import pandas as pd
import sqlite3
conn = sqlite3.connect('1-5.db')
用外连接进行行列转换1(行 -> 列): 制作交叉表
怎么使用outer join，将row转换成column
下面的方式一，使用的是外连接的方法。但是效果却是最差的。
# create Course Table and insert data
conn.execute("""
CREATE TABLE IF NOT EXISTS Courses (
name VARCHAR(10) NOT NULL,
course VARCHAR(10),
PRIMARY KEY(name, course)
)
""")
conn.execute("""DELETE FROM Courses;""")
conn.execute("""
INSERT INTO Courses(name, course)
VALUES
('赤井', 'SQL入门'),
('赤井', 'UNIX基础'),
('铃木', 'SQL入门'),
('工藤', 'SQL入门'),
('工藤', 'Java中级'),
('吉田', 'UNIX基础'),
('渡边', 'SQL入门')
""")
pd.read_sql_query('SELECT * FROM Courses', conn)
| | name | course |
|---|
| 0 |
赤井 |
SQL入门 |
|---|
| 1 |
赤井 |
UNIX基础 |
|---|
| 2 |
铃木 |
SQL入门 |
|---|
| 3 |
工藤 |
SQL入门 |
|---|
| 4 |
工藤 |
Java中级 |
|---|
| 5 |
吉田 |
UNIX基础 |
|---|
| 6 |
渡边 |
SQL入门 |
|---|
# 方式一: 每个要转换的列都使用一
pd.read_sql_query("""
SELECT c0.name,
CASE WHEN c1.name IS NOT NULL THEN 'O' ELSE 'X' END AS 'SQL入门',
CASE WHEN c2.name IS NOT NULL THEN 'O' ELSE 'X' END AS 'UNIX基础',
CASE WHEN c3.name is NOT NULL THEN 'O' ELSE 'X' END AS 'Java中级'
FROM (SELECT DISTINCT name FROM Courses) C0
LEFT OUTER JOIN
(SELECT name FROM Courses WHERE course='SQL入门') C1
ON c0.name = c1.name
LEFT OUTER JOIN
(SELECT name FROM Courses WHERE course='UNIX基础') C2
ON c0.name = c2.name
LEFT OUTER JOIN
(SELECT name FROM Courses WHERE course='Java中级') c3
ON c0.name = c3.name;
""", conn)
| | name | SQL入门 | UNIX基础 | Java中级 |
|---|
| 0 |
吉田 |
X |
O |
X |
|---|
| 1 |
工藤 |
O |
X |
O |
|---|
| 2 |
渡边 |
O |
X |
X |
|---|
| 3 |
赤井 |
O |
O |
X |
|---|
| 4 |
铃木 |
O |
X |
X |
|---|
# 方式二: 使用标量子查询代替外连接
# 优点是，加入新的列，只需要修改SELECT下的子查询。而方式一需要修改SELECT和FROM两处.
# 缺点是，在SELECT中使用标量子查询，开销比较大。因为需要对每一行都进行一次或多次子查询.
pd.read_sql_query("""
SELECT c0.name,
(SELECT 'O'
FROM Courses AS c1
WHERE course='SQL入门' AND c0.name = c1.name) AS 'SQL入门',
(SELECT 'O'
FROM Courses AS c2
WHERE course='UNIX基础' AND c0.name = c2.name) AS 'UNIX基础',
(SELECT 'O'
FROM Courses AS c3
WHERE course='Java中级' AND c0.name = c3.name) AS 'Java中级'
FROM (SELECT DISTINCT name FROM Courses) AS c0;
""", conn).replace(to_replace=[None], value='X')
| | name | SQL入门 | UNIX基础 | Java中级 |
|---|
| 0 |
吉田 |
X |
O |
X |
|---|
| 1 |
工藤 |
O |
X |
O |
|---|
| 2 |
渡边 |
O |
X |
X |
|---|
| 3 |
赤井 |
O |
O |
X |
|---|
| 4 |
铃木 |
O |
X |
X |
|---|
# 方式三: 嵌套使用CASE表单是
pd.read_sql_query("""
SELECT name,
CASE WHEN SUM(CASE WHEN course='SQL入门' THEN 1 ELSE 0 END) = 1 THEN 'O' ELSE 'X' END AS 'SQL入门',
CASE WHEN SUM(CASE WHEN course='UNIX基础' THEN 1 ELSE 0 END) = 1 THEN 'O' ELSE 'X' END AS 'UNIX基础',
