How can I implement a tree in Python?
Asked Answered
M

20

340

How can I implement a general tree in Python? Is there a built-in data structure for this?

Meadowlark answered 1/3, 2010 at 18:24 Comment(3)
geeksforgeeks.org/binarytree-module-in-pythonRena
It's not actually clear what was originally meant by "general" in the question. It could mean "arbitrary number of children at each node", "arbitrary type of data stored in the nodes", or possibly other things.Goingson
Each n-ary tree can be represented by binary tree.Gruchot
D
391

I recommend anytree (I am the author).

Example:

from anytree import Node, RenderTree

udo = Node("Udo")
marc = Node("Marc", parent=udo)
lian = Node("Lian", parent=marc)
dan = Node("Dan", parent=udo)
jet = Node("Jet", parent=dan)
jan = Node("Jan", parent=dan)
joe = Node("Joe", parent=dan)

print(udo)
Node('/Udo')
print(joe)
Node('/Udo/Dan/Joe')

for pre, fill, node in RenderTree(udo):
    print("%s%s" % (pre, node.name))
Udo
├── Marc
│   └── Lian
└── Dan
    ├── Jet
    ├── Jan
    └── Joe

print(dan.children)
(Node('/Udo/Dan/Jet'), Node('/Udo/Dan/Jan'), Node('/Udo/Dan/Joe'))

anytree has also a powerful API with:

  • simple tree creation
  • simple tree modification
  • pre-order tree iteration
  • post-order tree iteration
  • resolve relative and absolute node paths
  • walking from one node to an other.
  • tree rendering (see example above)
  • node attach/detach hookups
Dobby answered 2/9, 2016 at 12:52 Comment(5)
This does not answer the question but is merely advertisement for your own project, yet for some reason is the top result in google. The question was how to implement a tree, not how to use your implementation.Anagnorisis
His open source implementation is an excellent example to learn fromDepartmentalism
Part of the question was "Is there a built-in data structure for this?", so presumably OP was considering to reuse an existing implementation.Gibrian
After Using it seems like it's the best option that i've found out there, Very good!Thrust
@EmanuelP as long as it's free software, I wouldn't call it "advertisement".Holifield
I
156

Python doesn't have the quite the extensive range of "built-in" data structures as Java does. However, because Python is dynamic, a general tree is easy to create. For example, a binary tree might be:

class Tree:
    def __init__(self):
        self.left = None
        self.right = None
        self.data = None

You can use it like this:

root = Tree()
root.data = "root"
root.left = Tree()
root.left.data = "left"
root.right = Tree()
root.right.data = "right"

If you need an arbitrary number of children per node, then use a list of children:

class Tree:
    def __init__(self, data):
        self.children = []
        self.data = data

left = Tree("left")
middle = Tree("middle")
right = Tree("right")
root = Tree("root")
root.children = [left, middle, right]
Interlink answered 1/3, 2010 at 18:29 Comment(8)
This doesn't really explain much about making a useful tree implementation.Celebration
The question is tagged with Python3, there's no need to derive class Tree from object thenAldridge
@Aldridge Deriving from object is sometimes just a guideline: If a class inherits from no other base classes, explicitly inherit from object. This also applies to nested classes. See Google Python Style GuidePotboy
@platzhirsch: Please read and quote the guideline completely: Google explicitly points out that this is required for Python 2 code to work as expected and recommended to improve compatibility with Py3. Here we're talking about Py3 code. There's no need to do extra, legacy typing.Aldridge
That's a binary tree, not a general one as asked for.Deprecatory
@Greg Hewgill Are you able to provide an example where the "binary" is generic, e.g. the size of splits per node is unknown? In form of a List.Judgment
@JanHackenberg I have used this simple one in the past: #3754165Schou
Can you also please tell how can one build a tree with arbitrary number of children?Lingual
W
81

A generic tree is a node with zero or more children, each one a proper (tree) node. It isn't the same as a binary tree, they're different data structures, although both shares some terminology.

