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Simplify identifier searches
This commit removes code that deduplicated search results for identifier searches. If it was the case that multiple books have the same identifier, in theory this would produce better search results, but in practice this doesn't happen very much, is probably worth seeing when it does, and worsens the performance of identifier search overall.
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commit
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1 changed files with 8 additions and 32 deletions
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@ -4,7 +4,7 @@ from functools import reduce
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import operator
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from django.contrib.postgres.search import SearchRank, SearchQuery
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from django.db.models import OuterRef, Subquery, F, Q
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from django.db.models import F, Q
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from bookwyrm import models
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from bookwyrm import connectors
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@ -17,8 +17,13 @@ def search(query, min_confidence=0, filters=None, return_first=False):
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filters = filters or []
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if not query:
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return []
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results = None
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# first, try searching unqiue identifiers
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# unique identifiers never have spaces, title/author usually do
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if not " " in query:
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results = search_identifiers(query, *filters, return_first=return_first)
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# if there were no identifier results...
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if not results:
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# then try searching title/author
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results = search_title_author(
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@ -35,24 +40,10 @@ def isbn_search(query):
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# If the ISBN has only 9 characters, prepend missing zero
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query = query.strip().upper().rjust(10, "0")
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filters = [{f: query} for f in ["isbn_10", "isbn_13"]]
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results = models.Edition.objects.filter(
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return models.Edition.objects.filter(
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reduce(operator.or_, (Q(**f) for f in filters))
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).distinct()
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# when there are multiple editions of the same work, pick the default.
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# it would be odd for this to happen.
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default_editions = models.Edition.objects.filter(
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parent_work=OuterRef("parent_work")
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).order_by("-edition_rank")
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results = (
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results.annotate(default_id=Subquery(default_editions.values("id")[:1])).filter(
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default_id=F("id")
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)
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or results
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)
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return results
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def format_search_result(search_result):
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"""convert a book object into a search result object"""
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@ -88,22 +79,7 @@ def search_identifiers(query, *filters, return_first=False):
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results = models.Edition.objects.filter(
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*filters, reduce(operator.or_, (Q(**f) for f in or_filters))
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).distinct()
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if results.count() <= 1:
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if return_first:
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return results.first()
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return results
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# when there are multiple editions of the same work, pick the default.
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# it would be odd for this to happen.
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default_editions = models.Edition.objects.filter(
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parent_work=OuterRef("parent_work")
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).order_by("-edition_rank")
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results = (
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results.annotate(default_id=Subquery(default_editions.values("id")[:1])).filter(
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default_id=F("id")
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)
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or results
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)
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if return_first:
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return results.first()
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return results
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