CASE WHEN SUM(CASE WHEN course='Java中级' THEN 1 ELSE 0 END) = 1 THEN 'O' ELSE 'X' END AS 'Java中级'
FROM Courses
GROUP BY name
""", conn)
| | name | SQL入门 | UNIX基础 | Java中级 |
|---|
| 0 |
吉田 |
X |
O |
X |
|---|
| 1 |
工藤 |
O |
X |
O |
|---|
| 2 |
渡边 |
O |
X |
X |
|---|
| 3 |
赤井 |
O |
O |
X |
|---|
| 4 |
铃木 |
O |
X |
X |
|---|
用外连接进行行列转换2(列 -> 行): 汇总重复项于一列
# create Personnel Table and insert data
conn.execute("""
CREATE TABLE IF NOT EXISTS Personnel (
employee VARCHAR(10) NOT NULL PRIMARY KEY,
child_1 VARCHAR(10),
child_2 VARCHAR(10),
child_3 VARCHAR(10)
)
""")
conn.execute("""DELETE FROM Personnel;""")
conn.execute("""
INSERT INTO Personnel(employee, child_1, child_2, child_3)
VALUES
('赤井', '一郎', '二郎', '三郎'),
('工藤', '春子', '夏子', null),
('铃木', '夏子', null, null),
('吉田', null, null, null)
""")
pd.read_sql_query('SELECT * FROM Personnel', conn)
| | employee | child_1 | child_2 | child_3 |
|---|
| 0 |
赤井 |
一郎 |
二郎 |
三郎 |
|---|
| 1 |
工藤 |
春子 |
夏子 |
None |
|---|
| 2 |
铃木 |
夏子 |
None |
None |
|---|
| 3 |
吉田 |
None |
None |
None |
|---|
# 这个方法，会把child为null的结果也返回。
# 但是如果排除掉child为null的结果，那么吉田这个employee因为没有孩子，就不会出现在结果表里面
pd.read_sql_query("""
SELECT employee, child_1 AS child FROM Personnel
UNION ALL
SELECT employee, child_2 AS child FROM Personnel
UNION ALL
SELECT employee, child_3 AS child FROM Personnel
""", conn)
| | employee | child |
|---|
| 0 |
赤井 |
一郎 |
|---|
| 1 |
工藤 |
春子 |
|---|
| 2 |
铃木 |
夏子 |
|---|
| 3 |
吉田 |
None |
|---|
| 4 |
赤井 |
二郎 |
|---|
| 5 |
工藤 |
夏子 |
|---|
| 6 |
铃木 |
None |
|---|
| 7 |
吉田 |
None |
|---|
| 8 |
赤井 |
三郎 |
|---|
| 9 |
工藤 |
None |
|---|
| 10 |
铃木 |
None |
|---|
| 11 |
吉田 |
None |
|---|
conn.execute("""
CREATE VIEW IF NOT EXISTS Children(child) AS
SELECT child_1 FROM Personnel
UNION
SELECT child_2 FROM Personnel
UNION
SELECT child_3 FROM Personnel
""")
pd.read_sql_query("""
SELECT EMP.employee, CHILDREN.child
FROM Personnel EMP
LEFT OUTER JOIN Children
ON Children.child IN (EMP.child_1, EMP.child_2, EMP.child_3)
""", conn)
| | employee | child |
|---|
| 0 |
赤井 |
一郎 |
|---|
| 1 |
赤井 |
三郎 |
|---|
| 2 |
赤井 |
二郎 |
|---|
| 3 |
工藤 |
夏子 |
|---|
| 4 |
工藤 |
春子 |
|---|
| 5 |
铃木 |
夏子 |
|---|
| 6 |
吉田 |
None |
|---|
# 习题1-5-2: 求每个雇员的孩子数量
pd.read_sql_query("""
SELECT EMP.employee, COUNT(CHILDREN.child) AS child_cnt
FROM Personnel EMP
LEFT OUTER JOIN Children
ON Children.child IN (EMP.child_1, EMP.child_2, EMP.child_3)
GROUP BY EMP.employee
""", conn)
| | employee | child_cnt |
|---|
| 0 |
吉田 |
0 |
|---|
| 1 |
工藤 |
2 |
|---|
| 2 |
赤井 |
3 |
|---|
| 3 |
铃木 |
1 |
|---|
在交叉表里制作嵌套式表侧栏
