48

我正在尝试用 Dot 和 GraphViz 绘制家谱。

这是我目前拥有的：

# 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] ;
}

如果我通过点 ( dot simpsons.dot -Tsvg > simpsons.svg) 运行它，我会得到以下布局： 原创，由 dot/graphviz 制作

但是，我希望边缘更像“家谱”：两个已婚人士之间的 T 形路口，T 的垂直线再次分支成一个倒置的 T 形路口，每个人都有小细分孩子们，就像这个模型一样，在 KolourPaint 中完成：

我想达到什么

我必须使用什么点语法来实现这一点？

4

6 回答 6

21

尝试以下操作：

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]
}

产生：

http://dl.dropbox.com/u/72629/simpsons.png

于 2010-02-16T19:29:34.833 回答
14

我认为您不能采用任意的家谱并自动生成一个点文件，它在 GraphViz 中总是看起来不错。

但我认为，如果您：

  • 使用提到的排名=相同的其他答案来获得 OP 所需的“T”连接
  • 使用 Brian Blank 所做的排序技巧来防止奇怪的线条
  • 假设没有第二次婚姻和同父异母的兄弟姐妹
  • 仅绘制符合以下规则的树的子集：
    • 让 S 成为“中心”人
    • 如果 S 有兄弟姐妹，请确保 S 在所有兄弟姐妹的右边。
    • 如果 S 有配偶并且配偶有兄弟姐妹，请确保配偶在他/她所有兄弟姐妹的左侧。
    • 请勿展示 S 或 S 配偶的侄子、侄女、阿姨或叔叔
    • 不要显示兄弟姐妹的配偶
    • 不要显示配偶兄弟姐妹的配偶
    • 显示 S 的孩子，但不显示他们的配偶或孩子
    • 显示 S 的父母和配偶的父母

这将最终一次显示不超过 3 代，其中 S 在中间代。

在下图中 S=Homer（根据 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;
}

这会通过 GraphViz 生成以下树（带有我在 Power Point 中添加的注释）： 在此处输入图像描述

于 2013-12-29T05:47:48.197 回答
12

Gramps (www.gramps-project.org) 为有或没有婚姻节点的家谱生成点文件。还有一种方法可以在 Gramps 界面本身中看到这一点。http://gramps-project.org/wiki/index.php?title=Graph_View 所以我想说，看一下 Gramps 创建的家谱的输出

于 2010-03-08T08:24:27.083 回答
6

尽管您无法控制节点放置，但我发现您可以通过以不同顺序对节点进行排序来帮助放置节点。我重新排序了一些节点，如下所示，并得到了一个没有产生交叉的图表。

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：

家谱布局

我不完全理解它为什么起作用，但这是我所做更改的思考过程。

  1. 我把克兰西/杰克琳放在亚伯拉罕/莫娜之前，认为他们站在错误的一边。这改变了画面，但仍然不完美。
  2. 我首先放置 Homer/Marge，认为软件必须首先考虑这些部分，并且可能会放置与 Homer/Marge 相关的所有其他节点。这进一步提供了帮助，但仍然不完美。
  3. Herb 仍然放错了位置，所以我把 Herb 放在第一位，这样 graphviz 可能会考虑先放置 Herb。

它有效，但我仍然无法设计一种算法来确保没有重叠边缘的一致树。我觉得如果没有这些提示，graphviz 应该做得更好。我不知道使用的算法，但如果他们考虑使用目标函数来最小化或消除重叠边缘，那么应该可以设计出更好的算法。

于 2012-03-04T03:06:59.187 回答
3

在 graphviz 中做到这一点相当简单。您需要几种语法模式：（i）表示线对线连接的语法（上图中的“T”连接）；(ii) 强制执行层次结构的语法（即，在垂直轴上的同一平面上的同一代节点）。更容易展示：

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];

}

上面的代码将生成下面的图表（我省略了我用来为节点着色的代码）。我在左侧（g1->g2 ....）留下了可见的“指南”，只是为了向您展示我如何在同等等级的节点之间强制执行位置，您可能希望使其在您自己的情节中不可见。最后，标签以“ph”开头的节点是“T-junctions”的占位符节点。

替代文字

于 2010-02-16T11:20:22.277 回答
2

我快到了，灵感来自graphviz-interest 邮件列表和doug 的回答上的一个旧回复。

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：

替代文字

所以，看起来不错，除了荷马周围的奇怪边缘。如果我能找到一种方法将亚伯拉罕、莫娜和赫伯移动到图片的左侧，那么我将有一个完美对齐的图片。

关于如何实现这一目标的任何想法？

于 2010-02-16T11:29:36.087 回答