Lesson 23 · Ranked retrieval - Build a Search Engine

Ranked retrieval

Today you turn a map of document scores into an ordered list of the best few results - sorted by score, ties broken predictably, cut to the top k. This is the moment search returns a ranking, not just a set.

The goal

Sort scored documents by score descending, break ties by document id, and return the top k.

Start here - the target

TO DO

Scenario: Ordering results and taking the best

Giventhe document scores {"d1": 2.0, "d2": 5.0, "d3": 5.0}

Whenyou take the top 2 ranked results

Thenthey are [("d2", 5.0), ("d3", 5.0)]

Andranking all three gives [("d2", 5.0), ("d3", 5.0), ("d1", 2.0)]

Background

Users do not want a set of matching documents; they want the best ones, in order. Ranked retrieval sorts the scored candidates by score, highest first, and returns only the top k. Everything upstream - candidates, weighting, summing - existed to produce the numbers this step now orders.

Ties need a rule, or results wobble between runs. Break them by document id so the order is deterministic: d2 and d3 both score 5.0, but d2 comes first. Wire this onto your query scorer and you have end-to-end search - hand it query terms, get back a ranked list. Try it on a handful of documents; this is the first lesson the engine truly searches.

Make it work

def top_k(scores, k):
    # highest score first; ties -> smaller id first
    ordered = sorted(scores.items(), key=lambda kv: (-kv[1], kv[0]))
    return ordered[:k]