Rendering a tree in python using anytree and graphviz, without merging common nodes

Question

I am creating a tree from a list ["abc", "abd", "aec", "add", "adcf"] using anytree package of python3. In this tree first character of each list element - a is a root, and subsequently, other characters are added as their children. When I render tree it looks like:

a
├── b
│   ├── c
│   └── d
├── e
│   └── c
└── d
    ├── d
    └── c
        └── f

But when I render the tree to picture using to_picture method, the image is -

I don't want the common nodes to be merged, as it is adding unwanted paths to my tree.

You want something like this? https://i.stack.imgur.com/uRwYG.png — Tarun Lalwani, Mar 22 '18 at 07:28
@Rojan.. Thanks for the concern. You may refer to the answer of Tarun Lalwani for the code. — Himadri, Mar 24 '18 at 10:21

score 11 · Accepted Answer · answered Mar 23 '18 at 05:33

The node are arranged in graphviz using their id. In your case the graph is generated with just node names and then graphviz created the loop edge as you get it.

What you really want is different id for each node and a label associated with the same. The DotExporter class has a attribute named nodeattrfunc to which we can pass a function or a lambda and generate attributes for the nodes

Below is how you would do it based on your array

import anytree
from anytree import Node, RenderTree

data = ["abc", "abd", "aec", "add", "adcf"]
from anytree.exporter import DotExporter

nodes = {}
first_node = None

for elem in data:
    parent_node = None
    parent_node_name = ""
    for i, val in enumerate(elem):
        if i not in nodes:
            nodes[i] = {}
        key = parent_node_name + val
        if key not in nodes[i]:
            print("Creating new node for ", key)
            nodes[i][key] = Node(key, parent=parent_node, display_name=val)

        if first_node is None:
            first_node = nodes[i][key]
        parent_node = nodes[i][key]
        parent_node_name = val

print(nodes)
DotExporter(nodes[0]["a"],
            nodeattrfunc=lambda node: 'label="{}"'.format(node.display_name)).to_dotfile("graph.txt")
DotExporter(nodes[0]["a"],
            nodeattrfunc=lambda node: 'label="{}"'.format(node.display_name)).to_picture("graph.png")

This will generate the below dot file

digraph tree {
    "a" [label="a"];
    "ab" [label="b"];
    "bc" [label="c"];
    "bd" [label="d"];
    "ae" [label="e"];
    "ec" [label="c"];
    "ad" [label="d"];
    "dd" [label="d"];
    "dc" [label="c"];
    "cf" [label="f"];
    "a" -> "ab";
    "a" -> "ae";
    "a" -> "ad";
    "ab" -> "bc";
    "ab" -> "bd";
    "ae" -> "ec";
    "ad" -> "dd";
    "ad" -> "dc";
    "dc" -> "cf";
}

And the below graph

Hey Tarun, great answer. I'm having trouble with generating the picture, can you take a look at https://stackoverflow.com/questions/52493775/rendering-a-tree-in-python-with-anytree-and-graphviz-cant-open-file. Thx in advance. — luQ, Oct 11 '18 at 08:21

score 1 · Answer 2 · answered Mar 22 '18 at 16:42

1

As anytree to_picture() function use graphviz to generate the file.

From here = similar question, you can read :

graphviz uses the node id as label. If distinct nodes need to have the same label, the label has to be defined explicitly.

answered Mar 22 '18 at 16:42

A. STEFANI

6,707
1
23
48

score 1 · Answer 3 · answered May 03 '22 at 03:14

1

I was about to implement the recommended fix but found that using UniqueDotExporter dealt with the repeated ids perfectly!

answered May 03 '22 at 03:14

notsamdonald

11
1

Rendering a tree in python using anytree and graphviz, without merging common nodes

3 Answers3

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