333 lines
13 KiB
Python
333 lines
13 KiB
Python
#!/usr/bin/env python3
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"""
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行课进度与退款率关系分析
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分析维度：
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1. 按用户完成课时数分桶，计算每桶的退款率
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2. 退款用户的退款前已完成课时数分布
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3. 不同课程等级(L1/L2)的行课进度与退款率
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"""
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import os
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import sys
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import psycopg2
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import pandas as pd
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from collections import defaultdict
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# ── 数据库 ──
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DB_HOST = "bj-postgres-16pob4sg.sql.tencentcdb.com"
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DB_PORT = 28591
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DB_USER = "ai_member"
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DB_NAME = "vala_bi"
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def get_password():
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pw = os.environ.get("PG_ONLINE_PASSWORD", "")
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if pw:
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return pw
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secrets_path = os.path.join(os.path.dirname(os.path.abspath(__file__)), "..", "secrets.env")
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if os.path.exists(secrets_path):
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with open(secrets_path) as f:
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for line in f:
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if line.startswith("PG_ONLINE_PASSWORD="):
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return line.strip().split("=", 1)[1].strip("'\"")
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raise RuntimeError("PG_ONLINE_PASSWORD not found")
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def get_conn():
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return psycopg2.connect(host=DB_HOST, port=DB_PORT, user=DB_USER, password=get_password(), dbname=DB_NAME, connect_timeout=60)
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def main():
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conn = get_conn()
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cur = conn.cursor()
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print("=" * 60)
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print("行课进度与退款率关系分析")
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print("=" * 60)
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# ── 1. 获取所有付费用户（排除测试账号）──
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print("\n[1/5] 获取付费用户...")
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cur.execute("""
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SELECT o.account_id, o.trade_no, o.pay_success_date, o.pay_amount_int, o.order_status,
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o.goods_name, o.key_from
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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.deleted_at IS NULL
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AND o.pay_success_date IS NOT NULL
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AND o.order_status IN (3, 4)
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ORDER BY o.account_id, o.pay_success_date
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""")
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orders = cur.fetchall()
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print(f" 共 {len(orders)} 条订单")
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# ── 2. 获取退款信息 ──
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print("\n[2/5] 获取退款信息...")
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trade_nos = list(set(o[1] for o in orders if o[1]))
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refund_set = set() # trade_nos that were refunded
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for i in range(0, len(trade_nos), 500):
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batch = trade_nos[i:i+500]
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ph = ",".join(["%s"] * len(batch))
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cur.execute(f"""
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SELECT DISTINCT trade_no
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FROM bi_refund_order
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WHERE trade_no IN ({ph})
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AND status = 3
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AND deleted_at IS NULL
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""", batch)
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for (tn,) in cur.fetchall():
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refund_set.add(tn)
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print(f" 退款订单 trade_no 数: {len(refund_set)}")
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# ── 3. 获取所有角色 ──
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print("\n[3/5] 获取用户角色...")
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account_ids = list(set(o[0] for o in orders))
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# 分批
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all_chars = []
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for i in range(0, len(account_ids), 1000):
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batch = account_ids[i:i+1000]
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ph = ",".join(["%s"] * len(batch))
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cur.execute(f"""
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SELECT id AS character_id, account_id, nickname
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FROM bi_vala_app_character
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WHERE account_id IN ({ph})
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AND nickname IS NOT NULL AND nickname != ''
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AND deleted_at IS NULL
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""", batch)
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all_chars.extend(cur.fetchall())
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print(f" 共 {len(all_chars)} 个角色")
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char_account_map = {} # character_id -> account_id
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account_chars = defaultdict(list) # account_id -> [character_ids]
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for cid, aid, nick in all_chars:
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char_account_map[cid] = aid
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account_chars[aid].append(cid)
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# ── 4. 获取课时完成记录（所有8个分表）──
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print("\n[4/5] 获取课时完成记录...")
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char_ids = list(set(c[0] for c in all_chars))
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char_lesson_count = defaultdict(int) # character_id -> 完成课时数
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char_first_done = {} # character_id -> 首次完成时间
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char_last_done = {} # character_id -> 最近完成时间
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for tbl_idx in range(8):
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table = f"bi_user_chapter_play_record_{tbl_idx}"
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for i in range(0, len(char_ids), 2000):
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batch = char_ids[i:i+2000]
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ph = ",".join(["%s"] * len(batch))
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try:
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cur.execute(f"""
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SELECT user_id, chapter_id, MIN(created_at), MAX(created_at), COUNT(*)
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FROM {table}
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WHERE user_id IN ({ph})
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AND play_status = 1
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AND deleted_at IS NULL
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GROUP BY user_id, chapter_id
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""", batch)
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for user_id, ch_id, first_at, last_at, cnt in cur.fetchall():
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char_lesson_count[user_id] += 1
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if user_id not in char_first_done or first_at < char_first_done[user_id]:
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char_first_done[user_id] = first_at
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if user_id not in char_last_done or last_at > char_last_done[user_id]:
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char_last_done[user_id] = last_at
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except Exception as e:
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print(f" 警告: {table} 查询失败: {e}")
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print(f" 有行课记录的角色: {len(char_lesson_count)}")
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# ── 5. 构建分析数据 ──
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print("\n[5/5] 构建分析数据...")
