SQL PageRank實現

Question

那裏有很好的SQL PageRank實現嗎？我看過http://www.databasedevelop.com/197517/，但它缺少可讀性和正確性（T-SQL）語法。

雖然我們在這，但有人知道上面的鏈接正在使用什麼樣的SQL？什麼SQL使用'是'在隨機的地方，'哪裏'什麼都沒有，奇怪的AT關鍵字，等等？

Answer 1

根據您的SQL Server版本，可以查看OFFSET_FETCH窗口函數。這裏有很多頁面排名的應用程序。當然，這將需要2012年。

我也用SSIS和一個NTILE（）分割臨時表來完成分頁的能力，使用OFFSET_FETCH的缺席。通常使用類似記錄計數除以我想在頁面中看到的最大數量作爲NTILE調用的種子。

無論出於何種原因，我甚至無法打開您的鏈接，所以希望這是您要求的。

MSDN - OFFSET_FETCH

MSDN - NTILE

Answer 2

我只是在SQL中實現的PageRank算法。該算法在以下進展中起作用。

1.計算PageRank初始等級值;

2.Joining 的PageRank表邊緣表以發射秩值至鄰接的節點，並且使用聚集函數總和爲「收集」接收到的值。然後，將結果保存到寺廟表TmpRank;

交換的PageRank和的內容TmpRank，並轉到步驟2，直到收斂條件被滿足或達到最大重複時間。

下面是代碼：

-- 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;