家族树布局与dot/graphviz

Question

我正在尝试用dot和graphviz绘制一棵家谱。

这就是我目前拥有的：

# just graph set-up
digraph simpsons {
ratio = "auto"
mincross = 2.0

# draw some nodes
"Abraham"   [shape=box, regular=1, color="blue"] ;
"Mona"      [shape=box, regular=1, color="pink"] ;
"Clancy"    [shape=box, regular=1, color="blue"] ;
"Jackeline" [shape=box, regular=1, color="pink"] ;
"Herb"      [shape=box, regular=1, color="blue"] ;
"Homer"     [shape=box, regular=1, color="blue"] ;
"Marge"     [shape=box, regular=1, color="pink"] ;
"Patty"     [shape=box, regular=1, color="pink"] ;
"Selma"     [shape=box, regular=1, color="pink"] ;
"Bart"      [shape=box, regular=1, color="blue"] ;
"Lisa"      [shape=box, regular=1, color="pink"] ;
"Maggie"    [shape=box, regular=1, color="pink"] ;
"Ling"      [shape=box, regular=1, color="blue"] ;
# creating tiny nodes w/ no label, no color
"ParentsHomer" [shape=diamond,style=filled,label="",height=.1,width=.1] ;
"ParentsMarge" [shape=diamond,style=filled,label="",height=.1,width=.1] ;
"ParentsBart"  [shape=diamond,style=filled,label="",height=.1,width=.1] ;

# draw the edges
"Abraham"      -> "ParentsHomer" [dir=none, weight=1] ;
"Mona"         -> "ParentsHomer" [dir=none, weight=1] ;
"ParentsHomer" -> "Homer"        [dir=none, weight=2] ;
"ParentsHomer" -> "Herb"         [dir=none, weight=2] ;
"Clancy"       -> "ParentsMarge" [dir=none, weight=1] ;
"Jackeline"    -> "ParentsMarge" [dir=none, weight=1] ;
"ParentsMarge" -> "Marge"        [dir=none, weight=2] ;
"ParentsMarge" -> "Patty"        [dir=none, weight=2] ;
"ParentsMarge" -> "Selma"        [dir=none, weight=2] ;
"Homer"        -> "ParentsBart"  [dir=none, weight=1] ;
"Marge"        -> "ParentsBart"  [dir=none, weight=1] ;
"ParentsBart"  -> "Bart"         [dir=none, weight=2] ;
"ParentsBart"  -> "Lisa"         [dir=none, weight=2] ;
"ParentsBart"  -> "Maggie"       [dir=none, weight=2] ;
"Selma"        -> "Ling"         [dir=none, weight=2] ;
}

如果我通过dot运行它（dot simpsons.dot -Tsvg > simpsons.svg），我得到以下布局：

但是，我希望边缘更像“家谱” - 类似：两个已婚者之间的T型结合，t的垂直线再次分支在一个颠倒的T型结中，每个t型结合，每个分区都有小细分像这个模型一样，孩子在Kolourpaint上完成：

我必须使用的DOT语法是什么？

Answer 1

这是另一个解决方案：

digraph simpsons {  
  subgraph Generation0 {
    rank = same
    Abraham [shape = box, color = blue]
    Mona [shape = box, color = pink]
    AbrahamAndMona [shape = point]
    Abraham -> AbrahamAndMona [dir = none]
    AbrahamAndMona -> Mona [dir = none]

    Clancy [shape = box, color = blue]
    Jackeline [shape = box, color = pink]
    ClancyAndJackeline [shape = point]
    Clancy -> ClancyAndJackeline [dir = none]
    ClancyAndJackeline -> Jackeline [dir = none]
  }

  subgraph Generation0Sons {
    rank = same
    AbrahamAndMonaSons [shape = point]
    HerbSon [shape = point]
    HomerSon [shape = point]
    HerbSon -> AbrahamAndMonaSons [dir = none]
    HomerSon -> AbrahamAndMonaSons [dir = none]

    MargeSon [shape = point]
    PattySon [shape = point]
    SelmaSon [shape = point]
    MargeSon -> PattySon [dir = none] 
    PattySon -> SelmaSon [dir = none] 
  }

  AbrahamAndMona -> AbrahamAndMonaSons [dir = none]
  ClancyAndJackeline -> PattySon [dir = none]

  subgraph Generation1 {
    rank  =  same
    Herb [shape = box, color = blue] 
    Homer [shape = box, color = blue] 
    Marge [shape = box, color = pink] 
    Patty [shape = box, color = pink] 
    Selma [shape = box, color = pink] 

    HomerAndMarge [shape = point]
    Homer -> HomerAndMarge [dir = none]
    Marge -> HomerAndMarge [dir = none]
  }

