# Write your MySQL query statement below
1148. 文章浏览 I
Views 表：
+---------------+---------+ | Column Name | Type | +---------------+---------+ | article_id | int | | author_id | int | | viewer_id | int | | view_date | date | +---------------+---------+ 此表无主键，因此可能会存在重复行。 此表的每一行都表示某人在某天浏览了某位作者的某篇文章。 请注意，同一人的 author_id 和 viewer_id 是相同的。
请编写一条 SQL 查询以找出所有浏览过自己文章的作者，结果按照 id 升序排列。
查询结果的格式如下所示：
Views 表： +------------+-----------+-----------+------------+ | article_id | author_id | viewer_id | view_date | +------------+-----------+-----------+------------+ | 1 | 3 | 5 | 2019-08-01 | | 1 | 3 | 6 | 2019-08-02 | | 2 | 7 | 7 | 2019-08-01 | | 2 | 7 | 6 | 2019-08-02 | | 4 | 7 | 1 | 2019-07-22 | | 3 | 4 | 4 | 2019-07-21 | | 3 | 4 | 4 | 2019-07-21 | +------------+-----------+-----------+------------+ 结果表： +------+ | id | +------+ | 4 | | 7 | +------+
原站题解
mysql 解法, 执行用时: 414 ms, 内存消耗: 0 B, 提交时间: 2022-06-02 10:00:24
# Write your MySQL query statement below select distinct author_id as id from views where author_id=viewer_id order by id asc;
mysql 解法, 执行用时: 425 ms, 内存消耗: 0 B, 提交时间: 2022-05-27 11:40:25
# Write your MySQL query statement below select distinct author_id as id from views where author_id=viewer_id order by id asc;
pythondata 解法, 执行用时: 320 ms, 内存消耗: 60.5 MB, 提交时间: 2023-08-07 16:33:18
import pandas as pd
def article_views(views: pd.DataFrame) -> pd.DataFrame:
df = views[views['author_id'] == views['viewer_id']]
df.drop_duplicates(subset=['author_id'], inplace=True)
df.sort_values(by=['author_id'], inplace=True)
df.rename(columns={'author_id':'id'}, inplace=True)
df = df[['id']]
return df