上周， 我寫了關(guān)于中間性中心性算法以及使用graphstream 理解它的嘗試 ，在閱讀源代碼時(shí)，我意識(shí)到我可以使用neo4j的所有最短路徑算法將某些東西放在一起。

概括地說(shuō)，中間性中心度算法用于確定圖中節(jié)點(diǎn)的負(fù)載和重要性。

在與Jen討論這一點(diǎn)時(shí)，她指出，計(jì)算整個(gè)圖上節(jié)點(diǎn)之間的中間性通常是沒(méi)有意義的。 但是，了解在您感興趣的較小子圖中哪個(gè)節(jié)點(diǎn)最重要可能很有用。

在這種情況下，我有興趣在一個(gè)非常小的有向圖中確定節(jié)點(diǎn)的中間性：

讓我們簡(jiǎn)要回顧一下算法：

[中間性中心]等于從所有頂點(diǎn)到經(jīng)過(guò)該節(jié)點(diǎn)的所有其他頂點(diǎn)的最短路徑數(shù)。

這意味著我們排除了直接在兩個(gè)節(jié)點(diǎn)之間而不經(jīng)過(guò)任何其他路徑的任何路徑，這是我最初沒(méi)有掌握的。

如果我們手工確定適用的路徑，我們將得到以下結(jié)果：

A -> B: Direct Path Exists A -> C: B A -> D: E A -> E: Direct Path Exists B -> A: No Path Exists B -> C: Direct Path Exists B -> D: E or C B -> E: Direct Path Exists C -> A: No Path Exists C -> B: No Path Exists C -> D: Direct Path Exists C -> E: No Path Exists D -> A: No Path Exists D -> B: No Path Exists D -> C: No Path Exists D -> E: No Path Exists E -> A: No Path Exists E -> B: No Path Exists E -> C: No Path Exists E -> D: Direct Path Exists

給出以下中間性中心值：

A: 0 B: 1 C: 0.5 D: 0 E: 1.5

我們可以針對(duì)最新版本的graphstream （考慮方向）編寫測(cè)試，以確認(rèn)我們的手動(dòng)算法：

@Testpublic void calculateBetweennessCentralityOfMySimpleGraph() {Graph graph = new SingleGraph("Tutorial 1");Node A = graph.addNode("A");Node B = graph.addNode("B");Node E = graph.addNode("E");Node C = graph.addNode("C");Node D = graph.addNode("D");graph.addEdge("AB", A, B, true);graph.addEdge("BE", B, E, true);graph.addEdge("BC", B, C, true);graph.addEdge("ED", E, D, true);graph.addEdge("CD", C, D, true);graph.addEdge("AE", A, E, true);BetweennessCentrality bcb = new BetweennessCentrality();bcb.computeEdgeCentrality(false);bcb.betweennessCentrality(graph);System.out.println("A="+ A.getAttribute("Cb"));System.out.println("B="+ B.getAttribute("Cb"));System.out.println("C="+ C.getAttribute("Cb"));System.out.println("D="+ D.getAttribute("Cb"));System.out.println("E="+ E.getAttribute("Cb"));}

輸出是預(yù)期的：

A=0.0 B=1.0 C=0.5 D=0.0 E=1.5

我想看看是否可以使用neo4j做同樣的事情，所以我使用以下cypher語(yǔ)句在空白數(shù)據(jù)庫(kù)中創(chuàng)建了圖形：

CREATE (A {name: "A"}) CREATE (B {name: "B"}) CREATE (C {name: "C"}) CREATE (D {name: "D"}) CREATE (E {name: "E"})CREATE A-[:TO]->E CREATE A-[:TO]->B CREATE B-[:TO]->C CREATE B-[:TO]->E CREATE C-[:TO]->D CREATE E-[:TO]->D

然后，我編寫了一個(gè)查詢，該查詢找到了圖中所有節(jié)點(diǎn)集之間的最短路徑：

MATCH p = allShortestPaths(source-[r:TO*]->destination) WHERE source <> destination RETURN NODES(p)