  HerbSon -> Herb [dir = none]
  HomerSon -> Homer [dir = none]
  MargeSon -> Marge [dir = none]
  PattySon -> Patty [dir = none]
  SelmaSon -> Selma [dir = none]

  subgraph Generation1Sons {
    rank  =  same
    BartSon [shape = point] 
    LisaSon [shape = point] 
    MaggieSon [shape = point] 

    BartSon -> LisaSon [dir = none]
    LisaSon -> MaggieSon [dir = none]
  }

  HomerAndMarge -> LisaSon [dir = none]

  subgraph Generation2 {
    rank  =  same
    Bart [shape = box, color = blue] 
    Lisa [shape = box, color = pink] 
    Maggie [shape = box, color = pink] 
    Ling [shape = box, color = blue] 
  }

  Selma -> Ling [dir = none]
  BartSon -> Bart [dir = none]
  LisaSon -> Lisa [dir = none]
  MaggieSon -> Maggie [dir = none]
}

结果：

Answer 2

gramps（www.gramps-project.org）生成家庭树的点文件，无论有没有结婚节点。也有一种方法可以在Gramps接口本身中看到这一点。 http://gramps-project.org/wiki/index.php?title=graph_view所以我要说的是，查看由Gramps创建的家谱的输出

Answer 3

我认为您不能拿一棵任意的家谱并自动生成一个点文件，在GraphViz中总是看起来不错。

但我想你 能够 如果您的话，请始终使它看起来不错：

• 使用等级=相同的其他答案，以获取OP所需的“ T”连接
• 使用订购技巧布莱恩·布兰克（Brian Blank）
• 没有第二次婚姻和半兄弟姐妹
• 仅绘制遵守以下规则的树的一个子集：
  • 让S成为“中心”人
  • 如果S有兄弟姐妹，请确保所有人都正确。
  • 如果S有配偶，并且配偶有兄弟姐妹，请确保配偶在所有兄弟姐妹的左边。
  • 不要展示S或S配偶的侄子，侄女，阿姨或叔叔
  • 不要表现出兄弟姐妹的配偶
  • 不要表现出配偶的兄弟姐妹的配偶
  • 展示S的孩子，但没有给他们的配偶或孩子
  • 展示S的父母和配偶的父母

最终将一次显示不超过3代，而S则在中期。

在下图中S =荷马（Brian Blank的版本稍作修改）：

digraph G {
  edge [dir=none];
  node [shape=box];
  graph [splines=ortho];

  "Herb"      [shape=box, regular=0, color="blue", style="filled" fillcolor="lightblue"] ;
  "Homer"     [shape=box, regular=0, color="blue", style="bold, filled" fillcolor="lightblue"] ;
  "Marge"     [shape=oval, regular=0, color="red", style="filled" fillcolor="pink"] ;
  "Clancy"    [shape=box, regular=0, color="blue", style="filled" fillcolor="lightblue"] ;
  "Jackeline" [shape=oval, regular=0, color="red", style="filled" fillcolor="pink"] ;
  "Abraham"   [shape=box, regular=0, color="blue", style="filled" fillcolor="lightblue"] ;
  "Mona"      [shape=oval, regular=0, color="red", style="filled" fillcolor="pink"] ;
  "Patty"     [shape=oval, regular=0, color="red", style="filled" fillcolor="pink"] ;
  "Selma"     [shape=oval, regular=0, color="red", style="filled" fillcolor="pink"] ;
  "Bart"      [shape=box, regular=0, color="blue", style="filled" fillcolor="lightblue"] ;
  "Lisa"      [shape=oval, regular=0, color="red", style="filled" fillcolor="pink"] ;
  "Maggie"    [shape=oval, regular=0, color="red", style="filled" fillcolor="pink"] ;

  a1 [shape=diamond,label="",height=0.25,width=0.25];
  b1 [shape=circle,label="",height=0.01,width=0.01];
  b2 [shape=circle,label="",height=0.01,width=0.01];
  b3 [shape=circle,label="",height=0.01,width=0.01];
  {rank=same; Abraham -> a1 -> Mona};
  {rank=same; b1 -> b2 -> b3};
  {rank=same; Herb; Homer};
  a1 -> b2
  b1 -> Herb
  b3 -> Homer

  p1 [shape=diamond,label="",height=0.25,width=0.25];
  q1 [shape=circle,label="",height=0.01,width=0.01];
  q2 [shape=circle,label="",height=0.01,width=0.01];
  q3 [shape=circle,label="",height=0.01,width=0.01];
  {rank=same; Homer -> p1 -> Marge};
  {rank=same; q1 -> q2 -> q3};
  {rank=same; Bart; Lisa; Maggie};
  p1 -> q2;
  q1 -> Bart;
  q2 -> Lisa;
  q3 -> Maggie;