There isn't any builtin data structure for generic trees in Python, but it's easily implemented with classes.

class Tree(object):
    "Generic tree node."
    def __init__(self, name='root', children=None):
        self.name = name
        self.children = []
        if children is not None:
            for child in children:
                self.add_child(child)
    def __repr__(self):
        return self.name
    def add_child(self, node):
        assert isinstance(node, Tree)
        self.children.append(node)
#    *
#   /|\
#  1 2 +
#     / \
#    3   4
t = Tree('*', [Tree('1'),
               Tree('2'),
               Tree('+', [Tree('3'),
                          Tree('4')])])
Wenwenceslaus answered 18/1, 2015 at 21:40 Comment(3)
Amazing, this can be easily used as a graph too! The only problem that i saw is: How can I differ the left node from the right node?Besprinkle
By children index. Left will always be children[0] in that case.Fencesitter
and right always be len(self.children)Manson
S
46

You can try:

from collections import defaultdict
def tree(): return defaultdict(tree)
users = tree()
users['harold']['username'] = 'hrldcpr'
users['handler']['username'] = 'matthandlersux'

As suggested here: https://gist.github.com/2012250

Suite answered 25/4, 2012 at 19:39 Comment(1)
if you want to extend to an an arbitrary amount of levels check: https://mcmap.net/q/98072/-how-can-i-implement-a-tree-in-pythonAmara
Y
42
class Node:
    """
    Class Node
    """
    def __init__(self, value):
        self.left = None
        self.data = value
        self.right = None

class Tree:
    """
    Class tree will provide a tree as well as utility functions.
    """

    def createNode(self, data):
        """
        Utility function to create a node.
        """
        return Node(data)

    def insert(self, node , data):
        """
        Insert function will insert a node into tree.
        Duplicate keys are not allowed.
        """
        #if tree is empty , return a root node
        if node is None:
            return self.createNode(data)
        # if data is smaller than parent , insert it into left side
        if data < node.data:
            node.left = self.insert(node.left, data)
        elif data > node.data:
            node.right = self.insert(node.right, data)

        return node


    def search(self, node, data):
        """
        Search function will search a node into tree.
        """
        # if root is None or root is the search data.
        if node is None or node.data == data:
            return node

        if node.data < data:
            return self.search(node.right, data)
        else:
            return self.search(node.left, data)



    def deleteNode(self,node,data):
        """
        Delete function will delete a node into tree.
        Not complete , may need some more scenarion that we can handle
        Now it is handling only leaf.
        """

        # Check if tree is empty.
        if node is None:
            return None

        # searching key into BST.
        if data < node.data:
            node.left = self.deleteNode(node.left, data)
        elif data > node.data:
            node.right = self.deleteNode(node.right, data)
        else: # reach to the node that need to delete from BST.
            if node.left is None and node.right is None:
                del node
            if node.left == None:
                temp = node.right
                del node
                return  temp
            elif node.right == None:
                temp = node.left
                del node
                return temp

        return node

    def traverseInorder(self, root):
        """
        traverse function will print all the node in the tree.
        """
        if root is not None:
            self.traverseInorder(root.left)
            print(root.data)
            self.traverseInorder(root.right)

    def traversePreorder(self, root):
        """
        traverse function will print all the node in the tree.
        """
        if root is not None:
            print(root.data)
            self.traversePreorder(root.left)
            self.traversePreorder(root.right)

    def traversePostorder(self, root):
        """
        traverse function will print all the node in the tree.
        """
        if root is not None:
            self.traversePostorder(root.left)
            self.traversePostorder(root.right)
            print(root.data)


def main():
    root = None
    tree = Tree()
    root = tree.insert(root, 10)
    print(root)
    tree.insert(root, 20)
    tree.insert(root, 30)
    tree.insert(root, 40)
    tree.insert(root, 70)
    tree.insert(root, 60)
    tree.insert(root, 80)

    print("Traverse Inorder")
    tree.traverseInorder(root)

    print("Traverse Preorder")
    tree.traversePreorder(root)

    print("Traverse Postorder")
    tree.traversePostorder(root)


if __name__ == "__main__":
    main()
Yakut answered 9/4, 2015 at 7:7 Comment(2)
Can you add just some notes to introduce your code and your implementation?Secondclass
Thanks for the complete Binary Tree implementation with utility functions. Since it is Python 2, I created a gist for Binary Tree implementation (Py3) for people needing a Python 3 version.Kelby
D
18

There aren't trees built in, but you can easily construct one by subclassing a Node type from List and writing the traversal methods. If you do this, I've found bisect useful.

There are also many implementations on PyPi that you can browse.

If I remember correctly, the Python standard lib doesn't include tree data structures for the same reason that the .NET base class library doesn't: locality of memory is reduced, resulting in more cache misses. On modern processors it's usually faster to just bring a large chunk of memory into the cache, and "pointer rich" data structures negate the benefit.

