Question

如何从具有命名顶点的边创建Mathematica图？ EG：

http://pastebin.com/Se1Nhe3M

我已经尝试了上述和几种变体，但Combinatorica从未非常接受它。显然，Graph []想要坐标我希望Combinatorica能够找到自己的位置。

AddVertex到EmptyGraph [0]（或其他）也失败了。

GraphUtilities不是一个选项，因为我想做相当复杂的事情我的图表分析。

这似乎是一个简单的问题。 Graphviz可以轻松创建图表有命名顶点的边缘，所以我确定Mathematica也可以吗？

我读过：

ShowGraph[ { {e1,e2}, {e1, e3} }, {e1,e2,e3} ]; // what is the problem here?

但它似乎对我的具体案例没有帮助。

Answer 1

如果您有Mathematica 7，请尝试使用内置GraphPlot：

GraphPlot[{"Conga" -> "Egypt", "Egypt" -> "Conga", 
  "Conga" -> "Sarah Desert", "Sarah Desert" -> "Conga", 
  "Egypt" -> "Europe", "Europe" -> "Egypt", "Egypt" -> "Arabia", 
  "Arabia" -> "Egypt", "Egypt" -> "Sarah Desert", 
  "Sarah Desert" -> "Egypt", "UK" -> "Europe", "Europe" -> "UK", 
  "UK" -> "Iceland", "Iceland" -> "UK", "UK" -> "Greenland", 
  "Greenland" -> "UK", "Europe" -> "Arabia", "Arabia" -> "Europe", 
  "Europe" -> "Germany", "Germany" -> "Europe", "Europe" -> "Iceland",
   "Iceland" -> "Europe", "Europe" -> "Sarah Desert", 
  "Sarah Desert" -> "Europe", "Germany" -> "Russia", 
  "Russia" -> "Germany", "Germany" -> "Arabia", "Arabia" -> "Germany",
   "Germany" -> "Iceland", "Iceland" -> "Germany", 
  "Germany" -> "Irakistan", "Irakistan" -> "Germany", 
  "Austr(al)ia" -> "China", "China" -> "Austr(al)ia", 
  "Arabia" -> "Irakistan", "Irakistan" -> "Arabia", 
  "Canada" -> "More Russia", "More Russia" -> "Canada", 
  "Canada" -> "USA", "USA" -> "Canada", 
  "Canada" -> "Andy's Mountains", "Andy's Mountains" -> "Canada", 
  "More Russia" -> "Russia", "Russia" -> "More Russia", 
  "More Russia" -> "China", "China" -> "More Russia", 
  "More Russia" -> "Irakistan", "Irakistan" -> "More Russia", 
  "China" -> "Irakistan", "Irakistan" -> "China", 
  "USA" -> "Greenland", "Greenland" -> "USA", 
  "USA" -> "Andy's Mountains", "Andy's Mountains" -> "USA", 
  "Brazil" -> "Sarah Desert", "Sarah Desert" -> "Brazil", 
  "Brazil" -> "Andy's Mountains", "Andy's Mountains" -> "Brazil", 
  "Russia" -> "Irakistan", "Irakistan" -> "Russia"}, 
 DirectedEdges -> True]

这将为您提供以下内容，例如：

alt text http://img638.imageshack.us/img638/7803/plotm.png

布局，顶点和边缘标签和样式等有很多选项。

Answer 2

如果粘滞点是表示为字符串的节点，而重型图分析函数希望它们是整数，则可以考虑将字符串映射为整数，反之亦然：

nodes = DeleteDuplicates@Flatten[graph /. Rule -> List]

{"Conga", "Egypt", "Sarah Desert", "Europe", "Arabia", "UK", "Iceland", 
 "Greenland", "Germany", "Russia", "Irakistan", "Austr(al)ia", "China", "Canada",
 "More Russia", "USA", "Andy's Mountains", "Brazil"}

现在你有了节点列表。接下来是与整数的映射：

each[{i_, s_}, Transpose[{Range@Length@nodes, nodes}],
  numify[s] = i;
  namify[i] = s]

您现在可以轻松地将节点转换为整数或从整数转换：

numify["Europe"]

4

namify[4]

"Europe"

像这样转换整个图：

graph /. s_String -> numify[s]

请注意，each是以下实用程序函数，在此处讨论：ForEach loop in Mathematica

SetAttributes[each, HoldAll];               (* each[pattern, list, body]      *)
each[pat_, lst_, bod_] := ReleaseHold[      (*  converts pattern to body for  *)
  Hold[Cases[Evaluate@lst, pat:>bod];]];    (*   each element of list.        *)