  x1 [shape=diamond,label="",height=0.25,width=0.25];
  y1 [shape=circle,label="",height=0.01,width=0.01];
  y2 [shape=circle,label="",height=0.01,width=0.01];
  y3 [shape=circle,label="",height=0.01,width=0.01];
  {rank=same; Clancy -> x1 -> Jackeline};
  {rank=same; y1 -> y2 -> y3};
  {rank=same; Patty; Selma; Marge};
  x1 -> y2;
  y1 -> Marge;
  y2 -> Patty;
  y3 -> Selma;
}

这将通过GraphViz产生以下树（带有电源点的注释，请注释）：

Answer 4

尽管您无法控制节点放置，但我发现您可以通过以不同的顺序订购节点来帮助节点放置。我重新排序了一些节点，如下所示，并得到了没有产生交叉跨度的图。

以下代码：

digraph G {
  edge [dir=none];
  node [shape=box];

  "Herb"      [shape=box, regular=1, color="blue"] ;
  "Homer"     [shape=box, regular=1, color="blue"] ;
  "Marge"     [shape=box, regular=1, color="pink"] ;
  "Clancy"    [shape=box, regular=1, color="blue"] ;
  "Jackeline" [shape=box, regular=1, color="pink"] ;
  "Abraham"   [shape=box, regular=1, color="blue"] ;
  "Mona"      [shape=box, regular=1, color="pink"] ;
  "Patty"     [shape=box, regular=1, color="pink"] ;
  "Selma"     [shape=box, regular=1, color="pink"] ;
  "Bart"      [shape=box, regular=1, color="blue"] ;
  "Lisa"      [shape=box, regular=1, color="pink"] ;
  "Maggie"    [shape=box, regular=1, color="pink"] ;
  "Ling"      [shape=box, regular=1, color="blue"] ;

  a1 [shape=circle,label="",height=0.01,width=0.01];
  b1 [shape=circle,label="",height=0.01,width=0.01];
  b2 [shape=circle,label="",height=0.01,width=0.01];
  b3 [shape=circle,label="",height=0.01,width=0.01];
  {rank=same; Abraham -> a1 -> Mona};
  {rank=same; b1 -> b2 -> b3};
  {rank=same; Herb; Homer};
  a1 -> b2
  b1 -> Herb
  b3 -> Homer

  p1 [shape=circle,label="",height=0.01,width=0.01];
  q1 [shape=circle,label="",height=0.01,width=0.01];
  q2 [shape=circle,label="",height=0.01,width=0.01];
  q3 [shape=circle,label="",height=0.01,width=0.01];
  {rank=same; Homer -> p1 -> Marge};
  {rank=same; q1 -> q2 -> q3};
  {rank=same; Bart; Lisa; Maggie};
  p1 -> q2;
  q1 -> Bart;
  q2 -> Lisa;
  q3 -> Maggie;

  x1 [shape=circle,label="",height=0.01,width=0.01];
  y1 [shape=circle,label="",height=0.01,width=0.01];
  y2 [shape=circle,label="",height=0.01,width=0.01];
  y3 [shape=circle,label="",height=0.01,width=0.01];
  {rank=same; Clancy -> x1 -> Jackeline};
  {rank=same; y1 -> y2 -> y3};
  {rank=same; Marge; Patty; Selma};
  {rank=same; Bart; Ling}
  x1 -> y2;
  y1 -> Marge;
  y2 -> Patty;
  y3 -> Selma;
  Selma -> Ling;
}

现在产生这个：

我不完全理解为什么它起作用，但这是我所做的更改的思考过程。

1. 我在亚伯拉罕/蒙娜（Abraham/Mona）之前以克兰西/杰克琳（Jackeline）的身份认为他们在错误的一面。这改变了图片，但仍然并不完美。
2. 我让荷马/玛格首先认为该软件必须首先考虑这些作品，也许可以将所有其他节点相对于荷马/Marge放置。这进一步有所帮助，但仍然并不完美。
3. Herb仍然放错了位置，所以我将Herb放在首位，以便Graphviz可以考虑将Herb放置在首先。

它起作用了，但我仍然无法设计一种算法，该算法可以确保没有重叠边缘的树木一致。我觉得没有这些提示，Graphviz应该做得更好。我不知道所使用的算法，但是如果它们考虑一个目标函数以最大程度地减少或消除重叠的边缘，则应该可以设计出更好的算法。

