我知道有一些相关的问题,但我想知道是否有更好的解决方案来强制graphviz保持节点位置的定义顺序。
这就是我的问题:我有两个子图,每个子图有5个节点。每个节点都连接到另一个图的每个节点。我希望节点按1-1到1-5的顺序排列。但graphviz总是把它们混为一谈。
下面是图表,我使用的是DOT:
digraph G {
rankdir=LR;
ranksep=4.0;
subgraph cluster_1 {
rank=same;
label="Nr:1";
"1-1" [width=1, shape=circle, style=filled, fillcolor="#E3A869", label="1-1"];
"1-2" [width=1, shape=circle, style=filled, fillcolor="#E3A869", label="1-2"];
"1-3" [width=1, shape=circle, style=filled, fillcolor="#E3A869", label="1-3"];
"1-4" [width=1, shape=circle, style=filled, fillcolor="#E3A869", label="1-4"];
"1-5" [width=1, shape=circle, style=filled, fillcolor="#E3A869", label="1-5"];
}
subgraph cluster_2 {
rank=same;
label="Nr:2";
"2-1" [width=1, shape=circle, style=filled, fillcolor="#E3A869", label="2-1"];
"1-1" -> "2-1" [color=blue, labelfontcolor="#009933", fontsize="10.0", penwidth=1];
"1-2" -> "2-1" [color=blue, labelfontcolor="#009933", fontsize="10.0", penwidth=1];
"1-3" -> "2-1" [color=blue, labelfontcolor="#009933", fontsize="10.0", penwidth=1];
"1-4" -> "2-1" [color=blue, labelfontcolor="#009933", fontsize="10.0", penwidth=1];
"1-5" -> "2-1" [color=blue, labelfontcolor="#009933", fontsize="10.0", penwidth=1];
"2-2" [width=1, shape=circle, style=filled, fillcolor="#E3A869", label="2-2"];
"1-1" -> "2-2" [color=blue, labelfontcolor="#009933", fontsize="10.0", penwidth=1];
"1-2" -> "2-2" [color=blue, labelfontcolor="#009933", fontsize="10.0", penwidth=1];
"1-3" -> "2-2" [color=blue, labelfontcolor="#009933", fontsize="10.0", penwidth=1];
"1-4" -> "2-2" [color=blue, labelfontcolor="#009933", fontsize="10.0", penwidth=1];
"1-5" -> "2-2" [color=blue, labelfontcolor="#009933", fontsize="10.0", penwidth=1];
"2-3" [width=1, shape=circle, style=filled, fillcolor="#E3A869", label="2-3"];
"1-1" -> "2-3" [color=blue, labelfontcolor="#009933", fontsize="10.0", penwidth=1];
"1-2" -> "2-3" [color=blue, labelfontcolor="#009933", fontsize="10.0", penwidth=1];
"1-3" -> "2-3" [color=blue, labelfontcolor="#009933", fontsize="10.0", penwidth=1];
"1-4" -> "2-3" [color=blue, labelfontcolor="#009933", fontsize="10.0", penwidth=1];
"1-5" -> "2-3" [color=blue, labelfontcolor="#009933", fontsize="10.0", penwidth=1];
"2-4" [width=1, shape=circle, style=filled, fillcolor="#E3A869", label="2-4"];
"1-1" -> "2-4" [color=blue, labelfontcolor="#009933", fontsize="10.0", penwidth=1];
"1-2" -> "2-4" [color=blue, labelfontcolor="#009933", fontsize="10.0", penwidth=1];
"1-3" -> "2-4" [color=blue, labelfontcolor="#009933", fontsize="10.0", penwidth=1];
"1-4" -> "2-4" [color=blue, labelfontcolor="#009933", fontsize="10.0", penwidth=1];
"1-5" -> "2-4" [color=blue, labelfontcolor="#009933", fontsize="10.0", penwidth=1];
"2-5" [width=1, shape=circle, style=filled, fillcolor="#E3A869", label="2-5"];
"1-1" -> "2-5" [color=blue, labelfontcolor="#009933", fontsize="10.0", penwidth=1];
"1-2" -> "2-5" [color=blue, labelfontcolor="#009933", fontsize="10.0", penwidth=1];
"1-3" -> "2-5" [color=blue, labelfontcolor="#009933", fontsize="10.0", penwidth=1];
"1-4" -> "2-5" [color=blue, labelfontcolor="#009933", fontsize="10.0", penwidth=1];
"1-5" -> "2-5" [color=blue, labelfontcolor="#009933", fontsize="10.0", penwidth=1];
}

发布于 2013-04-03 03:03:46
使用一些不可见边对子图中的节点进行排序,避免LR等级方向及其怪癖,使用不影响可见蓝色节点连接的节点排名(constraint=false)的最大直边边( straight ,splines=false),应用默认样式<代码>E29以最大限度减少重复和重新排序图形标记的获得以下结果:
digraph G {
nodesep=4.0;
splines=false;
node[width=1, shape=circle, style=filled, fillcolor="#E3A869"];
edge[style=invis];
subgraph cluster_1 {
label="Nr:1";
"1-1" -> "1-2" -> "1-3" -> "1-4" -> "1-5";
}
subgraph cluster_2 {
label="Nr:2";
"2-1" -> "2-2" -> "2-3" -> "2-4" -> "2-5";
}
edge[style=solid, color=blue, penwidth=1, constraint=false];
"1-1" -> "2-1";
"1-2" -> "2-1";
"1-3" -> "2-1";
"1-4" -> "2-1";
"1-5" -> "2-1";
"1-1" -> "2-2";
"1-2" -> "2-2";
"1-3" -> "2-2";
"1-4" -> "2-2";
"1-5" -> "2-2";
"1-1" -> "2-3";
"1-2" -> "2-3";
"1-3" -> "2-3";
"1-4" -> "2-3";
"1-5" -> "2-3";
"1-1" -> "2-4";
"1-2" -> "2-4";
"1-3" -> "2-4";
"1-4" -> "2-4";
"1-5" -> "2-4";
"1-1" -> "2-5";
"1-2" -> "2-5";
"1-3" -> "2-5";
"1-4" -> "2-5";
"1-5" -> "2-5";
}

https://stackoverflow.com/questions/15762014
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