SQL PageRank 实现

Question

有没有好的 SQL PageRank 实现？我查看了http://www.databasedevelop.com/197517/，但它缺乏易读性和正确的（T-SQL）语法。

当我们这样做时，有人知道上面的链接使用的是哪种 SQL 吗？什么 SQL 在随机位置使用“is”、后面没有任何内容的“where”、奇怪的AT关键字等等？

Answer 1

我刚刚用 SQL 实现了 PageRank 算法。该算法的工作进展如下。

1.计算PageRank初始排名值；

2.将PageRank表与Edge表连接以向相邻节点发出排名值，并使用聚合函数sum来“收集”接收到的值。然后，将结果保存到寺庙表TmpRank；

3. 交换PageRank和TmpRank的内容，转步骤 2，直到满足收敛条件或达到最大迭代时间。

这是代码：

-- The graph data and algorithm source from the book "Mining of Massive Datasets", P175, http://infolab.stanford.edu/~ullman/mmds/book.pdf
-- This script has been verified the correctness in SQL Server 2017 Linux Version.
DROP TABLE Node;
DROP TABLE Edge;
DROP TABLE OutDegree;
DROP TABLE PageRank;
CREATE TABLE Node(id int PRIMARY KEY);
CREATE TABLE Edge(src int,dst int, PRIMARY KEY (src, dst));
CREATE TABLE OutDegree(id int PRIMARY KEY, degree int);
CREATE TABLE PageRank(id int PRIMARY KEY, rank float);
CREATE TABLE TmpRank(id int PRIMARY KEY, rank float);

--delete all records
DELETE FROM Node;
DELETE FROM Edge;
DELETE FROM OutDegree;
DELETE FROM PageRank;
DELETE FROM TmpRank;

--init basic tables
INSERT INTO Node VALUES (0);
INSERT INTO Node VALUES (1);
INSERT INTO Node VALUES (2);
INSERT INTO Node VALUES (3);

INSERT INTO Edge VALUES (0, 1);
INSERT INTO Edge VALUES (0, 2);
INSERT INTO Edge VALUES (0, 3);
INSERT INTO Edge VALUES (1, 0);
INSERT INTO Edge VALUES (1, 3);
INSERT INTO Edge VALUES (2, 2);
INSERT INTO Edge VALUES (3, 1);
INSERT INTO Edge VALUES (3, 2);

--compute out-degree
INSERT INTO OutDegree
SELECT Node.id, COUNT(Edge.src) --Count(Edge.src) instead of Count(*) for count no out-degree edge
FROM Node LEFT OUTER JOIN Edge
ON Node.id = Edge.src
GROUP BY Node.id;

--WARN: There's no special process for node with out-degree, This may cause wrong result
--      Please to make sure every node in graph has out-degree

DECLARE @ALPHA float = 0.8;
DECLARE @Node_Num int;
SELECT @Node_Num = COUNT(*) FROM Node;

--PageRank Init Value
INSERT INTO PageRank
SELECT Node.id, rank = (1 - @ALPHA) / @Node_Num
FROM Node INNER JOIN OutDegree
ON Node.id = OutDegree.id

/*
--For Debugging
SELECT * FROM Node;
SELECT * FROM Edge;
SELECT * FROM OutDegree;
SELECT * FROM PageRank;
SELECT * FROM TmpRank;
*/

DECLARE @Iteration int = 0;

WHILE @Iteration < 50
BEGIN
--Iteration Style
    SET @Iteration = @Iteration + 1

    INSERT INTO TmpRank
    SELECT Edge.dst, rank = SUM(@ALPHA * PageRank.rank / OutDegree.degree) + (1 - @ALPHA) / @Node_Num
    FROM PageRank
    INNER JOIN Edge ON PageRank.id = Edge.src
    INNER JOIN OutDegree ON PageRank.id = OutDegree.id
    GROUP BY Edge.dst

    DELETE FROM PageRank;
    INSERT INTO PageRank
    SELECT * FROM TmpRank;
    DELETE FROM TmpRank;
END

SELECT * FROM PageRank;