Answer 5

在GraphViz中这样做非常简单。您需要几个语法模式：（i）语法表示线到线连接（上面的图中的“ t” - 接口）； （ii）执行层次结构的语法（即，在垂直轴上同一平面上的同一一代节点）。更容易显示：

digraph G {
    nodesep=0.6;
    edge [arrowsize=0.3];

    "g1" -> "g2" -> "g3" -> "g4"

    { rank = same;
        "g1"; "King"; "ph1"; "Queen";
    };

    { rank = same; 
        "g2"; "ph2"; "ph2L"; "ph2R"; "ph2LL"; "ph2RR"
    };

    { rank = same;
        "g3"; "ps1"; "ps2"; "pr1"; "pr2"
    };

    "King" -> "ph1" [arrowsize=0.0];
    "ph1" -> "Queen" [arrowsize=0.0];

    "ph1" -> "ph2" [arrowsize=0.0];
    "ph2LL" -> "ph2L" [arrowsize=0.0];
    "ph2L" -> "ph2" [arrowsize=0.0];
    "ph2" -> "ph2R" [arrowsize=0.0];
    "ph2R" -> "ph2RR" [arrowsize=0.0];

    "ph2LL" -> "ps1" [arrowsize=0.0];
    "ph2L"-> "pr1" [arrowsize=0.0];
    "ph2R" -> "ps2" [arrowsize=0.0];
    "ph2RR" -> "pr2" [arrowsize=0.0];

}

上面的代码将产生下面的图（我省略了用于为节点上色的代码）。我在左侧（g1-> g2 ....）留下了“指南”，只是为了向您展示我如何在等级相等的节点之间执行位置，您可能希望在自己的图中使其不可见。最后，以“ pH”开头的标签的节点是“ t junctions”的占位符节点。

Answer 6

我快到了，受到启发 GraphViz-Interest MailingList上的旧响应 和 道格的回答.

以下代码：

digraph G {
  edge [dir=none];
  node [shape=box];

  "Abraham"   [shape=box, regular=1, color="blue"] ;
  "Mona"      [shape=box, regular=1, color="pink"] ;
  "Clancy"    [shape=box, regular=1, color="blue"] ;
  "Jackeline" [shape=box, regular=1, color="pink"] ;
  "Herb"      [shape=box, regular=1, color="blue"] ;
  "Homer"     [shape=box, regular=1, color="blue"] ;
  "Marge"     [shape=box, regular=1, color="pink"] ;
  "Patty"     [shape=box, regular=1, color="pink"] ;
  "Selma"     [shape=box, regular=1, color="pink"] ;
  "Bart"      [shape=box, regular=1, color="blue"] ;
  "Lisa"      [shape=box, regular=1, color="pink"] ;
  "Maggie"    [shape=box, regular=1, color="pink"] ;
  "Ling"      [shape=box, regular=1, color="blue"] ;

  a1 [shape=circle,label="",height=0.01,width=0.01];
  b1 [shape=circle,label="",height=0.01,width=0.01];
  b2 [shape=circle,label="",height=0.01,width=0.01];
  b3 [shape=circle,label="",height=0.01,width=0.01];
  {rank=same; Abraham -> a1 -> Mona};
  {rank=same; b1 -> b2 -> b3};
  {rank=same; Herb; Homer};
  a1 -> b2
  b1 -> Herb
  b3 -> Homer

  p1 [shape=circle,label="",height=0.01,width=0.01];
  q1 [shape=circle,label="",height=0.01,width=0.01];
  q2 [shape=circle,label="",height=0.01,width=0.01];
  q3 [shape=circle,label="",height=0.01,width=0.01];
  {rank=same; Homer -> p1 -> Marge};
  {rank=same; q1 -> q2 -> q3};
  {rank=same; Bart; Lisa; Maggie};
  p1 -> q2;
  q1 -> Bart;
  q2 -> Lisa;
  q3 -> Maggie;

  x1 [shape=circle,label="",height=0.01,width=0.01];
  y1 [shape=circle,label="",height=0.01,width=0.01];
  y2 [shape=circle,label="",height=0.01,width=0.01];
  y3 [shape=circle,label="",height=0.01,width=0.01];
  {rank=same; Clancy -> x1 -> Jackeline};
  {rank=same; y1 -> y2 -> y3};
  {rank=same; Marge; Patty; Selma};
  {rank=same; Bart; Ling}
  x1 -> y2;
  y1 -> Marge;
  y2 -> Patty;
  y3 -> Selma;
  Selma -> Ling;
}

现在产生这个：

因此，除了荷马周围的那个奇怪的边缘外，看起来不错。如果我能找到一种将亚伯拉罕，蒙娜和草药转移到图片左侧的方法，那么我将有一个完美的图片。

关于如何实现这一目标的任何想法？