasing words that are already lower case (by precondition)
* compiling and evaluating regular expressions for simple substring matches
* branching to an empty else statement
In summary, this solution does not look particularly speed-o
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Second-Tatapouftatapute
The text is lowercased for each word in the words set.
It should be faster to do this only once.
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This is a bit scary in that the loop might never terminate or the fibo list might have a typo. Mission accomplished.
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You might want to remove the debug output and the boilerplate comment (replace this...).
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A set of words is iterable. There is no need to transform it to a list first.
Neither do we need to use an index i since we don't care about it in the loop.
Your solution can thus be simplified like this:
def count_words(text, words):
N = 0
text = text.lower()
for
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