# 创建相关的表
conn.execute("""
CREATE TABLE IF NOT EXISTS TblAge (
age_class integer PRIMARY KEY,
age_range varchar(20)
);
""")
conn.execute("""
CREATE TABLE IF NOT EXISTS TblSex (
sex_cd char(1) PRIMARY KEY,
sex char(1)
);
""")
conn.execute("""
CREATE TABLE IF NOT EXISTS TblPop (
pref_name VARCHAR(10),
age_class integer,
sex_cd char(1),
population integer,
PRIMARY KEY (pref_name, age_class, sex_cd),
FOREIGN KEY (age_class) REFERENCES TblAge(age_class),
FOREIGN KEY (sex_cd) REFERENCES TblSex(sex_cd)
);
""")
# 确保没有数据
conn.execute("""DELETE FROM TblPop""")
conn.execute("""DELETE FROM TblSex""")
conn.execute("""DELETE FROM TblAge""")
# 插入数据
conn.execute("""
INSERT INTO TblAge(age_class, age_range) VALUES
(1, '21岁 ~ 30岁'),
(2, '31岁 ~ 40岁'),
(3, '41岁 ~ 50岁')
""")
conn.execute("""
INSERT INTO TblSex(sex_cd, sex) VALUES
('m', '男'),
('f', '女')
""")
conn.execute("""
INSERT INTO TblPop(pref_name, age_class, sex_cd, population) VALUES
('秋田', 1, 'm', 400),
('秋田', 3, 'm', 1000),
('秋田', 1, 'f', 800),
('秋田', 3, 'f', 1000),
('青森', 1, 'm', 700),
('青森', 1, 'f', 500),
('青森', 3, 'f', 800),
('东京', 1, 'm', 1500),
('东京', 1, 'f', 1200),
('千叶', 1, 'm', 900),
('千叶', 1, 'f', 1000),
('千叶', 3, 'f', 900)
""")
<sqlite3.Cursor at 0x113c47180>
pd.read_sql("""
SELECT * FROM TblAge
""", conn)
| | age_class | age_range |
|---|
| 0 |
1 |
21岁 ~ 30岁 |
|---|
| 1 |
2 |
31岁 ~ 40岁 |
|---|
| 2 |
3 |
41岁 ~ 50岁 |
|---|
pd.read_sql("""
SELECT * FROM TblSex
""", conn)
pd.read_sql("""
SELECT * FROM TblPop
""", conn)
| | pref_name | age_class | sex_cd | population |
|---|
| 0 |
秋田 |
1 |
m |
400 |
|---|
| 1 |
秋田 |
3 |
m |
1000 |
|---|
| 2 |
秋田 |
1 |
f |
800 |
|---|
| 3 |
秋田 |
3 |
f |
1000 |
|---|
| 4 |
青森 |
1 |
m |
700 |
|---|
| 5 |
青森 |
1 |
f |
500 |
|---|
| 6 |
青森 |
3 |
f |
800 |
|---|
| 7 |
东京 |
1 |
m |
1500 |
|---|
| 8 |
东京 |
1 |
f |
1200 |
|---|
| 9 |
千叶 |
1 |
m |
900 |
|---|
| 10 |
千叶 |
1 |
f |
1000 |
|---|
| 11 |
千叶 |
3 |
f |
900 |
|---|
# 下面这种，以两张表作为外连接操作的方式。
# 有时候并不能满足需求，比如不会显示age_class=2的情况，因为没有对应的人口数据
pd.read_sql("""
SELECT MASTER1.age_class AS age_class,
MASTER2.sex_cd AS sex_cd,
DATA.pop_tohoku AS pop_tohoku,
DATA.pop_kanto AS pop_kanto
FROM (SELECT age_class, sex_cd,
SUM(CASE WHEN pref_name IN ('青森', '秋田') THEN population ELSE NULL END) AS pop_tohoku,
SUM(CASE WHEN pref_name IN ('东京', '千叶') THEN population ELSE NULL END) AS pop_kanto
FROM TblPop
GROUP BY age_class, sex_cd) AS DATA
LEFT OUTER JOIN TblAge MASTER1
ON MASTER1.age_class = DATA.age_class
LEFT OUTER JOIN TblSex MASTER2
ON MASTER2.sex_cd = DATA.sex_cd
""", conn)