Disrate answered 1/3, 2010 at 18:28 Comment(2)
FYI: The interwebs are plastered with hatred against Boost. Apparently it's supposed to be a HUGE pain to deal with, especially since support for it has been discontinued. So I would recommend staying away from thatPetrozavodsk
Thanks. I haven't had any trouble personally, but I don't want to mislead so I've removed that reference.Disrate
V
17

I implemented a rooted tree as a dictionary {child:parent}. So for instance with the root node 0, a tree might look like that:

tree={1:0, 2:0, 3:1, 4:2, 5:3}

This structure made it quite easy to go upward along a path from any node to the root, which was relevant for the problem I was working on.

Volitant answered 26/9, 2013 at 23:38 Comment(2)
This is the way I was considering doing it, until I saw the answer. Although since a tree is a parent with two children, and if you want to go down, you can do {parent:[leftchild,rightchild]}.Dissever
Another way is to use lists of lists where the first (or more) element in the list is the node value, and the following nested two lists represent its left and right subtrees (or more for n-ary tree).Phoney
R
10

Greg Hewgill's answer is great but if you need more nodes per level you can use a list|dictionary to create them: And then use method to access them either by name or order (like id)

class node(object):
    def __init__(self):
        self.name=None
        self.node=[]
        self.otherInfo = None
        self.prev=None
    def nex(self,child):
        "Gets a node by number"
        return self.node[child]
    def prev(self):
        return self.prev
    def goto(self,data):
        "Gets the node by name"
        for child in range(0,len(self.node)):
            if(self.node[child].name==data):
                return self.node[child]
    def add(self):
        node1=node()
        self.node.append(node1)
        node1.prev=self
        return node1

Now just create a root and build it up: ex:

tree=node()  #create a node
tree.name="root" #name it root
tree.otherInfo="blue" #or what ever 
tree=tree.add() #add a node to the root
tree.name="node1" #name it

    root
   /
child1

tree=tree.add()
tree.name="grandchild1"

       root
      /
   child1
   /
grandchild1

tree=tree.prev()
tree=tree.add()
tree.name="gchild2"

          root
           /
        child1
        /    \
grandchild1 gchild2

tree=tree.prev()
tree=tree.prev()
tree=tree.add()
tree=tree.name="child2"

              root
             /   \
        child1  child2
       /     \
grandchild1 gchild2


tree=tree.prev()
tree=tree.goto("child1") or tree=tree.nex(0)
tree.name="changed"

              root
              /   \
         changed   child2
        /      \
  grandchild1  gchild2

That should be enough for you to start figuring out how to make this work

Relish answered 4/5, 2014 at 0:13 Comment(1)
There's something is missing in this answer, I was trying this solution for the past 2 days and I think you have some logical flow in the object addition method. I will submit my answer to this question, please check that out and let me know if I can help.Axial
A
9
class Tree(dict):
    """A tree implementation using python's autovivification feature."""
    def __missing__(self, key):
        value = self[key] = type(self)()
        return value

    #cast a (nested) dict to a (nested) Tree class
    def __init__(self, data={}):
        for k, data in data.items():
            if isinstance(data, dict):
                self[k] = type(self)(data)
            else:
                self[k] = data

works as a dictionary, but provides as many nested dicts you want. Try the following:

your_tree = Tree()

your_tree['a']['1']['x']  = '@'
your_tree['a']['1']['y']  = '#'
your_tree['a']['2']['x']  = '$'
your_tree['a']['3']       = '%'
your_tree['b']            = '*'

will deliver a nested dict ... which works as a tree indeed.

{'a': {'1': {'x': '@', 'y': '#'}, '2': {'x': '$'}, '3': '%'}, 'b': '*'}

... If you have already a dict, it will cast each level to a tree:

d = {'foo': {'amy': {'what': 'runs'} } }
tree = Tree(d)

print(d['foo']['amy']['what']) # returns 'runs'
d['foo']['amy']['when'] = 'now' # add new branch

In this way, you can keep edit/add/remove each dict level as you wish. All the dict methods for traversal etc, still apply.

Amara answered 5/4, 2017 at 16:54 Comment(2)
Is there a reason why you chose to extend dict instead of defaultdict? From my tests, extending defaultdict instead of dict and then adding self.default_factory = type(self) to the top of init should function the same way.Nador
i'm probably missing something here, how do you navigate this structure ? it seems very difficult to go from children to parents, for example, or siblingsSmallage
W
9

If someone needs a simpler way to do it, a tree is only a recursively nested list (since set is not hashable) :

[root, [child_1, [[child_11, []], [child_12, []]], [child_2, []]]]

Where each branch is a pair: [ object, [children] ]
and each leaf is a pair: [ object, [] ]

But if you need a class with methods, you can use anytree.