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# 按 account_id 聚合
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# 每个用户的：订单列表、退款订单列表、所有角色的总完成课时数
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user_orders = defaultdict(list) # account_id -> [(trade_no, pay_date, amount, order_status, goods_name, key_from)]
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for o in orders:
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aid, tn, pd_, amt, os_, gn, kf = o
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user_orders[aid].append({
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"trade_no": tn,
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"pay_date": pd_,
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"amount": amt / 100.0,
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"order_status": os_,
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"goods_name": gn,
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"key_from": kf,
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"is_refunded": tn in refund_set,
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})
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# 用户维度分析
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rows = []
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for aid in account_ids:
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my_orders = user_orders.get(aid, [])
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if not my_orders:
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continue
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# 总完成课时数（所有角色汇总）
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my_chars = account_chars.get(aid, [])
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total_lessons = sum(char_lesson_count.get(cid, 0) for cid in my_chars)
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# 首个角色首次行课时间
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first_lesson = None
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for cid in my_chars:
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if cid in char_first_done:
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if first_lesson is None or char_first_done[cid] < first_lesson:
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first_lesson = char_first_done[cid]
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# 最近行课时间
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last_lesson = None
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for cid in my_chars:
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if cid in char_last_done:
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if last_lesson is None or char_last_done[cid] > last_lesson:
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last_lesson = char_last_done[cid]
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# 订单分析
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total_orders = len(my_orders)
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refunded_orders = sum(1 for o in my_orders if o["is_refunded"])
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total_gmv = sum(o["amount"] for o in my_orders)
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total_refund = sum(o["amount"] for o in my_orders if o["is_refunded"])
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gsv = total_gmv - total_refund
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# 是否全部退款
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all_refunded = (refunded_orders == total_orders and total_orders > 0)
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# 首次购买时间
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first_pay = min(o["pay_date"] for o in my_orders if o["pay_date"])
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# 退款时间（取最早的退款订单）
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refund_orders_list = [o for o in my_orders if o["is_refunded"]]
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first_refund_date = None
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if refund_orders_list:
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first_refund_date = min(o["pay_date"] for o in refund_orders_list)
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# 退款前完成课时数（退款日期之前完成的课时）
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# 这里用简化方式：取退款日期
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lessons_before_refund = None
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if first_refund_date:
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# 计算退款前完成课时数
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cnt = 0
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for cid in my_chars:
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# 需要查分表，这里简化：如果首次行课时间在退款之前，则计入
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if cid in char_first_done and char_first_done[cid] < first_refund_date:
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cnt += char_lesson_count.get(cid, 0)
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lessons_before_refund = cnt
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rows.append({
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"用户ID": aid,
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"角色数": len(my_chars),
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"总完成课时数": total_lessons,
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"首次行课时间": first_lesson,
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"最近行课时间": last_lesson,
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"首次购买时间": first_pay,
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"订单数": total_orders,
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"退款订单数": refunded_orders,
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"GMV": round(total_gmv, 2),
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"GSV": round(gsv, 2),
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"退款金额": round(total_refund, 2),
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"是否全部退款": "是" if all_refunded else "否",
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"是否退过款": "是" if refunded_orders > 0 else "否",
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})
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df = pd.DataFrame(rows)
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conn.close()
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# ── 6. 分析输出 ──
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print("\n" + "=" * 60)
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print("分析结果")
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print("=" * 60)
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# 6.1 按完成课时数分桶
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print("\n【维度1】按完成课时数分桶的退款率")
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def lesson_bucket(n):
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if n == 0:
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return "0课时"
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elif n <= 3:
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return "1-3课时"
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elif n <= 7:
