原文地址python
此文为 SQL to Pandas 系列第二篇sql
可先阅读post
【Python】当数据离开了 SQL 该怎么活下去(一)测试
本篇将解构下面的 SQL 查询句式, 使用 Pandas 进行实现ui
SQL 查询句式spa
SELECT DISTINCT [字段]
FROM [表] JOIN [bin] ON [链接条件]
WHERE [过滤条件]
GROUP BY [字段]
HAVING [条件]
ORDER BY [字段] DESC
LIMIT [个数] OFFSET [个数]
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df_comments = pd.read_sql(sql="select * from comments", con=conn)
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数据预览code
df_comments
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id | student_id | content | |
---|---|---|---|
0 | 1 | 1 | 测试评论1 |
1 | 2 | 5 | 测试评论5 |
2 | 3 | 2 | 测试评论2 |
3 | 4 | 3 | 测试评论3 |
4 | 5 | 1 | 测试评论11 |
5 | 6 | 9 | 测试评论9 |
(INNER) JOINcdn
SQLci
SELECT
*
FROM
student
INNER JOIN comments ON student.id = comments.student_id;
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Pandasget
pd.merge(df, df_comments, left_on='id', right_on='student_id')
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LEFT (OUTER) JOIN
SQL
SELECT
*
FROM
student
LEFT JOIN comments ON student.id = comments.student_id;
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Pandas
pd.merge(df, df_comments, left_on='id', right_on='student_id', how='left')
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RIGHT (OUTER) JOIN
SQL
SELECT
*
FROM
student
RIGHT JOIN comments ON student.id = comments.student_id;
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Pandas
pd.merge(df, df_comments, left_on='id', right_on='student_id', how='right')
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UNION
SQL
SELECT * FROM student where city ='北京'
UNION
SELECT * FROM student where sex ='男';
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Pandas
pd.concat([df[df.city == '北京'], df[df.sex == '男']]).drop_duplicates().reset_index()
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UNION ALL
SQL
SELECT * FROM student where city ='北京'
UNION ALL
SELECT * FROM student where sex ='男';
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Pandas
pd.concat([df[df.city == '北京'], df[df.sex == '男']]).reset_index()
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本篇将解构下面的 SQL 查询句式, 使用 Pandas 进行实现
SQL 建立句式
CREATE TABLE [表名] (
[列名] [类型],
[列名] [类型],
....
);
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SQL 插入句式
INSERT INTO [表名] VALUES ([值], [值], ...);
INSERT INTO [表名] ([列名],[列名] ...) VALUES ([值], [值], ...);
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SQL 更新句式
UPDATE [表名]
SET [列名] = [值], [列名] = [值]
WHERE [过滤条件];
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SQL 删除句式
DELETE FROM [表名] WHERE [过滤条件];
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SQL
CREATE TABLE student (
id INT ( 11 ) NOT NULL AUTO_INCREMENT,
name VARCHAR ( 10 ) COLLATE utf8mb4_general_ci DEFAULT NULL,
age date DEFAULT NULL,
sex VARCHAR ( 10 ) COLLATE utf8mb4_general_ci DEFAULT NULL,
city VARCHAR ( 255 ) CHARACTER
SET utf8mb4 COLLATE utf8mb4_general_ci DEFAULT NULL,
money DOUBLE ( 255, 2 ) DEFAULT NULL,
);
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Pandas
pd.DataFrame(columns=['id', 'name', 'sex', 'city', 'money'])
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SQL
INSERT INTO student (id, name, age, sex, city, money )
VALUES
(1, '张三', '2017-12-20', '女', '天津', 20.00 );
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Pandas
# 第一种
df.loc[-1] = [1, '张三', '女', '天津', 20.00]
df.index = df.index + 1
df = df.sort_index()
# 第二种
temp_pd = pd.DataFrame({'id': [1], 'name': ['张三'], 'sex': ['女'], 'city': ['天津'], 'money': [20.00]})
df = pd.concat([df,temp_pd], ignore_index=True)
df.reset_index()
# 第三种
temp_pd = pd.DataFrame([[1,'张三1', '女', '天津', 20.00]], columns=df.columns)
df = pd.concat([df, temp_pd])
df.reset_index()
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SQL
UPDATE student SET money = 300 WHERE id = 1;
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Pandas
df.loc[df.id == 1, 'money'] = 300
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SQL
-- 测试时会因外键报错,此处忽略,仅讨论句法
DELETE FROM student WHERE id = 1;
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Pandas
df = df.loc[df.id != 1]
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