Answer 3

由于你专门询问了Combinatorica，因为我总是犹豫是否开始试图打包一个包的内部细节，所以这可能会对你有所帮助：

使用＆lt;＆lt;＆lt;＆lt;＆lt;＆lt;＆lt;或如果你的版本需要这个需要。然后使用您的数据：

edges = {{“Conga” - ＆gt; “埃及”}，{“埃及” - ＆gt; “Conga”}，{“Conga” - ＆gt; “莎拉沙漠”}， {“莎拉沙漠” - ＆gt; “Conga”}，{“埃及” - ＆gt; “Europe”}，{“Europe” - ＆gt; “埃及”}， {“埃及” - ＆gt; “Arabia”}，{“Arabia” - ＆gt; “埃及”}，{“埃及” - ＆gt; “莎拉沙漠”}， {“莎拉沙漠” - ＆gt; “Egypt”}，{“UK” - ＆gt; “Europe”}，{“Europe” - ＆gt; “英国”}， {“UK” - ＆gt; “Iceland”}，{“Iceland” - ＆gt; “UK”}，{“UK” - ＆gt; “格陵兰”}， {“格陵兰岛” - ＆gt; “UK”}，{“Europe” - ＆gt; “Arabia”}，{“Arabia” - ＆gt; “欧洲”}， {“欧洲” - ＆gt; “德国”}，{“德国” - ＆gt; “Europe”}，{“Europe” - ＆gt; “冰岛”}， {“冰岛” - ＆gt; “Europe”}，{“Europe” - ＆gt; “莎拉沙漠”}， {“莎拉沙漠” - ＆gt; “Europe”}，{“Germany” - ＆gt; “俄罗斯”}，{“俄罗斯” - ＆gt; “德国”}， {“德国” - ＆gt; “Arabia”}，{“Arabia” - ＆gt; “德国”}，{“德国” - ＆gt; “冰岛”}， {“冰岛” - ＆gt; “德国”}，{“德国” - ＆gt; “伊拉克斯坦”}，{“伊拉克斯坦” - ＆gt; “德国”}， {“Austr（al）ia” - ＆gt; “中国”}，{“中国” - ＆gt; “Austr（al）ia”}，{“Arabia” - ＆gt; “Irakistan”}， {“伊拉克斯坦” - ＆gt; “Arabia”}，{“Canada” - ＆gt; “更多俄罗斯”}，{“更多俄罗斯” - ＆gt; “加拿大”}， {“加拿大” - ＆gt; “USA”}，{“USA” - ＆gt; “Canada”}，{“Canada” - ＆gt; “安迪的山脉”}， {“Andy's Mountains” - ＆gt; “Canada”}，{“More Russia” - ＆gt; “俄国”}， {“俄罗斯” - ＆gt; “更多俄罗斯”}，{“更多俄罗斯” - ＆gt; “中国”}，{“中国” - ＆gt; “更多俄罗斯”}， {“更多俄罗斯” - ＆gt; “伊拉克斯坦”}，{“伊拉克斯坦” - ＆gt; “更多俄罗斯”}， {“中国” - ＆gt; “伊拉克斯坦”}，{“伊拉克斯坦” - ＆gt; “China”}，{“USA” - ＆gt; “格陵兰”}， {“格陵兰岛” - ＆gt; “USA”}，{“USA” - ＆gt; “安迪的山脉”}， {“Andy's Mountains” - ＆gt; “USA”}，{“Brazil” - ＆gt; “莎拉沙漠”}， {“莎拉沙漠” - ＆gt; “Brazil”}，{“Brazil” - ＆gt; “安迪的山脉”}， {“Andy's Mountains” - ＆gt; “Brazil”}，{“Russia” - ＆gt; “Irakistan”}， {“伊拉克斯坦” - ＆gt; “俄罗斯”}} /规则[from_，对_] - ＆GT; {从，到};

labels = {“Canada”，“USA”，“Greenland”，“Brazil”，“Andy's Mountains”，“UK”，“Iceland”，“Germany”，“Europe”，“Russia”，“More Russia “，”伊拉克斯坦“，”阿拉伯“，”中国“，”澳大利亚“，”埃及“，”莎拉沙漠“，”康加“};

numberededges =分区[拼合[边缘/ .Thread [规则[标签，范围[长度[标签]]]]]，2];

ShowGraph [AddEdges [EmptyGraph [长度[标签]，numberededges] VertexLabel-＆GT;标签，＆PlotRange- GT;全部]

现在，如果自动格式化没有破坏这一点，那么我想你可能已经准备好了。

我没有将边缘指向此，我想你可能想要这样做。希望这足以让你开始。

所有这些都是基于Pemmaraju和Skiena在“Computational Discrete Mathematics：Combinatorics and Graph Theory with Mathematica”中来回寻呼几分钟。我相信任何试图使用Combinatorica但没有在他们面前的人只是坚果。我希望我们可以说服他们推出一个新版本，修复一些拼写错误，并让他们更容易让那些不了解Combinatorica的人能够使用它来开始。

Answer 4

您可以将边缘转换为邻接矩阵，然后使用Combinatorica的FromAdjacencyMatrix函数。

(* Converts lists of edges into adjacency matrix, saving "nodename->column #" mapping into global variable nodeMap *)
graph2mat[edges_] := Module[{nodes, mat, n},
   nodes = Sequence @@@ edges // Union // Sort;
   nodeMap = (# -> (Position[nodes, #] // Flatten // First)) & /@ 
     nodes;
   mat = (({#1, #2} -> 1) & @@@ (edges /. nodeMap)) // SparseArray // 
     Normal;
   n = Max[Length[#] & /@ {mat, Transpose[mat]}];
   PadRight[mat, {n, n}]
   ];

g = FromAdjacencyMatrix[graph2mat[{"a" -> "b", "b" -> "a"}]];
reverseNodeMap = Reverse /@ nodeMap;
MaximumClique[g] /. reverseNodeMap

Answer 5

Combinatorica Graph []对象不易处理。你试过www.sagenb.org吗？ Sage的Graph对象非常酷。

使用具有命名顶点的边创建Mathematica / Combinatorica图

5 个答案: