Heaps for Top-K Retrieval

import heapq

# Python's heapq is a MIN-HEAP by default
# The smallest element is always at index 0

# Push and pop — O(log n)
h = []
heapq.heappush(h, 5)
heapq.heappush(h, 1)
heapq.heappush(h, 3)
print(h[0])  # 1 — minimum is always at root

smallest = heapq.heappop(h)  # removes and returns 1
print(smallest, h[0])  # 1, 3

# Build heap from list — O(n) using heapify
data = [5, 1, 3, 2, 4]
heapq.heapify(data)  # in-place, O(n)
print(data[0])  # 1

# Push and pop atomically — more efficient than separate push + pop
result = heapq.heappushpop(data, 0)  # pushes 0, pops minimum
print(result)  # 0 (smaller than current min)

result = heapq.heapreplace(data, 100)  # pops minimum, then pushes 100
print(result)  # the value that was popped

# MAX-HEAP trick: store negated values
max_heap = []
scores = [0.82, 0.91, 0.74, 0.88, 0.65]

for score in scores:
    heapq.heappush(max_heap, -score)  # negate!

# Largest score is at the root (most negative = most positive original)
largest = -heapq.heappop(max_heap)
print(f"Largest score: {largest}")  # 0.91

# For (score, value) pairs — negate only the score
candidates = [
    (0.82, "doc_a"),
    (0.91, "doc_b"),
    (0.74, "doc_c"),
]

max_heap_pairs = []
for score, doc_id in candidates:
    heapq.heappush(max_heap_pairs, (-score, doc_id))

score, doc_id = heapq.heappop(max_heap_pairs)
print(f"Most similar: {doc_id} with score {-score:.2f}")  # doc_b with 0.91

import heapq

scores = {
    "doc_a": 0.82,
    "doc_b": 0.91,
    "doc_c": 0.74,
    "doc_d": 0.88,
    "doc_e": 0.65,
    "doc_f": 0.95,
    "doc_g": 0.71,
}

# Top 3 by score value
top3 = heapq.nlargest(3, scores.items(), key=lambda x: x[1])
print("Top 3:", top3)
# [('doc_f', 0.95), ('doc_b', 0.91), ('doc_d', 0.88)]

# Bottom 2 by score (least similar)
bottom2 = heapq.nsmallest(2, scores.items(), key=lambda x: x[1])
print("Bottom 2:", bottom2)

# When to use nlargest vs sorted:
# - k << n: nlargest is O(n log k), sorted is O(n log n) — use nlargest
# - k is close to n: just sort the full list — O(n log n) with smaller constant
# Rule of thumb: if k < n // 10, use nlargest/nsmallest

import heapq
import math
from dataclasses import dataclass

@dataclass
class Document:
    doc_id: str
    embedding: list[float]
    text: str


def dot_product(a: list[float], b: list[float]) -> float:
    return sum(x * y for x, y in zip(a, b))


def magnitude(v: list[float]) -> float:
    return math.sqrt(sum(x * x for x in v))


def cosine_similarity(a: list[float], b: list[float]) -> float:
    """
    Cosine similarity between two vectors.
    Returns value in [-1, 1]. Higher = more similar.
    Returns 0.0 for zero vectors to avoid division by zero.
    """
    mag_a = magnitude(a)
    mag_b = magnitude(b)
    if mag_a == 0.0 or mag_b == 0.0:
        return 0.0
    return dot_product(a, b) / (mag_a * mag_b)


def top_k_similar_documents(
    query_embedding: list[float],
    documents: list[Document],
    k: int,
) -> list[tuple[float, Document]]:
    """
    Find the k documents most similar to the query embedding.
    
    Algorithm: min-heap of size k
    - We maintain a min-heap of the k best (score, doc) pairs seen so far
    - For each document, compute similarity
    - If heap has fewer than k elements, push
    - If similarity > heap minimum, pop minimum and push new doc
    - At the end, pop all k elements (they come out in ascending order, so reverse)
    
    Time: O(n log k)  — n docs, each heap op is O(log k)
    Space: O(k)       — heap only ever holds k elements
    
    vs. sort-all approach:
    Time: O(n + n log n) = O(n log n)
    Space: O(n)
    
    For k=10, n=100,000: heap is ~5x faster
    """
    if k <= 0:
        return []
    
    # Min-heap: (score, tie_breaker, document)
    # We use a tie_breaker integer to avoid comparing Document objects
    heap: list[tuple[float, int, Document]] = []
    
    for idx, doc in enumerate(documents):
        score = cosine_similarity(query_embedding, doc.embedding)
        
        if len(heap) < k:
            # Heap not full yet — always push
            heapq.heappush(heap, (score, idx, doc))
        elif score > heap[0][0]:
            # New score beats the current minimum — replace it
            heapq.heapreplace(heap, (score, idx, doc))
        # Otherwise: score is not in top k, discard
    
    # Extract results in descending order (most similar first)
    results = []
    while heap:
        score, _, doc = heapq.heappop(heap)
        results.append((score, doc))
    
    results.reverse()  # heappop gives ascending order, we want descending
    return results


