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one-line tree in python

One-line Tree in Python

Using Python's built-in defaultdict we can easily define a tree data structure:

def tree(): return defaultdict(tree)

That's it!

It simply says that a tree is a dict whose default values are trees.

(If you're following along at home, make sure to from collections import defaultdict)

(Also: Hacker News reader @zbuc points out that this is called autovivification. Cool!)



Now we can create JSON-esque nested dictionaries without explicitly creating sub-dictionaries—they magically come into existence as we reference them:

users = tree()
users['harold']['username'] = 'hrldcpr'
users['handler']['username'] = 'matthandlersux'

We can print this as json with print(json.dumps(users)) and we get the expected:

{"harold": {"username": "hrldcpr"}, "handler": {"username": "matthandlersux"}}

Without assignment

We can even create structure with no assignment at all, since merely referencing an entry creates it:

taxonomy = tree()
taxonomy['Plantae']['Solanales']['Convolvulaceae']['Ipomoea']['sweet potato']

We'll prettyprint this time, which requires us to convert to standard dicts first:

def dicts(t): return {k: dicts(t[k]) for k in t}

Now we can prettyprint the structure with pprint(dicts(taxonomy)):

{'Animalia': {'Chordata': {'Mammalia': {'Carnivora': {'Canidae': {'Canis': {'coyote': {},
                                                                            'dog': {}}},
                                                      'Felidae': {'Felis': {'cat': {}},
                                                                  'Panthera': {'lion': {}}}}}}},
 'Plantae': {'Solanales': {'Convolvulaceae': {'Ipomoea': {'sweet potato': {}}},
                           'Solanaceae': {'Solanum': {'potato': {},
                                                      'tomato': {}}}}}}

So the substructures we referenced now exist as dicts, with empty dicts at the leaves.


This tree can be fun to iteratively walk through, again because structure comes into being simply by referring to it.

For example, suppose we are parsing a list of new animals to add to our taxonomy above, so we want to call a function like:

    'Animalia>Chordata>Mammalia>Cetacea>Balaenopteridae>Balaenoptera>blue whale'.split('>'))

We can implement this simply as:

def add(t, path):
  for node in path:
    t = t[node]

Again we are never assigning to the dictionary, but just by referencing the keys we have created our new structure:

{'Animalia': {'Chordata': {'Mammalia': {'Carnivora': {'Canidae': {'Canis': {'coyote': {},
                                                                            'dog': {}}},
                                                      'Felidae': {'Felis': {'cat': {}},
                                                                  'Panthera': {'lion': {}}}},
                                        'Cetacea': {'Balaenopteridae': {'Balaenoptera': {'blue whale': {}}}}}}},
 'Plantae': {'Solanales': {'Convolvulaceae': {'Ipomoea': {'sweet potato': {}}},
                           'Solanaceae': {'Solanum': {'potato': {},
                                                      'tomato': {}}}}}}


This probably isn't very useful, and it makes for some perplexing code (see add() above).

But if you like Python then I hope this was fun to think about or worthwhile to understand.

There was a good discussion of this gist on Hacker News.

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drasko commented May 18, 2014

In case someone needs it off-the-shelf, code to create JSON representation of the file tree using this technique:

import os
import collections
import json

def tree(): return collections.defaultdict(tree)

def add(t, keys):
  for key in keys:
    t = t[key]

s = tree()

# Set the directory you want to start from
path = 'path/to/directory'

for root, dirs, files in os.walk(path):
    for f in files:
        l = root.split('/')

        add(s, l)

print(json.dumps(s, indent=4, sort_keys=True))

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kzinglzy commented Jun 7, 2014

Wow! It is amazing. Thanks

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amazing. Thanks!

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adugin commented Dec 26, 2014

def tree(levels=0, func=None):
    if levels > 0 and func:
        return defaultdict(reduce(lambda f, i: lambda: defaultdict(f), xrange(levels-1), func))
        return defaultdict(tree)


>>> d = tree()
>>> d[1][2][3][4][5] = 6
>>> d = tree(3, float)
>>> d
defaultdict(<function <lambda> at 0x0280E3F0>, {})
>>> d[1][2][3]
>>> d[7][8][9] += 10
>>> d[7][8][9]
>>> d = tree(0, int)
>>> d
defaultdict(<function tree at 0x0280C830>, {})
>>> d = tree(1, int)
>>> d
defaultdict(<type 'int'>, {})
>>> d[1]

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adugin commented Dec 30, 2014

def tree(func=lambda: tree(), depth=0):
    return defaultdict(reduce(lambda f, i: lambda: defaultdict(f), xrange(depth-1), func))

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Same with ordered tree:

from collections import OrderedDict, defaultdict

def tree():
    Autovivication ordered hash
    return OrderedDict(defaultdict(tree))

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n3h3m commented May 17, 2016

I guess this attracts more appreciation than it deserves. In the reality it will break for a very basic case.

>>> from collections import defaultdict
>>> def tree(): return defaultdict(tree)
>>> t = tree()
>>> t['a']['b']['c'] = 'hello'
>>> t['a']['b']['c']['d'] = 'world'
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: 'str' object does not support item assignment

I believe I don't have to explain why it fails.

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ionFreeman commented Aug 9, 2016

@nehemiahjacob Yes, you can break it by assigning to it. Don't assign to it.

t = tree()

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sshadmand commented Dec 3, 2016

This is great! Question though: How would I programmatically create a dynamic set list of keys into a tree with depth?

For example, I have a variable passed in as a list of terms: ['a', 'b', 'c'], and, based on that list I want to create a tree where each item is the child of the previous item.

Based on your example, the output would be equivalent to the output generated by this assignment:

t = tree()

But again, I want to create that tree dynamically based on the list of keys I am sent.

Any ideas? :)


I found the answer on S.O. here:

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There is a powerful python tree library

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martinym commented Dec 27, 2017

There's a very good answer discussion on stackoverflow about the different ways of doing this and in particular one very simple one that doesn't use defaultdict. It's very portable (Python 2.5+ & 3.x) and another potentially very nice thing about it is the resulting subclass instances print just like a normal dicts.

If you're too lazy to look (TL;DR), here's the code (all of it):

class Vividict(dict):
    def __missing__(self, key):
        value = self[key] = type(self)() # retain local pointer to value
        return value                     # faster to return than dict lookup

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jedie commented May 3, 2018

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dvizzini commented Aug 24, 2022

Added a wrapper:

class TreeBuilder:
    """Returns mutable trees structure, allowing for the following:
    tree_builder = utils.TreeBuilder()
    tree_builder["harold"]["username"] = "hrldcpr"
    tree_builder['handler']['username'] = 'matthandlersux'
    users = tree_builder.to_dict()


    def __init__(self):
        self._builder = self._tree()

    def __getitem__(self, key):
        return self._builder[key]

    def _tree(cls) -> dict:
        return defaultdict(cls._tree)

    def _to_dict(cls, node):
        if isinstance(node, dict):
            return {k: cls._to_dict(node[k]) for k in node}
        return node

    def to_dict(self) -> dict:
        """Returns trees are either values or a builtin python dict."""
        return self._to_dict(self._builder)

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