I expect that the issues and PRs I've posted here are going to my own fork and no further, but if anyone out there is reading, your feedback is welcome.
In Parser.getKeywords()
:
# ...
uniqueWords = list(set(words))
keywords = [{'word': word, 'count': words.count(word)} for word in uniqueWords]
keywords = sorted(keywords, key=lambda x: -x['count'])
# ... returns: ...
keywords[0] = {'word': 'foo', 'count': 100} # rank: 1st (most common keyword)
keywords[1] = {'word': 'eggs', 'count': 25} # rank: 9th
keywords[2] = {'word': 'bacon', 'count': 32} # rank: 8th
# ...
keywords[10] = {'word': 'spam', 'count': 25} # rank: 9th
keywords[11] = {'word': 'bar', 'count': 19} # rank: 13th
keywords[12] = {'word': 'ham', 'count': 25} # rank: 9th
# ...
set()
does not return an ordered enumerable, so uniqueWords
is an unordered list. Iterating it to build keywords
means this list is also unordered, so the return order of sorted()
on the count key alone is inconsistent.
Down in Summarizer.summarize()
it means sentence scores are also inconsistent:
# ...
(keywords, wordCount) = self.parser.getKeywords(text)
topKeywords = self.getTopKeywords(keywords[:10], wordCount, source, category)
# ... iterating sentences ...
sbsFeature = self.sbs(words, topKeywords, keywordList)
dbsFeature = self.dbs(words, topKeywords, keywordList)
# ... calculate sentence score based on these features ...
# ...
For consistency, when the summarize
method calls getTopKeywords()
, should it still pass a fixed list of ten keywords that's been sorted with more tiebreakers…
# replace: keywords = sorted(keywords, key=lambda x: -x['count'])
keywords = sorted(keywords, key=lambda x: (-x['count'], -len(x['word'], x['word']))
…every keyword that ranks 10th or better, i.e. this comprehension…
topKeywordSlice = [kw for kw in keywords if kw['count'] >= keywords[9]['count']]
… or maybe both?
Computationally, the advanced sort seems unnecessarily expensive, but I don't know if there's a rationale for exactly ten top keywords. What's the best way to make this work?