Tree data as a nested list

One of the nice things about Adjacency Lists as a method of storing tree structures is that there is not much redundancy: you only store a reference to the parent, and that’s it.

It does mean that getting that data in a nested object is a bit complicated. I’ve written before about getting data out of a database: I’ll revisit that again I’m sure, but for now, I’m going to deal with data that has the following shape: that is, has been built up into a Materialized Path:

[
  {
    "node": 1,
    "ancestors": [],
    "label": "Australia"
  },
  {
    "node": 2,
    "ancestors": [1],
    "label": "South Australia"
  },
  {
    "node": 3,
    "ancestors": [1],
    "label": "Victoria"
  },
  {
    "node": 4,
    "ancestors": [1, 2],
    "label": "South-East"
  },
  {
    "node": 5,
    "ancestors": [1, 3],
    "label": "Western Districts"
  },
  {
    "node": 6,
    "ancestors": [],
    "label": "New Zealand"
  },
  {
    "node": 7,
    "ancestors": [1, 2],
    "label": "Barossa Valley"
  },
  {
    "node": 8,
    "ancestors": [1, 2],
    "label": "Riverland"
  }
]

From here, we want to build up something that looks like:

  • Australia
    • South Australia
      • Barossa Valley
      • Riverland
      • South East
    • Victoria
      • Western Districts
  • New Zealand

Or, a nested python data structure:

[
  ('Australia', [
    ('South Australia', [
      ('Barossa Valley', []),
      ('Riverland', []),
      ('South-East', [])
    ]),
    ('Victoria', [
      ('Western Districts', [])
    ])
  ]),
  ('New Zealand', [])
]

You’ll see that each node is a 2-tuple, and each set of siblings is a list. Even a node with no children still gets an empty list.

We can build up this data structure in two steps: based on the fact that a dict, as key-value pairs, matches a 2-tuple. That is, we will start by creating:

{
  1: {
    2: {
      4: {},
      7: {},
      8: {},
    },
    3: {
      5: {},
    }
  },
  6: {},
}

You might be reaching for python’s defaultdict class at this point, but there is a slightly nicer way:

class Tree(dict):
    def __missing__(self, key):
        value = self[key] = type(self)()
        return value

(Note: This class, and the seed of the idea, came from this answer on StackOverflow).

We can also create a recursive method on this class that creates a node and all of it’s ancestors:

    def insert(self, key, ancestors):
        if ancestors:
            self[ancestors[0]].insert(key, ancestors[1:])
        else:
          self[key]
>>> tree = Tree()
>>> for node in data:
...     tree.insert(node['node'], node['ancestors'])
>>> print tree
{1: {2: {8: {}, 4: {}, 7: {}}, 3: {5: {}}}, 6: {}}

Looking good.

Let’s make another method that allows us to actually insert the labels (and apply a sort, if we want):

    def label(self, label_dict, sort_key=lambda x: x[0]):
        return sorted([
          (label_dict.get(key), value.label(label_dict, sort_key))
          for key, value in self.items()
        ], key=sort_key)

We also need to build up the simple key-value store to pass as label_dict, but that’s pretty easy.

Let’s look at the full code: first the complete class:

class Tree(dict):
    """Simple Tree data structure

    Stores data in the form:

    {
        "a": {
            "b": {},
            "c": {},
        },
        "d": {
            "e": {},
        },
    }

    And can be nested to any depth.
    """

    def __missing__(self, key):
        value = self[key] = type(self)()
        return value

    def insert(self, node, ancestors):
        """Insert the supplied node, creating all ancestors as required.

        This expects a list (possibly empty) containing the ancestors,
        and a value for the node.
        """
        if not ancestors:
            self[node]
        else:
            self[ancestors[0]].insert(node, ancestors[1:])

    def label(self, labels, sort_key=lambda x: x[0]):
        """Return a nested 2-tuple with just the supplied labels.

        Realistically, the labels could be any type of object.
        """
        return sorted([
            (
                labels.get(key),
                value.label(labels, sort_key)
            ) for key, value in self.items()
        ], key=sort_key)

Now, using it:

>>> tree = Tree()
>>> labels = {}
>>>
>>> for node in data:
>>>     tree.insert(node['node'], node['ancestors'])
>>>     labels[node['node']] = node['label']
>>>
>>> from pprint import pprint
>>> pprint(tree.label(labels))

[('Australia',
  [('South Australia',
    [('Barossa Valley', []), ('Riverland', []), ('South-East', [])]),
   ('Victoria', [('Western Districts', [])])]),
 ('New Zealand', [])]

Awesome. Now use your template rendering of choice to turn this into a nicely formatted list.

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