Woolery answered 10/2, 2019 at 7:58 Comment(0)
H
8

I've published a Python 3 tree implementation on my site: https://web.archive.org/web/20120723175438/www.quesucede.com/page/show/id/python_3_tree_implementation

Here's the code:

import uuid

def sanitize_id(id):
    return id.strip().replace(" ", "")

(_ADD, _DELETE, _INSERT) = range(3)
(_ROOT, _DEPTH, _WIDTH) = range(3)

class Node:

    def __init__(self, name, identifier=None, expanded=True):
        self.__identifier = (str(uuid.uuid1()) if identifier is None else
                sanitize_id(str(identifier)))
        self.name = name
        self.expanded = expanded
        self.__bpointer = None
        self.__fpointer = []

    @property
    def identifier(self):
        return self.__identifier

    @property
    def bpointer(self):
        return self.__bpointer

    @bpointer.setter
    def bpointer(self, value):
        if value is not None:
            self.__bpointer = sanitize_id(value)

    @property
    def fpointer(self):
        return self.__fpointer

    def update_fpointer(self, identifier, mode=_ADD):
        if mode is _ADD:
            self.__fpointer.append(sanitize_id(identifier))
        elif mode is _DELETE:
            self.__fpointer.remove(sanitize_id(identifier))
        elif mode is _INSERT:
            self.__fpointer = [sanitize_id(identifier)]

class Tree:

    def __init__(self):
        self.nodes = []

    def get_index(self, position):
        for index, node in enumerate(self.nodes):
            if node.identifier == position:
                break
        return index

    def create_node(self, name, identifier=None, parent=None):

        node = Node(name, identifier)
        self.nodes.append(node)
        self.__update_fpointer(parent, node.identifier, _ADD)
        node.bpointer = parent
        return node

    def show(self, position, level=_ROOT):
        queue = self[position].fpointer
        if level == _ROOT:
            print("{0} [{1}]".format(self[position].name,
                                     self[position].identifier))
        else:
            print("\t"*level, "{0} [{1}]".format(self[position].name,
                                                 self[position].identifier))
        if self[position].expanded:
            level += 1
            for element in queue:
                self.show(element, level)  # recursive call

    def expand_tree(self, position, mode=_DEPTH):
        # Python generator. Loosly based on an algorithm from 'Essential LISP' by
        # John R. Anderson, Albert T. Corbett, and Brian J. Reiser, page 239-241
        yield position
        queue = self[position].fpointer
        while queue:
            yield queue[0]
            expansion = self[queue[0]].fpointer
            if mode is _DEPTH:
                queue = expansion + queue[1:]  # depth-first
            elif mode is _WIDTH:
                queue = queue[1:] + expansion  # width-first

    def is_branch(self, position):
        return self[position].fpointer

    def __update_fpointer(self, position, identifier, mode):
        if position is None:
            return
        else:
            self[position].update_fpointer(identifier, mode)

    def __update_bpointer(self, position, identifier):
        self[position].bpointer = identifier

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

    def __setitem__(self, key, item):
        self.nodes[self.get_index(key)] = item

    def __len__(self):
        return len(self.nodes)

    def __contains__(self, identifier):
        return [node.identifier for node in self.nodes
                if node.identifier is identifier]

if __name__ == "__main__":

    tree = Tree()
    tree.create_node("Harry", "harry")  # root node
    tree.create_node("Jane", "jane", parent = "harry")
    tree.create_node("Bill", "bill", parent = "harry")
    tree.create_node("Joe", "joe", parent = "jane")
    tree.create_node("Diane", "diane", parent = "jane")
    tree.create_node("George", "george", parent = "diane")
    tree.create_node("Mary", "mary", parent = "diane")
    tree.create_node("Jill", "jill", parent = "george")
    tree.create_node("Carol", "carol", parent = "jill")
    tree.create_node("Grace", "grace", parent = "bill")
    tree.create_node("Mark", "mark", parent = "jane")

    print("="*80)
    tree.show("harry")
    print("="*80)
    for node in tree.expand_tree("harry", mode=_WIDTH):
        print(node)
    print("="*80)
Hunan answered 7/9, 2011 at 13:47 Comment(0)
B
7

I've implemented trees using nested dicts. It is quite easy to do, and it has worked for me with pretty large data sets. I've posted a sample below, and you can see more at Google code

  def addBallotToTree(self, tree, ballotIndex, ballot=""):
    """Add one ballot to the tree.

    The root of the tree is a dictionary that has as keys the indicies of all 
    continuing and winning candidates.  For each candidate, the value is also
    a dictionary, and the keys of that dictionary include "n" and "bi".
    tree[c]["n"] is the number of ballots that rank candidate c first.
    tree[c]["bi"] is a list of ballot indices where the ballots rank c first.

