261 lines
11 KiB
Python
261 lines
11 KiB
Python
#!/usr/bin/env python3
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"""
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对于最近14天无完课行为的付费用户，统计他们「最后一次完课」所在的 Unit 分布（按 L1/L2 拆分）。
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时间范围：2026-05-09 ~ 2026-05-22
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"""
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import psycopg2
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import psycopg2.extras
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from collections import defaultdict
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conn = psycopg2.connect(
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host="bj-postgres-16pob4sg.sql.tencentcdb.com",
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port=28591,
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user="ai_member",
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password="LdfjdjL83h3h3^$&**YGG*",
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dbname="vala_bi"
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)
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conn.set_session(autocommit=True)
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cur = conn.cursor(cursor_factory=psycopg2.extras.DictCursor)
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START_DATE = "2026-05-09"
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END_DATE = "2026-05-22"
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print("=" * 70)
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print("📊 无完课用户最后一次完课 Unit 分布分析")
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print("=" * 70)
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# ═══════════════════════════════════════════════════════════
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# Step 1: 获取所有付费用户
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# ═══════════════════════════════════════════════════════════
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print("\n[1/6] 获取付费用户...")
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cur.execute("""
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SELECT DISTINCT o.account_id
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FROM bi_vala_order o
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INNER JOIN bi_vala_app_account a ON o.account_id = a.id AND a.status = 1
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WHERE o.order_status IN (3, 4)
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AND o.pay_success_date IS NOT NULL
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AND o.deleted_at IS NULL
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""")
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paid_accounts = {row[0] for row in cur.fetchall()}
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print(f" 付费用户总数: {len(paid_accounts)}")
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# ═══════════════════════════════════════════════════════════
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# Step 2: L1/L2 分类
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# ═══════════════════════════════════════════════════════════
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print("\n[2/6] 分类用户等级...")
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L1_GOODS = {57, 60, 63}
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L2_GOODS = {31, 32, 33, 54}
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L1L2_GOODS = {61}
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cur.execute("""
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SELECT o.account_id, o.goods_id
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FROM bi_vala_order o
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WHERE o.account_id = ANY(%s)
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AND o.order_status IN (3, 4)
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AND o.deleted_at IS NULL
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""", (list(paid_accounts),))
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user_goods = defaultdict(set)
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for row in cur.fetchall():
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user_goods[row[0]].add(row[1])
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only_l1, only_l2, both_l1l2, other = set(), set(), set(), set()
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for aid in paid_accounts:
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g = user_goods.get(aid, set())
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h1, h2, h12 = bool(g & L1_GOODS), bool(g & L2_GOODS), bool(g & L1L2_GOODS)
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if h12 or (h1 and h2):
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both_l1l2.add(aid)
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elif h1:
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only_l1.add(aid)
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elif h2:
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only_l2.add(aid)
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else:
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other.add(aid)
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# ═══════════════════════════════════════════════════════════
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# Step 3: 获取最近14天有完课的用户
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# ═══════════════════════════════════════════════════════════
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print("\n[3/6] 查询最近14天完课行为...")
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cur.execute("SELECT id, account_id FROM bi_vala_app_character WHERE account_id = ANY(%s) AND deleted_at IS NULL",
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(list(paid_accounts),))
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char_to_account = {}
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account_chars = defaultdict(list)
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for row in cur.fetchall():
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char_to_account[row[0]] = row[1]
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account_chars[row[1]].append(row[0])
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all_char_ids = set(char_to_account.keys())
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active_char_ids = set()
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for shard in range(8):
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cur.execute(f"""
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SELECT DISTINCT user_id FROM bi_user_chapter_play_record_{shard}
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WHERE play_status = 1 AND updated_at >= %s AND updated_at < %s::date + interval '1 day'
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""", (START_DATE, END_DATE))
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for r in cur.fetchall():
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if r[0] in all_char_ids:
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active_char_ids.add(r[0])
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active_accounts = {char_to_account[c] for c in active_char_ids}
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inactive_accounts = paid_accounts - active_accounts
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print(f" 无完课付费用户: {len(inactive_accounts)}")
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# ═══════════════════════════════════════════════════════════
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# Step 4: 获取无完课用户的所有角色
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# ═══════════════════════════════════════════════════════════
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print("\n[4/6] 获取无完课用户的角色...")
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inactive_chars = {}
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for aid in inactive_accounts:
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for cid in account_chars.get(aid, []):
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inactive_chars[cid] = aid
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print(f" 无完课用户角色数: {len(inactive_chars)}")
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# ═══════════════════════════════════════════════════════════
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# Step 5: 查询每个角色最后一次完课记录（遍历8张分表）
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# ═══════════════════════════════════════════════════════════
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print("\n[5/6] 查询各角色最后一次完课记录...")
