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Use relative list order ranking in openlibrary search
Set OpenLibrary search condifidence based on the provided result order, just using 1/(list index), so the first has rank 1, the second 0.5, the third 0.33, et cetera.
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1 changed files with 7 additions and 1 deletions
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@ -153,12 +153,17 @@ class Connector(AbstractConnector):
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return f"{self.covers_url}/b/id/{image_name}"
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def parse_search_data(self, data, min_confidence):
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for search_result in data.get("docs"):
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for idx, search_result in enumerate(data.get("docs")):
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# build the remote id from the openlibrary key
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key = self.books_url + search_result["key"]
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author = search_result.get("author_name") or ["Unknown"]
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cover_blob = search_result.get("cover_i")
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cover = self.get_cover_url([cover_blob], size="M") if cover_blob else None
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# OL doesn't provide confidence, but it does sort by an internal ranking, so
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# this confidence value is relative to the list position
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confidence = 1 / (idx + 1)
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yield SearchResult(
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title=search_result.get("title"),
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key=key,
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@ -166,6 +171,7 @@ class Connector(AbstractConnector):
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connector=self,
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year=search_result.get("first_publish_year"),
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cover=cover,
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confidence=confidence,
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)
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def parse_isbn_search_data(self, data):
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