SQL to Pandas 速查表（二）

介绍

此文为 SQL to Pandas 系列第二篇

可先阅读

【Python】SQL to Pandas 速查表（一）

补充上篇 JOIN 内容

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]

读取评论数据

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

数据预览

df_comments

JOIN

(INNER) JOIN

SQL

SELECT
    * 
FROM
    student
    INNER JOIN comments ON student.id = comments.student_id;

Pandas

pd.merge(df, df_comments, left_on='id', right_on='student_id')

LEFT (OUTER) JOIN

SQL

SELECT
    * 
FROM
    student
    LEFT JOIN comments ON student.id = comments.student_id;

Pandas

pd.merge(df, df_comments, left_on='id', right_on='student_id', how='left')

RIGHT (OUTER) JOIN

SQL

SELECT
    * 
FROM
    student
    RIGHT JOIN comments ON student.id = comments.student_id;

Pandas

pd.merge(df, df_comments, left_on='id', right_on='student_id', how='right')

UNION

SQL

SELECT * FROM student where city ='北京' 
UNION 
SELECT * FROM student where sex ='男';

Pandas

pd.concat([df[df.city == '北京'], df[df.sex == '男']]).drop_duplicates().reset_index()

UNION ALL

SQL

SELECT * FROM student where city ='北京' 
UNION ALL 
SELECT * FROM student where sex ='男';

Pandas

pd.concat([df[df.city == '北京'], df[df.sex == '男']]).reset_index()

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],
   ....
);

SQL 插入句式

INSERT INTO [表名] VALUES ([值], [值], ...);
INSERT INTO [表名] ([列名],[列名] ...) VALUES ([值], [值], ...);

SQL 更新句式

UPDATE [表名]
SET [列名] = [值], [列名] = [值]
WHERE [过滤条件];

SQL 删除句式

DELETE FROM [表名] WHERE [过滤条件];

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,

);

Pandas

pd.DataFrame(columns=['id', 'name', 'sex', 'city', 'money'])

INSERT

SQL

INSERT INTO student (id, name, age, sex, city, money )
VALUES
    (1, '张三', '2017-12-20', '女', '天津', 20.00 );

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()

UPDATE

SQL

UPDATE student SET money = 300 WHERE id = 1;

Pandas

df.loc[df.id == 1, 'money'] = 300

DELETE

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;

Pandas

df = df.loc[df.id != 1]

本系列文章

【Python】SQL to Pandas 速查表（一）

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