# Demonstration
import random
random.seed(42)

def random_embedding(dim: int = 8) -> list[float]:
    v = [random.gauss(0, 1) for _ in range(dim)]
    mag = magnitude(v)
    return [x / mag for x in v]  # normalize to unit sphere

# Create 20 test documents with 8-dim embeddings
docs = [
    Document(
        doc_id=f"doc_{i:02d}",
        embedding=random_embedding(8),
        text=f"This is document number {i}.",
    )
    for i in range(20)
]

# Query: a random vector
query = random_embedding(8)

# Get top 5
top5 = top_k_similar_documents(query, docs, k=5)
print("Top 5 similar documents:")
for rank, (score, doc) in enumerate(top5, 1):
    print(f"  #{rank}: {doc.doc_id} — similarity = {score:.4f}")

# Verify: compare against brute-force sort
all_scored = sorted(
    [(cosine_similarity(query, d.embedding), d) for d in docs],
    key=lambda x: -x[0],
)
print("\nGround truth (brute-force sort):")
for rank, (score, doc) in enumerate(all_scored[:5], 1):
    print(f"  #{rank}: {doc.doc_id} — similarity = {score:.4f}")

heap_ids = [doc.doc_id for _, doc in top5]
sort_ids = [doc.doc_id for _, doc in all_scored[:5]]
assert heap_ids == sort_ids, f"Mismatch: {heap_ids} vs {sort_ids}"
print("\nResults match! Heap and brute-force agree on top 5.")

from typing import Iterator, Generator

def stream_documents(n: int, dim: int = 8) -> Iterator[Document]:
    """Simulate a streaming document iterator (e.g., from a database cursor)."""
    for i in range(n):
        yield Document(
            doc_id=f"stream_doc_{i:04d}",
            embedding=random_embedding(dim),
            text=f"Streamed document {i}.",
        )


def top_k_streaming(
    query_embedding: list[float],
    document_stream: Iterator[Document],
    k: int,
) -> list[tuple[float, Document]]:
    """
    Top-k from a streaming source — never loads all documents into memory.
    
    This works because heaps only need to hold k elements at a time,
    regardless of how many documents flow through the stream.
    
    Time: O(n log k)
    Space: O(k) — only k documents in memory at any point
    """
    heap: list[tuple[float, int, Document]] = []
    
    for idx, doc in enumerate(document_stream):
        score = cosine_similarity(query_embedding, doc.embedding)
        
        if len(heap) < k:
            heapq.heappush(heap, (score, idx, doc))
        elif score > heap[0][0]:
            heapq.heapreplace(heap, (score, idx, doc))
    
    results = sorted(heap, key=lambda x: -x[0])
    return [(score, doc) for score, _, doc in results]


# Test streaming
query = random_embedding(8)
stream = stream_documents(n=10_000, dim=8)
top10 = top_k_streaming(query, stream, k=10)
print(f"\nTop 10 from stream of 10,000 documents:")
for rank, (score, doc) in enumerate(top10, 1):
    print(f"  #{rank}: {doc.doc_id} — score {score:.4f}")

def merge_k_sorted_result_lists(
    result_lists: list[list[tuple[float, str]]],
) -> list[tuple[float, str]]:
    """
    Merge k sorted lists of (score, doc_id) into one sorted list (descending).
    
    Application: you query k different vector indices and want to merge results.
    
    Algorithm: use a max-heap initialized with the first element from each list.
    Each pop gives the global maximum; then push the next element from that list.
    
    Time: O(n log k) where n = total elements across all lists
    Space: O(k) for the heap
    """
    # Min-heap with negated scores for max-heap behavior
    # Elements: (-score, list_index, element_index, doc_id)
    heap = []
    
    # Initialize with first element from each non-empty list
    for list_idx, result_list in enumerate(result_lists):
        if result_list:
            score, doc_id = result_list[0]
            heapq.heappush(heap, (-score, list_idx, 0, doc_id))
    
    merged = []
    
    while heap:
        neg_score, list_idx, elem_idx, doc_id = heapq.heappop(heap)
        merged.append((-neg_score, doc_id))
        
        # Push the next element from the same list
        next_elem_idx = elem_idx + 1
        if next_elem_idx < len(result_lists[list_idx]):
            next_score, next_doc_id = result_lists[list_idx][next_elem_idx]
            heapq.heappush(heap, (-next_score, list_idx, next_elem_idx, next_doc_id))
    
    return merged


# Test: merge results from 3 vector indices
list_a = [(0.95, "doc_a1"), (0.80, "doc_a2"), (0.70, "doc_a3")]
list_b = [(0.92, "doc_b1"), (0.85, "doc_b2"), (0.60, "doc_b3")]
list_c = [(0.88, "doc_c1"), (0.75, "doc_c2"), (0.55, "doc_c3")]

merged = merge_k_sorted_result_lists([list_a, list_b, list_c])
print("\nMerged results from 3 indices:")
for rank, (score, doc_id) in enumerate(merged, 1):
    print(f"  #{rank}: {doc_id} — {score:.2f}")