Python根据另一个列表中不存在的值过滤一个列表

Question

试图用表B中找不到的2个值过滤表A上的查询结果。什么是正确的语法和方法？

 import pyodbc
 MDB = 'C:/db/db1.mdb'; DRV = '{Microsoft Access Driver (*.mdb)}'; PWD = 'pw'
 con = pyodbc.connect('DRIVER={};DBQ={};PWD={}'.format(DRV,MDB,PWD))
 cur = con.cursor()
 SQLA = 'SELECT * FROM TABLE1;' # your query goes here
 SQLB = 'SELECT * FROM TABLE2;' # your query goes here
 rows1 = cura.execute(SQLA).fetchall()
 rows2 = cura.execute(SQLB).fetchall()
 cur.close()
 con.close()
 for rit in rows1: 
     for git in rows2: 
        if (rit[1] and rit[2]) not in (git[1] and git[2]):
           print ((rit[1])  (rit[2]))

Answer 1

只需使用带有熟悉的LEFT JOIN... IS NULL / NOT EXISTS / NOT IN的纯SQL解决方案。 下面是等效查询，它们符合MS Access，但基于col1和col2返回TableA中的行，而不返回TableB中的行。

左联接...为空

SELECT a.*
FROM TABLEA a
LEFT JOIN TABLEB b
ON a.col1 = b.col1 AND a.col2 = b.col2
WHERE b.col1 IS NULL AND b.col2 IS NULL

不存在

SELECT a.*
FROM TABLEA a 
WHERE NOT EXISTS 
  (SELECT 1 FROM TABLEB b
   WHERE a.col1 = b.col1 AND a.col2 = b.col2)

不在

SELECT a.*
FROM TABLEA a
WHERE a.col1 NOT IN (SELECT col1 FROM TABLEB)
AND a.col2 NOT IN (SELECT col1 FROM TABLEB)

Answer 2

Parfait提供的SQL语句是首选的解决方案，但是如果您真的想使用双循环方法，则需要这样：

for rit in rows1:
    match_found = False
    for git in rows2: 
        if (rit[1] == git[1]) and (rit[2] == git[2]):
            match_found = True
            break
    if not match_found:
        print(rit)