From b7c46750d4ccb865fb572dfc3ada5adcf1cf527e Mon Sep 17 00:00:00 2001
From: =?UTF-8?q?=E7=8E=8B=E5=A0=83=E8=90=A5?= <1572506709@qq.com>
Date: Wed, 7 Oct 2026 23:38:11 +0800
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 1 file changed, 34 insertions(+)

diff --git a/written.md b/written.md
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--- a/written.md
+++ b/written.md
@@ -45,4 +45,38 @@ c = copy.deepcopy(a)
 
 原因：浅拷贝只复制最外层容器，内部元素仍是引用（指向同一对象）；深拷贝会递归复制每一层，得到完全独立的对象。
 
+第 2 题：字典与列表的综合应用
+
+```python
+logs = [
+    {"user": "张三", "action": "login", "level": "INFO"},
+    {"user": "李四", "action": "logout", "level": "INFO"},
+    {"user": "张三", "action": "error", "level": "ERROR"},
+    {"user": "王五", "action": "login", "level": "INFO"},
+    {"user": "李四", "action": "error", "level": "ERROR"},
+]
+```
+
+① 找出所有 level 为 "ERROR" 的日志：
+
+```python
+error_logs = [log for log in logs if log["level"] == "ERROR"]
+# [{'user': '张三', 'action': 'error', 'level': 'ERROR'},
+#  {'user': '李四', 'action': 'error', 'level': 'ERROR'}]
+```
+
+② 统计每个用户出现几次：
+
+```python
+user_count = {}
+for log in logs:
+    user = log["user"]
+    user_count[user] = user_count.get(user, 0) + 1
+# {'张三': 2, '李四': 2, '王五': 1}
+```
+
+也可以用 `collections.Counter` 一行完成：`Counter(log["user"] for log in logs)`。
+
+③ 为什么不能用 len(logs)：`len(logs)` 只返回日志的总条数（这里是 5），它不做任何分类统计。要按用户分别计数，必须用一个 for 循环逐条遍历 logs，取出每条里的 `"user"` 字段，再用字典以用户名为键累加次数。所以需要的是「for 遍历 + 字典累加」这种结构，而不是一个 `len()`。
+