    If candidate c is a winning candidate, then that portion of the tree is
    expanded to indicate the breakdown of the subsequently ranked candidates.
    In this situation, additional keys are added to the tree[c] dictionary
    corresponding to subsequently ranked candidates.
    tree[c]["n"] is the number of ballots that rank candidate c first.
    tree[c]["bi"] is a list of ballot indices where the ballots rank c first.
    tree[c][d]["n"] is the number of ballots that rank c first and d second.
    tree[c][d]["bi"] is a list of the corresponding ballot indices.

    Where the second ranked candidates is also a winner, then the tree is 
    expanded to the next level.  

    Losing candidates are ignored and treated as if they do not appear on the 
    ballots.  For example, tree[c][d]["n"] is the total number of ballots
    where candidate c is the first non-losing candidate, c is a winner, and
    d is the next non-losing candidate.  This will include the following
    ballots, where x represents a losing candidate:
    [c d]
    [x c d]
    [c x d]
    [x c x x d]

    During the count, the tree is dynamically updated as candidates change
    their status.  The parameter "tree" to this method may be the root of the
    tree or may be a sub-tree.
    """

    if ballot == "":
      # Add the complete ballot to the tree
      weight, ballot = self.b.getWeightedBallot(ballotIndex)
    else:
      # When ballot is not "", we are adding a truncated ballot to the tree,
      # because a higher-ranked candidate is a winner.
      weight = self.b.getWeight(ballotIndex)

    # Get the top choice among candidates still in the running
    # Note that we can't use Ballots.getTopChoiceFromWeightedBallot since
    # we are looking for the top choice over a truncated ballot.
    for c in ballot:
      if c in self.continuing | self.winners:
        break # c is the top choice so stop
    else:
      c = None # no candidates left on this ballot

    if c is None:
      # This will happen if the ballot contains only winning and losing
      # candidates.  The ballot index will not need to be transferred
      # again so it can be thrown away.
      return

    # Create space if necessary.
    if not tree.has_key(c):
      tree[c] = {}
      tree[c]["n"] = 0
      tree[c]["bi"] = []

    tree[c]["n"] += weight

    if c in self.winners:
      # Because candidate is a winner, a portion of the ballot goes to
      # the next candidate.  Pass on a truncated ballot so that the same
      # candidate doesn't get counted twice.
      i = ballot.index(c)
      ballot2 = ballot[i+1:]
      self.addBallotToTree(tree[c], ballotIndex, ballot2)
    else:
      # Candidate is in continuing so we stop here.
      tree[c]["bi"].append(ballotIndex)
Barbarese answered 1/3, 2010 at 21:13 Comment(0)
I
7

If you are already using the networkx library, then you can implement a tree using that.

NetworkX is a Python package for the creation, manipulation, and study of the structure, dynamics, and functions of complex networks.

As 'tree' is another term for a (normally rooted) connected acyclic graph, and these are called 'arborescences' in NetworkX.

You may want to implement a plane tree (aka ordered tree) where each sibling has a unique rank and this is normally done via labelling the nodes.

However, graph language looks different from tree language, and the means of 'rooting' an arborescence is normally done by using a directed graph so, while there are some really cool functions and corresponding visualisations available, it would probably not be an ideal choice if you are not already using networkx.

An example of building a tree:

import networkx as nx
G = nx.Graph()
G.add_edge('A', 'B')
G.add_edge('B', 'C')
G.add_edge('B', 'D')
G.add_edge('A', 'E')
G.add_edge('E', 'F')

The library enables each node to be any hashable object, and there is no constraint on the number of children each node has.

Ingeingeberg answered 17/3, 2020 at 20:26 Comment(0)
P
4

bigtree is a Python tree implementation that integrates with Python lists, dictionaries, and pandas DataFrame. It is pythonic, making it easy to learn and extendable to many types of workflows.

There are various components to bigtree, namely

  • Constructing Trees from list, dictionary, and pandas DataFrame
  • Traversing Tree
  • Modifying Tree (shift/copy nodes)
  • Search Tree
  • Helper methods (clone tree, prune tree, get the difference between two treess)
  • Export tree (print to console, export tree to dictionary, pandas DataFrame, image etc.)
  • Other tree structures: Binary Trees!
  • Other graph structures: Directed Acyclic Graphs (DAGs)!

What more can I say... ah yes it is also well-documented.

