【Python】SQL to Pandas 速查表（二）

本篇将解构下面的 SQL 查询句式, 使用 Pandas 进行实现

SQL 查询句式

SELECT DISTINCT [字段] 
FROM [表] JOIN [bin] ON [连接条件] 
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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读取评论数据

df_comments = pd.read_sql(sql="select * from comments", con=conn)
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数据预览

df_comments
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JOIN

(INNER) JOIN

SQL

SELECT
	* 
FROM
	student
	INNER 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')
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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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本篇内容

本篇将解构下面的 SQL 查询句式, 使用 Pandas 进行实现

SQL 创建句式

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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CREATE

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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INSERT

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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UPDATE

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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