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# char_id → (chapter_id, updated_at)
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latest_completion = {}
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for shard in range(8):
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table = f"bi_user_chapter_play_record_{shard}"
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cur.execute(f"""
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SELECT DISTINCT ON (user_id) user_id, chapter_id, updated_at
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FROM {table}
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WHERE play_status = 1
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ORDER BY user_id, updated_at DESC
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""")
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rows = cur.fetchall()
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print(f" {table}: {len(rows)} 条记录")
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for row in rows:
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cid = row[0]
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if cid not in inactive_chars:
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continue
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ch_id = row[1]
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ts = row[2]
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if cid not in latest_completion or ts > latest_completion[cid][1]:
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latest_completion[cid] = (ch_id, ts)
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print(f" 有历史完课记录的角色数: {len(latest_completion)}")
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no_history = len(inactive_chars) - len(latest_completion)
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print(f" 无任何完课记录的角色数: {no_history}")
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# ═══════════════════════════════════════════════════════════
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# Step 6: 映射 chapter_id → Unit，按用户等级分类聚合
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# ═══════════════════════════════════════════════════════════
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print("\n[6/6] 映射 chapter_id → Unit 并聚合...")
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# 获取课程结构映射
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cur.execute("SELECT id, course_level, course_unit FROM bi_level_unit_lesson")
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chapter_map = {} # chapter_id → (course_level, course_unit)
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for row in cur.fetchall():
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chapter_map[row[0]] = (row[1], row[2])
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# 聚合：user_level_type → { unit → count }
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# user_level_type: "仅L1", "仅L2", "L1+L2"
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unit_counts = defaultdict(lambda: defaultdict(int))
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no_chapter = defaultdict(int) # 有完课但 chapter_id 映射不上的
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for cid, (ch_id, ts) in latest_completion.items():
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aid = inactive_chars[cid]
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if aid in only_l1:
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user_type = "仅L1"
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elif aid in only_l2:
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user_type = "仅L2"
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elif aid in both_l1l2:
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user_type = "L1+L2"
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else:
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user_type = "其他"
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if ch_id in chapter_map:
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level, unit = chapter_map[ch_id]
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unit_counts[user_type][unit] += 1
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else:
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no_chapter[user_type] += 1
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# ═══════════════════════════════════════════════════════════
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# 输出结果
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# ═══════════════════════════════════════════════════════════
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print("\n" + "=" * 70)
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print("📈 统计结果")
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print("=" * 70)
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# 定义 Unit 排序
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def unit_sort_key(u):
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if u is None:
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return (99, 99)
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# e.g. "U00", "U01", "S1U00"
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import re
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m = re.match(r'(?:S(\d+))?U(\d+)', u)
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if m:
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s = int(m.group(1)) if m.group(1) else 0
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uu = int(m.group(2))
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return (s, uu)
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return (99, 99)
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for user_type in ["仅L1", "仅L2", "L1+L2"]:
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data = unit_counts[user_type]
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total_with_history = sum(data.values())
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print(f"\n{'─' * 50}")
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print(f" {user_type} 用户")
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print(f" 最后一次完课 Unit 分布（共 {total_with_history} 人有完课记录）:")
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print(f"{'─' * 50}")
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# 先按 Unit 排序
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sorted_units = sorted(data.items(), key=lambda x: unit_sort_key(x[0]))
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total_all = total_with_history + no_history
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for unit, cnt in sorted_units:
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bar = "█" * max(1, int(cnt / max(1, max(data.values())) * 30))
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print(f" {unit:>10s} {cnt:>5d} {bar}")
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if no_chapter.get(user_type, 0) > 0:
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print(f" {'(未知)':>10s} {no_chapter[user_type]:>5d} (chapter_id 映射失败)")
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# ─── 总体汇总 ───
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print(f"\n{'═' * 50}")
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print("📋 汇总")
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print(f"{'═' * 50}")
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total_inactive = len(inactive_accounts)
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for user_type in ["仅L1", "仅L2", "L1+L2"]:
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type_users = len([a for a in inactive_accounts if
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(user_type == "仅L1" and a in only_l1) or
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(user_type == "仅L2" and a in only_l2) or
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(user_type == "L1+L2" and a in both_l1l2)])
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with_history = sum(unit_counts[user_type].values())
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no_hist = 0
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# Count users in this type without any history
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type_chars = {c: a for c, a in inactive_chars.items() if
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(user_type == "仅L1" and a in only_l1) or
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(user_type == "仅L2" and a in only_l2) or
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(user_type == "L1+L2" and a in both_l1l2)}
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# For users with no history: count distinct account_ids among chars with no record
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chars_with_history = set(latest_completion.keys())
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accounts_with_history = set()
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accounts_no_history = set()
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for c, a in type_chars.items():
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if c in chars_with_history:
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accounts_with_history.add(a)
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else:
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accounts_no_history.add(a)
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# An account may have some chars with history and some without.
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# Count accounts that have ANY char with history vs those with NO chars having history
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no_history_accounts = accounts_no_history - accounts_with_history
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print(f" {user_type}: 共 {type_users} 人")
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print(f" 有完课记录: {len(accounts_with_history)} 人")
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print(f" 完全无完课: {len(no_history_accounts)} 人")
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# Top units
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data = unit_counts[user_type]
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if data:
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top = sorted(data.items(), key=lambda x: -x[1])[:5]
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top_str = ", ".join(f"{u}({c})" for u, c in top)
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print(f" Top5 Unit: {top_str}")
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print(f"\n 合计无完课付费用户: {total_inactive}")
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print(f" 注：用户数 {total_inactive} 为上轮口径，本次统计以角色维度查最后完课记录")
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cur.close()
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conn.close()
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print("\n✅ 完成")
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