Some examples:

from bigtree import list_to_tree, tree_to_dict, tree_to_dot

# Create tree from list, print tree
root = list_to_tree(["a/b/d", "a/c"])
print_tree(root)
# a
# ├── b
# │   └── d
# └── c

# Query tree
root.children
# (Node(/a/b, ), Node(/a/c, ))

# Export tree to dictionary / image
tree_to_dict(root)
# {
#     '/a': {'name': 'a'},
#     '/a/b': {'name': 'b'},
#     '/a/b/d': {'name': 'd'},
#     '/a/c': {'name': 'c'}
# }

graph = tree_to_dot(root, node_colour="gold")
graph.write_png("tree.png")

enter image description here

Source/Disclaimer: I'm the creator of bigtree ;)

Pilaf answered 1/12, 2022 at 17:24 Comment(1)
Brilliant! I can insert a list of file paths to get a tree!Favored
M
2

Hi you may give itertree a try (I'm the author).

The package goes in the direction of anytree package but with a bit different focus. The performance on huge trees (>100000 items) is much better and it deals with iterators to have effective filter mechanism.

>>>from itertree import *
>>>root=iTree('root')

>>># add some children:
>>>root.append(iTree('Africa',data={'surface':30200000,'inhabitants':1257000000}))
>>>root.append(iTree('Asia', data={'surface': 44600000, 'inhabitants': 4000000000}))
>>>root.append(iTree('America', data={'surface': 42549000, 'inhabitants': 1009000000}))
>>>root.append(iTree('Australia&Oceania', data={'surface': 8600000, 'inhabitants': 36000000}))
>>>root.append(iTree('Europe', data={'surface': 10523000 , 'inhabitants': 746000000}))
>>># you might use __iadd__ operator for adding too:
>>>root+=iTree('Antarktika', data={'surface': 14000000, 'inhabitants': 1100})

>>># for building next level we select per index:
>>>root[0]+=iTree('Ghana',data={'surface':238537,'inhabitants':30950000})
>>>root[0]+=iTree('Niger', data={'surface': 1267000, 'inhabitants': 23300000})
>>>root[1]+=iTree('China', data={'surface': 9596961, 'inhabitants': 1411780000})
>>>root[1]+=iTree('India', data={'surface': 3287263, 'inhabitants': 1380004000})
>>>root[2]+=iTree('Canada', data={'type': 'country', 'surface': 9984670, 'inhabitants': 38008005})    
>>>root[2]+=iTree('Mexico', data={'surface': 1972550, 'inhabitants': 127600000 })
>>># extend multiple items:
>>>root[3].extend([iTree('Australia', data={'surface': 7688287, 'inhabitants': 25700000 }), iTree('New Zealand', data={'surface': 269652, 'inhabitants': 4900000 })])
>>>root[4]+=iTree('France', data={'surface': 632733, 'inhabitants': 67400000 }))
>>># select parent per TagIdx - remember in itertree you might put items with same tag multiple times:
>>>root[TagIdx('Europe'0)]+=iTree('Finland', data={'surface': 338465, 'inhabitants': 5536146 })

The created tree can be rendered:

>>>root.render()
iTree('root')
     └──iTree('Africa', data=iTData({'surface': 30200000, 'inhabitants': 1257000000}))
         └──iTree('Ghana', data=iTData({'surface': 238537, 'inhabitants': 30950000}))
         └──iTree('Niger', data=iTData({'surface': 1267000, 'inhabitants': 23300000}))
     └──iTree('Asia', data=iTData({'surface': 44600000, 'inhabitants': 4000000000}))
         └──iTree('China', data=iTData({'surface': 9596961,  'inhabitants': 1411780000}))
         └──iTree('India', data=iTData({'surface': 3287263, 'inhabitants': 1380004000}))
     └──iTree('America', data=iTData({'surface': 42549000, 'inhabitants': 1009000000}))
         └──iTree('Canada', data=iTData({'surface': 9984670, 'inhabitants': 38008005}))
         └──iTree('Mexico', data=iTData({'surface': 1972550, 'inhabitants': 127600000}))
     └──iTree('Australia&Oceania', data=iTData({'surface': 8600000, 'inhabitants': 36000000}))
         └──iTree('Australia', data=iTData({'surface': 7688287, 'inhabitants': 25700000}))
         └──iTree('New Zealand', data=iTData({'surface': 269652, 'inhabitants': 4900000}))
     └──iTree('Europe', data=iTData({'surface': 10523000, 'inhabitants': 746000000}))
         └──iTree('France', data=iTData({'surface': 632733, 'inhabitants': 67400000}))
         └──iTree('Finland', data=iTData({'surface': 338465, 'inhabitants': 5536146}))
     └──iTree('Antarktika', data=iTData({'surface': 14000000, 'inhabitants': 1100}))