如果運(yùn)行，它將返回以下內(nèi)容：

==> +---------------------------------------------------------+ ==> | NODES(p) | ==> +---------------------------------------------------------+ ==> | [Node[1]{name:"A"},Node[2]{name:"B"}] | ==> | [Node[1]{name:"A"},Node[2]{name:"B"},Node[3]{name:"C"}] | ==> | [Node[1]{name:"A"},Node[5]{name:"E"},Node[4]{name:"D"}] | ==> | [Node[1]{name:"A"},Node[5]{name:"E"}] | ==> | [Node[2]{name:"B"},Node[3]{name:"C"}] | ==> | [Node[2]{name:"B"},Node[3]{name:"C"},Node[4]{name:"D"}] | ==> | [Node[2]{name:"B"},Node[5]{name:"E"},Node[4]{name:"D"}] | ==> | [Node[2]{name:"B"},Node[5]{name:"E"}] | ==> | [Node[3]{name:"C"},Node[4]{name:"D"}] | ==> | [Node[5]{name:"E"},Node[4]{name:"D"}] | ==> +---------------------------------------------------------+ ==> 10 rows

我們?nèi)栽诜祷毓?jié)點(diǎn)之間的直接鏈接，但是通過(guò)基于路徑中節(jié)點(diǎn)的數(shù)量過(guò)濾結(jié)果可以很容易地糾正這一點(diǎn)：

MATCH p = allShortestPaths(source-[r:TO*]->destination) WHERE source <> destination AND LENGTH(NODES(p)) > 2 RETURN EXTRACT(n IN NODES(p): n.name)==> +--------------------------------+ ==> | EXTRACT(n IN NODES(p): n.name) | ==> +--------------------------------+ ==> | ["A","B","C"] | ==> | ["A","E","D"] | ==> | ["B","C","D"] | ==> | ["B","E","D"] | ==> +--------------------------------+ ==> 4 rows

如果我們稍微調(diào)整密碼查詢，我們可以獲得每個(gè)源/目標(biāo)的最短路徑的集合：

MATCH p = allShortestPaths(source-[r:TO*]->destination) WHERE source <> destination AND LENGTH(NODES(p)) > 2 WITH EXTRACT(n IN NODES(p): n.name) AS nodes RETURN HEAD(nodes) AS source, HEAD(TAIL(TAIL(nodes))) AS destination, COLLECT(nodes) AS paths==> +------------------------------------------------------+ ==> | source | destination | paths | ==> +------------------------------------------------------+ ==> | "A" | "D" | [["A","E","D"]] | ==> | "A" | "C" | [["A","B","C"]] | ==> | "B" | "D" | [["B","C","D"],["B","E","D"]] | ==> +------------------------------------------------------+ ==> 3 rows

當(dāng)我們有一種使用cypher來(lái)對(duì)集合進(jìn)行切片的方法時(shí)，從這里獲得節(jié)點(diǎn)間的中間性得分并不難，但是現(xiàn)在使用通用的編程語(yǔ)言要容易得多。

在這種情況下，我使用了Ruby并提出了以下代碼：

require 'neography' neo = Neography::Rest.newquery = " MATCH p = allShortestPaths(source-[r:TO*]->destination)" query << " WHERE source <> destination AND LENGTH(NODES(p)) > 2" query << " WITH EXTRACT(n IN NODES(p): n.name) AS nodes" query << " RETURN HEAD(nodes) AS source, HEAD(TAIL(TAIL(nodes))) AS destination, COLLECT(nodes) AS paths"betweenness_centrality = { "A" => 0, "B" => 0, "C" => 0, "D" => 0, "E" => 0 }neo.execute_query(query)["data"].map { |row| row[2].map { |path| path[1..-2] } }.each do |potential_central_nodes| number_of_central_nodes = potential_central_nodes.sizepotential_central_nodes.each do |nodes|nodes.each { |node| betweenness_centrality[node] += (1.0 / number_of_central_nodes) }end endp betweenness_centrality

輸出以下內(nèi)容：

$ bundle exec ruby centrality.rb {"A"=>0, "B"=>1.0, "C"=>0.5, "D"=>0, "E"=>1.5}

它似乎可以完成任務(wù)，但是我敢肯定，在某些情況下，它無(wú)法處理成熟的庫(kù)需要處理的問(wèn)題。 作為一個(gè)實(shí)驗(yàn)，看看有什么可能，我認(rèn)為還算不錯(cuò)！

該圖位于neo4j控制臺(tái)上 ，以防有人感興趣。

參考： 圖處理：betweens中心性–來(lái)自Mark Needham博客博客上我們JCG合作伙伴 Mark Needham的neo4j密碼與圖流 。

翻譯自: https://www.javacodegeeks.com/2013/08/graph-processing-betweeness-centrality-neo4js-cypher-vs-graphstream.html

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