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hybrid search example code (for docs) #30

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hybrid search example code (for docs) #30

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hybrid search example code (for docs)
morgangallant Jun 18, 2024
f01f1b7
bit cleaner
morgangallant Jun 19, 2024
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122 changes: 122 additions & 0 deletions examples/hybrid_search.py
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import turbopuffer as tpuf
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Does this run as part of the test suite?



# Helper to convert the results to a dictionary with ranks
def results_to_ranks(results, reverse=False):
if reverse:
results = sorted(results, key=lambda item: item.dist, reverse=True)
return {item.id: rank for rank, item in enumerate(results, start=1)}


# Fuses two search result sets together into one
# Uses reciprocal rank fusion
def rank_fusion(bm25_results, vector_results, k=60):
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@sirupsen sirupsen Jun 19, 2024 •
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I think we may want to inline/copy-paste this in a few places, so I think we need 1 top-level comments and then the function to be as small as possible, e.g. something alone the lines of this (not tested):

def results_to_ranks(results, reverse=False):
    return {item.id: rank for rank, item in enumerate(sorted(results, key=lambda item: item.dist, reverse=reverse), start=1)}

def rank_fusion(bm25_results, vector_results, k=60):
    bm25_ranks, vector_ranks = results_to_ranks(bm25_results), results_to_ranks(vector_results, reverse=True)
    scores = {doc_id: (1 / (k + bm25_ranks[doc_id]) if doc_id in bm25_ranks else 0) +
                      (1 / (k + vector_ranks[doc_id]) if doc_id in vector_ranks else 0)
              for doc_id in set(bm25_ranks) | set(vector_ranks)}
    return [{"id": doc_id, "score": score} for doc_id, score in sorted(scores.items(), key=lambda item: item[1], reverse=True)]

The more code you have to paste, the more you'll question why we don't build this into this library (good reasons). The shorter it is, the more you'll want to tweak it (good).

# Compute the ranks of all docs
# Note: a higher score in bm25 denotes a better result, whereas a higher score
# with vector euclidean distance indicates a worse result. When computing RRF,
# we reverse the order of vector results so that higher = better.
bm25_ranks = results_to_ranks(bm25_results)
vector_ranks = results_to_ranks(vector_results, reverse=True)

print("bm25_ranks:", bm25_ranks)
print("vector ranks", vector_ranks)

# Get all the unique document IDs
doc_ids = set(bm25_ranks.keys()).union(set(vector_ranks.keys()))

# Calculate the RRF score for each document
scores = {}
for doc_id in doc_ids:
score = 0.0
if doc_id in bm25_ranks:
score += 1.0 / (k + bm25_ranks[doc_id])
if doc_id in vector_ranks:
score += 1.0 / (k + vector_ranks[doc_id])
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@sirupsen sirupsen Jun 19, 2024 •
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Hm, is classic RRF only rank-based? Nothing about scores? Probably fine to begin with

scores[doc_id] = score

# Fuse the results together and return
fused_results = sorted(scores.items(), key=lambda item: item[1], reverse=True)
return [{"id": doc_id, "score": score} for doc_id, score in fused_results]


def main():
ns = tpuf.Namespace("tpuf_python_hybrid_search")
try:
ns.delete_all() # For cleaning up from previous runs
except tpuf.NotFoundError:
pass

# Upsert 10 documents, containing both vectors and text
# Whale facts from: https://www.natgeokids.com/uk/discover/animals/sea-life/10-blue-whale-facts/
ns.upsert(
{
"ids": [1, 2, 3, 4, 5, 6, 7, 8, 9, 10],
"vectors": [
[0.1, 0.1],
[0.2, 0.2],
[0.3, 0.3],
[0.4, 0.4],
[0.5, 0.5],
[0.6, 0.6],
[0.7, 0.7],
[0.8, 0.8],
[0.9, 0.9],
[1.0, 1.0],
],
"attributes": {
"text": [
"blue whales can grow to over 30m long",
"pretty much everything about the blue whale is massive",
"blue whales can be found in all oceans, except for the arctic",
"blue whales eat tiny shrimp-like crustaceans called krill",
"to communicate with each other, blue whales make a series of super loud vocal sounds",
"blah",
"blah",
"blah",
"blah",
"blah",
]
},
"schema": {
"text": {
"type": "?string",
"bm25": {
"language": "english",
"stemming": True,
"case_sensitive": False,
"remove_stopwords": True,
},
},
},
"distance_metric": "euclidean_squared",
}
)
print("upsert documents successfully")
print("")

# First, do query w/ BM25 full-text search
bm25_results = ns.query({"top_k": 10, "rank_by": ("text", "BM25", "blue whale")})
print(
"search results for 'blue whale':",
[(item.id, item.dist) for item in bm25_results],
)
print("")

# Now, we also do a query with a vector
vector_results = ns.query(
top_k=10, vector=[1.0, 1.0], distance_metric="euclidean_squared"
)
print(
"search results for [1.0, 1.0]:",
[(item.id, item.dist) for item in vector_results],
)
print("")

# Fuse the results with Reciprocal rank fusion (RRF)
fused_results = rank_fusion(bm25_results, vector_results)
print("")
print("fused results:", fused_results)


if __name__ == "__main__":
main()
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