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return "4-7课时"
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elif n <= 15:
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return "8-15课时"
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elif n <= 30:
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return "16-30课时"
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elif n <= 60:
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return "31-60课时"
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else:
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return "60课时以上"
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df["课时桶"] = df["总完成课时数"].apply(lesson_bucket)
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bucket_order = ["0课时", "1-3课时", "4-7课时", "8-15课时", "16-30课时", "31-60课时", "60课时以上"]
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bucket_stats = df.groupby("课时桶").agg(
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用户数=("用户ID", "count"),
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退款用户数=("是否退过款", lambda x: (x == "是").sum()),
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全部退款用户数=("是否全部退款", lambda x: (x == "是").sum()),
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平均完成课时=("总完成课时数", "mean"),
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GMV总额=("GMV", "sum"),
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GSV总额=("GSV", "sum"),
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).reindex(bucket_order).fillna(0)
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bucket_stats["退款率(用户)"] = (bucket_stats["退款用户数"] / bucket_stats["用户数"] * 100).round(1)
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bucket_stats["全部退款率(用户)"] = (bucket_stats["全部退款用户数"] / bucket_stats["用户数"] * 100).round(1)
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bucket_stats["平均完成课时"] = bucket_stats["平均完成课时"].round(1)
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bucket_stats["GMV总额"] = bucket_stats["GMV总额"].round(2)
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bucket_stats["GSV总额"] = bucket_stats["GSV总额"].round(2)
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print(bucket_stats.to_string())
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# 6.2 退款用户的完成课时分布
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print("\n\n【维度2】退款用户的行课完成情况")
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refund_users = df[df["是否退过款"] == "是"]
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print(f"退款用户总数: {len(refund_users)}")
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print(f" 其中0课时的: {(refund_users['总完成课时数'] == 0).sum()} 人")
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print(f" 其中1-3课时的: {((refund_users['总完成课时数'] >= 1) & (refund_users['总完成课时数'] <= 3)).sum()} 人")
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print(f" 其中4-7课时的: {((refund_users['总完成课时数'] >= 4) & (refund_users['总完成课时数'] <= 7)).sum()} 人")
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print(f" 其中8-15课时的: {((refund_users['总完成课时数'] >= 8) & (refund_users['总完成课时数'] <= 15)).sum()} 人")
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print(f" 其中16课时以上的: {(refund_users['总完成课时数'] >= 16).sum()} 人")
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print(f" 退款用户平均完成课时: {refund_users['总完成课时数'].mean():.1f}")
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# 6.3 未退款用户 vs 退款用户对比
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print("\n\n【维度3】退款 vs 未退款用户行课对比")
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no_refund = df[df["是否退过款"] == "否"]
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print(f" 未退款用户数: {len(no_refund)}, 平均完成课时: {no_refund['总完成课时数'].mean():.1f}")
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print(f" 退款用户数: {len(refund_users)}, 平均完成课时: {refund_users['总完成课时数'].mean():.1f}")
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print(f" 全部退款用户数: {(df['是否全部退款'] == '是').sum()}, 平均完成课时: {df[df['是否全部退款'] == '是']['总完成课时数'].mean():.1f}")
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# 6.4 整体统计
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print("\n\n【维度4】整体统计")
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total_users = len(df)
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total_refund_users = len(refund_users)
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print(f" 总付费用户: {total_users}")
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print(f" 总退款用户: {total_refund_users} ({total_refund_users/total_users*100:.1f}%)")
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print(f" 全部退款用户: {(df['是否全部退款'] == '是').sum()}")
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print(f" 平均完成课时: {df['总完成课时数'].mean():.1f}")
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print(f" 总GMV: {df['GMV'].sum():.2f}")
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print(f" 总GSV: {df['GSV'].sum():.2f}")
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# ── 7. 输出 Excel ──
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output_dir = os.path.join(os.path.dirname(os.path.abspath(__file__)), "..", "output")
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os.makedirs(output_dir, exist_ok=True)
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output_path = os.path.join(output_dir, "行课进度与退款率分析.xlsx")
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with pd.ExcelWriter(output_path, engine="openpyxl") as writer:
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# Sheet1: 明细
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df_out = df.drop(columns=["课时桶"], errors="ignore")
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for col in ["首次行课时间", "最近行课时间", "首次购买时间"]:
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if col in df_out.columns:
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df_out[col] = pd.to_datetime(df_out[col]).dt.tz_localize(None)
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df_out.to_excel(writer, sheet_name="用户明细", index=False)
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# Sheet2: 分桶统计
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bucket_stats.to_excel(writer, sheet_name="分桶统计")
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# Sheet3: 退款用户分桶
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refund_bucket = refund_users.groupby("课时桶").agg(
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用户数=("用户ID", "count"),
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平均完成课时=("总完成课时数", "mean"),
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GMV=("GMV", "sum"),
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).reindex(bucket_order).fillna(0)
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refund_bucket["占比"] = (refund_bucket["用户数"] / total_refund_users * 100).round(1)
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refund_bucket.to_excel(writer, sheet_name="退款用户分布")
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print(f"\n✅ 报表已生成: {output_path}")
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if __name__ == "__main__":
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main()
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