refactor: streamline candidate generation and pruning in Apriori algo…

machine_learning/apriori_algorithm.py

@@ -11,8 +11,8 @@
 Examples: https://www.kaggle.com/code/earthian/apriori-association-rules-mining
 """
 
-from collections import Counter
 from itertools import combinations
+from collections import defaultdict
 
 
 def load_data() -> list[list[str]]:
@@ -25,78 +25,91 @@
     return [["milk"], ["milk", "butter"], ["milk", "bread"], ["milk", "bread", "chips"]]
 
 
-def prune(itemset: list, candidates: list, length: int) -> list:
-    """
-    Prune candidate itemsets that are not frequent.
-    The goal of pruning is to filter out candidate itemsets that are not frequent.  This
-    is done by checking if all the (k-1) subsets of a candidate itemset are present in
-    the frequent itemsets of the previous iteration (valid subsequences of the frequent
-    itemsets from the previous iteration).
-
-    Prunes candidate itemsets that are not frequent.
-
-    >>> itemset = ['X', 'Y', 'Z']
-    >>> candidates = [['X', 'Y'], ['X', 'Z'], ['Y', 'Z']]
-    >>> prune(itemset, candidates, 2)
-    [['X', 'Y'], ['X', 'Z'], ['Y', 'Z']]
-
-    >>> itemset = ['1', '2', '3', '4']
-    >>> candidates = ['1', '2', '4']
-    >>> prune(itemset, candidates, 3)
-    []
+# ---------- Helpers ----------
+
+def get_support(itemset, transactions):
+    """Compute support count of an itemset efficiently."""
+    return sum(1 for t in transactions if itemset.issubset(t))
+
+
+def generate_candidates(prev_frequent, k):
     """
-    itemset_counter = Counter(tuple(item) for item in itemset)
-    pruned = []
-    for candidate in candidates:
-        is_subsequence = True
-        for item in candidate:
-            item_tuple = tuple(item)
-            if (
-                item_tuple not in itemset_counter
-                or itemset_counter[item_tuple] < length - 1
-            ):
-                is_subsequence = False
-                break
-        if is_subsequence:
-            pruned.append(candidate)
-    return pruned
-
-
-def apriori(data: list[list[str]], min_support: int) -> list[tuple[list[str], int]]:
+    Generate candidate itemsets of size k from frequent itemsets of size k-1.
     """
-    Returns a list of frequent itemsets and their support counts.
+    prev_list = list(prev_frequent)
+    candidates = set()
 
-    >>> data = [['A', 'B', 'C'], ['A', 'B'], ['A', 'C'], ['A', 'D'], ['B', 'C']]
-    >>> apriori(data, 2)
-    [(['A', 'B'], 1), (['A', 'C'], 2), (['B', 'C'], 2)]
+    for i in range(len(prev_list)):
+        for j in range(i + 1, len(prev_list)):
+            union = prev_list[i] | prev_list[j]
+            if len(union) == k:
+                candidates.add(union)
 
-    >>> data = [['1', '2', '3'], ['1', '2'], ['1', '3'], ['1', '4'], ['2', '3']]
-    >>> apriori(data, 3)
-    []
+    return candidates
+
+
+def has_infrequent_subset(candidate, prev_frequent):
     """
-    itemset = [list(transaction) for transaction in data]
-    frequent_itemsets = []
-    length = 1
+    Apriori pruning: all (k-1)-subsets must be frequent.
+    """
+    for subset in combinations(candidate, len(candidate) - 1):
+        if frozenset(subset) not in prev_frequent:
+            return True
+    return False
+
+
+# ---------- Main Apriori ----------
+
+def apriori(data: list[list[str]], min_support: int):
+    transactions = [set(t) for t in data]
+
+    # 1. initial 1-itemsets
+    item_counts = defaultdict(int)
+    for t in transactions:
+        for item in t:
+            item_counts[frozenset([item])] += 1
+
+    frequent = {
+        itemset for itemset, count in item_counts.items()
+        if count >= min_support
+    }
+
+    all_frequents = [(list(i)[0], c) for i, c in item_counts.items() if c >= min_support]
+
+    k = 2
+
+    while frequent:
+        # 2. generate candidates
+        candidates = generate_candidates(frequent, k)
+
+        # 3. prune
+        candidates = {
+            c for c in candidates
+            if not has_infrequent_subset(c, frequent)
+        }
 
-    while itemset:
-        # Count itemset support
-        counts = [0] * len(itemset)
-        for transaction in data:
-            for j, candidate in enumerate(itemset):
-                if all(item in transaction for item in candidate):
-                    counts[j] += 1
+        # 4. count support
+        candidate_counts = defaultdict(int)
+        for t in transactions:
+            for c in candidates:
+                if c.issubset(t):
+                    candidate_counts[c] += 1
 
-        # Prune infrequent itemsets
-        itemset = [item for i, item in enumerate(itemset) if counts[i] >= min_support]
+        # 5. filter frequent
+        frequent = {
+            c for c, count in candidate_counts.items()
+            if count >= min_support
+        }
 
-        # Append frequent itemsets (as a list to maintain order)
-        for i, item in enumerate(itemset):
-            frequent_itemsets.append((sorted(item), counts[i]))
+        all_frequents.extend(
+            (sorted(list(c)), count)
+            for c, count in candidate_counts.items()
+            if count >= min_support
+        )
 
-        length += 1
-        itemset = prune(itemset, list(combinations(itemset, length)), length)
+        k += 1
 
-    return frequent_itemsets
+    return all_frequents
 
 
 if __name__ == "__main__":