| | age_class | sex_cd | pop_tohoku | pop_kanto |
|---|
| 0 |
1 |
f |
1300 |
2200.0 |
|---|
| 1 |
1 |
m |
1100 |
2400.0 |
|---|
| 2 |
3 |
f |
1800 |
900.0 |
|---|
| 3 |
3 |
m |
1000 |
NaN |
|---|
# 解决方式是：调整为一次外连接
# 外连接的表是 TblAge和TblSex 的笛卡尔积，也就是把它们用交叉连接(CROSS JOIN)组合起来,
# 对于不支持CROSS JOIN的库，分别把两个表用FROM引入，也是一样的.
# 同时，请注意下面把临时表放在了 LEFT OUTER JOIN 的左边，充分发挥OUTER JOIN的作用.
pd.read_sql("""
SELECT MASTER.age_class AS age_class,
MASTER.sex_cd AS sex_cd,
DATA.pop_tohoku AS pop_tohoku,
DATA.pop_kanto AS pop_kanto
FROM (SELECT age_class, sex_cd
FROM TblAge CROSS JOIN TblSex) AS MASTER
LEFT OUTER JOIN
(SELECT age_class, sex_cd,
SUM(CASE WHEN pref_name IN ('青森', '秋田') THEN population ELSE NULL END) AS pop_tohoku,
SUM(CASE WHEN pref_name IN ('东京', '千叶') THEN population ELSE NULL END) AS pop_kanto
FROM TblPop
GROUP BY age_class, sex_cd) AS DATA
ON MASTER.age_class = DATA.age_class
AND MASTER.sex_cd = DATA.sex_cd
""", conn)
| | age_class | sex_cd | pop_tohoku | pop_kanto |
|---|
| 0 |
1 |
f |
1300.0 |
2200.0 |
|---|
| 1 |
1 |
m |
1100.0 |
2400.0 |
|---|
| 2 |
2 |
f |
NaN |
NaN |
|---|
| 3 |
2 |
m |
NaN |
NaN |
|---|
| 4 |
3 |
f |
1800.0 |
900.0 |
|---|
| 5 |
3 |
m |
1000.0 |
NaN |
|---|
# 更优的解决方法是：把JOIN看作乘法运算
# 关键在于，TblPop可以看作就是DATA。然后 MASTER -> DATA 的关系，就是一对多关系了
pd.read_sql("""
SELECT MASTER.age_class AS age_class,
MASTER.sex_cd AS sex_cd,
SUM(CASE WHEN pref_name IN ('青森', '秋田') THEN population ELSE NULL END) AS pop_tohoku,
SUM(CASE WHEN pref_name IN ('东京', '千叶') THEN population ELSE NULL END) AS pop_kanto
FROM (SELECT age_class, sex_cd
FROM TblAge CROSS JOIN TblSex) AS MASTER
LEFT OUTER JOIN TblPop AS DATA
ON MASTER.age_class = DATA.age_class
AND MASTER.sex_cd = DATA.sex_cd
GROUP BY MASTER.age_class, MASTER.sex_cd
""", conn)
| | age_class | sex_cd | pop_tohoku | pop_kanto |
|---|
| 0 |
1 |
f |
1300.0 |
2200.0 |
|---|
| 1 |
1 |
m |
1100.0 |
2400.0 |
|---|
| 2 |
2 |
f |
NaN |
NaN |
|---|
| 3 |
2 |
m |
NaN |
NaN |
|---|
| 4 |
3 |
f |
1800.0 |
900.0 |
|---|
| 5 |
3 |
m |
1000.0 |
NaN |
|---|
用外连接做集合运算
各个数据库的集合运算实现都不尽相同，参差不齐。
集合运算符会尽心排序，所以可能带来性能上的问题。
因此，了解一下集合运算符的替代方案还是有意义的。
# 创建数据库
conn.execute("""
CREATE TABLE IF NOT EXISTS Class_A (
id integer PRIMARY KEY AUTOINCREMENT,
name varchar(10)
);
""")
conn.execute("""
CREATE TABLE IF NOT EXISTS Class_B (
id integer PRIMARY KEY AUTOINCREMENT,
name varchar(10)
)
""")
# 插入数据
conn.execute("""DELETE FROM Class_A;""")
conn.execute("""DELETE FROM Class_B;""")
conn.execute("""
INSERT INTO Class_A(id, name) VALUES
(1, '田中'),
(2, '铃木'),
(3, '伊集院')
""")
conn.execute("""
INSERT INTO Class_B(id, name) VALUES
(1, '田中'),
(2, '铃木'),
(4, '西院寺')
""")
<sqlite3.Cursor at 0x113f3c5e0>
pd.read_sql("""
SELECT * FROM Class_A
""", conn)
| | id | name |
|---|
| 0 |
1 |
田中 |
|---|
| 1 |
2 |
铃木 |
|---|
| 2 |
3 |
伊集院 |
|---|
pd.read_sql("""
SELECT * FROM Class_B
""", conn)
| | id | name |
|---|
| 0 |
1 |
田中 |
|---|
| 1 |
2 |
铃木 |
|---|
| 2 |
4 |
西院寺 |
|---|
用外连接求差集: A - B
pd.read_sql("""
SELECT A.id id, A.name AS A_name
FROM Class_A AS A LEFT OUTER JOIN Class_B AS B
ON A.id = B.id
WHERE B.name IS NULL
""", conn)
用外连接求差集: B - A
pd.read_sql("""
SELECT B.id id, B.name AS B_name
FROM Class_B AS B LEFT OUTER JOIN Class_A AS A
ON A.id = B.id
WHERE A.name IS NULL
""", conn)
用外连接求异或集
pd.read_sql("""
SELECT A.id id, A.name AS name
FROM Class_A AS A LEFT OUTER JOIN Class_B AS B
ON A.id = B.id
WHERE B.name IS NULL
UNION
SELECT B.id id, B.name AS name
FROM Class_B AS B LEFT OUTER JOIN Class_A AS A
ON A.id = B.id
WHERE A.name IS NULL
""", conn)