E.g. Filtering can be done like this:

>>>item_filter = Filter.iTFilterData(data_key='inhabitants', data_value=iTInterval(0, 20000000))
>>>iterator=root.iter_all(item_filter=item_filter)
>>>for i in iterator:
>>>    print(i)
iTree("'New Zealand'", data=iTData({'surface': 269652, 'inhabitants': 4900000}), subtree=[])
iTree("'Finland'", data=iTData({'surface': 338465, 'inhabitants': 5536146}), subtree=[])
iTree("'Antarktika'", data=iTData({'surface': 14000000, 'inhabitants': 1100}), subtree=[])
Monumental answered 14/5, 2021 at 7:48 Comment(0)
P
2

You can create a Tree data structure using the dataclasses module in Python.

The iter method can be used to make the Tree iterable, allowing you to traverse the Tree by changing the order of the yield statements.

The contains method can be used to check if a specific value is present in the Tree.

from dataclasses import dataclass

#               A
#              / \
#             B   C
#            / \   \
#           D   E   F
#          / \
#         G   H

@dataclass
class Node:
    data: str
    left: Node = None
    right: Node = None
    
    def __iter__(self):
        if self.left:
            yield from self.left
        
        yield self

        if self.right:
            yield from self.right

    def __contains__(self, other):
        for node in self:
            if node.data == other:
                return True
        return False
    

t = Node(
    'A', 
    Node(
        'B', 
        Node(
            'D', 
            Node('G'),
            Node('H'),
        ),
        Node('E'),
    ),  
    Node(
        'C', 
        right=Node('F'),
    ),
)
assert ('A' in t) is True
assert ('I' in t) is not True
for node in t:
    print(node.data, ' -> ', end='')
# G  -> D  -> H  -> B  -> E  -> A  -> C  -> F  -> 
Philately answered 8/2, 2023 at 21:7 Comment(0)
R
1

Another tree implementation loosely based off of Bruno's answer:

class Node:
    def __init__(self):
        self.name: str = ''
        self.children: List[Node] = []
        self.parent: Node = self

    def __getitem__(self, i: int) -> 'Node':
        return self.children[i]

    def add_child(self):
        child = Node()
        self.children.append(child)
        child.parent = self
        return child

    def __str__(self) -> str:
        def _get_character(x, left, right) -> str:
            if x < left:
                return '/'
            elif x >= right:
                return '\\'
            else:
                return '|'

        if len(self.children):
            children_lines: Sequence[List[str]] = list(map(lambda child: str(child).split('\n'), self.children))
            widths: Sequence[int] = list(map(lambda child_lines: len(child_lines[0]), children_lines))
            max_height: int = max(map(len, children_lines))
            total_width: int = sum(widths) + len(widths) - 1
            left: int = (total_width - len(self.name) + 1) // 2
            right: int = left + len(self.name)

            return '\n'.join((
                self.name.center(total_width),
                ' '.join(map(lambda width, position: _get_character(position - width // 2, left, right).center(width),
                             widths, accumulate(widths, add))),
                *map(
                    lambda row: ' '.join(map(
                        lambda child_lines: child_lines[row] if row < len(child_lines) else ' ' * len(child_lines[0]),
                        children_lines)),
                    range(max_height))))
        else:
            return self.name

And an example of how to use it:

tree = Node()
tree.name = 'Root node'
tree.add_child()
tree[0].name = 'Child node 0'
tree.add_child()
tree[1].name = 'Child node 1'
tree.add_child()
tree[2].name = 'Child node 2'
tree[1].add_child()
tree[1][0].name = 'Grandchild 1.0'
tree[2].add_child()
tree[2][0].name = 'Grandchild 2.0'
tree[2].add_child()
tree[2][1].name = 'Grandchild 2.1'
print(tree)

Which should output:

                        Root node                        
     /             /                      \              
Child node 0  Child node 1           Child node 2        
                   |              /              \       
             Grandchild 1.0 Grandchild 2.0 Grandchild 2.1
Retardation answered 15/8, 2018 at 21:41 Comment(2)
Clean and simple. You're missing some definitions in the __str__ function, namely accumulation and add. But the tree implementation itself works great.Pascual
It should be itertools.accumulate and operator.add, sorry for omitting that.Retardation
B
1

Treelib is convenient for the task as well. Documentation can be found treelib.

from treelib import Node, Tree
tree = Tree() # creating an object
tree.create_node("Harry", "harry")  # root node 
tree.create_node("Jane", "jane", parent="harry") #adding nodes
tree.create_node("Bill", "bill", parent="harry")
tree.create_node("Diane", "diane", parent="jane")
tree.create_node("Mary", "mary", parent="diane")
tree.create_node("Mark", "mark", parent="jane")
tree.show()

Harry
├── Bill
└── Jane
    ├── Diane
    │   └── Mary
    └── Mark
Breger answered 8/9, 2022 at 7:18 Comment(0)
S
0

Yet another example of creating trees in Python is using littletree (I am the author). I originally had a problem and solved it using nested dictionaries. At some point, I wanted something more treelike and I tried a few tree packages. I found most of them either a bit slow or difficult to extend to my own problem. Eventually I went back to my original approach and refactored it to a new tree library.

Its basic usage is as follows:

from littletree import Node

tree = Node(identifier="World")
tree["Asia"] = Node()
tree["Africa"] = Node()
tree["America"] = Node()
tree["Europe"] = Node()

tree.show()  # Print the tree to console
tree.to_image().show()  # Show tree as an image

Check here for its source and more examples.

Side answered 5/1, 2024 at 13:34 Comment(0)
A
-4

If you want to create a tree data structure then first you have to create the treeElement object. If you create the treeElement object, then you can decide how your tree behaves.

To do this following is the TreeElement class:

class TreeElement (object):

def __init__(self):
    self.elementName = None
    self.element = []
    self.previous = None
    self.elementScore = None
    self.elementParent = None
    self.elementPath = []
    self.treeLevel = 0

def goto(self, data):
    for child in range(0, len(self.element)):
        if (self.element[child].elementName == data):
            return self.element[child]

def add(self):

    single_element = TreeElement()
    single_element.elementName = self.elementName
    single_element.previous = self.elementParent
    single_element.elementScore = self.elementScore
    single_element.elementPath = self.elementPath
    single_element.treeLevel = self.treeLevel

    self.element.append(single_element)

    return single_element

Now, we have to use this element to create the tree, I am using A* tree in this example.

class AStarAgent(Agent):
# Initialization Function: Called one time when the game starts
def registerInitialState(self, state):
    return;

# GetAction Function: Called with every frame
def getAction(self, state):

    # Sorting function for the queue
    def sortByHeuristic(each_element):

        if each_element.elementScore:
            individual_score = each_element.elementScore[0][0] + each_element.treeLevel
        else:
            individual_score = admissibleHeuristic(each_element)

        return individual_score

    # check the game is over or not
    if state.isWin():
        print('Job is done')
        return Directions.STOP
    elif state.isLose():
        print('you lost')
        return Directions.STOP

    # Create empty list for the next states
    astar_queue = []
    astar_leaf_queue = []
    astar_tree_level = 0
    parent_tree_level = 0

    # Create Tree from the give node element
    astar_tree = TreeElement()
    astar_tree.elementName = state
    astar_tree.treeLevel = astar_tree_level
    astar_tree = astar_tree.add()

    # Add first element into the queue
    astar_queue.append(astar_tree)

    # Traverse all the elements of the queue
    while astar_queue:

        # Sort the element from the queue
        if len(astar_queue) > 1:
            astar_queue.sort(key=lambda x: sortByHeuristic(x))

        # Get the first node from the queue
        astar_child_object = astar_queue.pop(0)
        astar_child_state = astar_child_object.elementName

        # get all legal actions for the current node
        current_actions = astar_child_state.getLegalPacmanActions()

        if current_actions:

            # get all the successor state for these actions
            for action in current_actions:

                # Get the successor of the current node
                next_state = astar_child_state.generatePacmanSuccessor(action)

                if next_state:

                    # evaluate the successor states using scoreEvaluation heuristic
                    element_scored = [(admissibleHeuristic(next_state), action)]

                    # Increase the level for the child
                    parent_tree_level = astar_tree.goto(astar_child_state)
                    if parent_tree_level:
                        astar_tree_level = parent_tree_level.treeLevel + 1
                    else:
                        astar_tree_level += 1

                    # create tree for the finding the data
                    astar_tree.elementName = next_state
                    astar_tree.elementParent = astar_child_state
                    astar_tree.elementScore = element_scored
                    astar_tree.elementPath.append(astar_child_state)
                    astar_tree.treeLevel = astar_tree_level
                    astar_object = astar_tree.add()

                    # If the state exists then add that to the queue
                    astar_queue.append(astar_object)

                else:
                    # Update the value leaf into the queue
                    astar_leaf_state = astar_tree.goto(astar_child_state)
                    astar_leaf_queue.append(astar_leaf_state)

You can add/remove any elements from the object, but make the structure intect.

Axial answered 17/10, 2018 at 6:27 Comment(0)

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