Rendering a Tree in Python

Question

I had dataset like below:

A    B

1    2
5    3
2    5
3    

Below code gives me the following output:

def all_paths(table, root):
    # convert table structure to adjacency list
    children = {}
    for node, child in table:
        if child: 
             children[node] = children.setdefault(node, []) + [child]

        # generator for iteration over all paths from a certain node
        def recurse(path):
            yield path
            if path[-1] in children:
                for child in children[path[-1]]: # recursive calls for all child nodes
                    yield from recurse(path + [child])

        return recurse([root])

# Sample data in simple list format    
table = [
[1, 2],
[5, 3],
[2, 5],
[2, 6],
[2, 4],
[6, 7],
]

# Output all paths from node 1
for path in all_paths(table, 1):
    print(path)

Output: 
[1]
[1, 2]
[1, 2, 5]
[1, 2, 5, 3]
[1, 2, 6]
[1, 2, 6, 7]
[1, 2, 4]

But what i want here is to print output in render tree format like below:

1
└── 2
    |── 5
    |   └──3 
    |── 6
    |   └──7
    └── 4

I know python library Anytree is useful here but I don't know how to implement this code. Any help would be highly appreciated.

Answer 1

Your current output [[1], [1, 2], [1, 2, 5], [1, 2, 5, 3], [1, 2, 6], [1, 2, 6, 7], [1, 2, 4]] is a flattened list. The first step is to fold it to a tree structure. Then you can render through it via a simple decorator. (code below were tested with Python 3.6.1)

# to fold a tree you need first to get the leave for each flattened branch at proper levels
# toleavel([1,2,5]) --> [[[5]]] 
def toleave(branch):
  if branch[1:] == []:
    return [branch[0]]
  else:
    return [toleave(branch[1:])]

# fold the flattened tree
def fold(flattened_tree):
  if flattened_tree == []:
    return []
  else:
    return toleave(flattened_tree[0]) + fold(flattened_tree[1:])

# decorator for rendering
def render(f):
  render.level = -2
  indent = '   '
  def _f(*args):
    render.level += 1
    try:
        result = f(*args) 
        if not isinstance(result, list):
          print(render.level*indent, result)
    finally:
        render.level = -2
    return result
  return _f

# go over a tree and render it
@render
def tree_render(tree):
  if not isinstance(tree, list):
    return tree
  elif tree == []:
    return []
  else:
    return [tree_render(tree[0])] + [tree_render(tree[1:])]

flattened_tree = [[1], [1, 2], [1, 2, 5], [1, 2, 5, 3], [1, 2, 6], [1, 2, 6, 7], [1, 2, 4]]
tree_render(fold(flattened_tree))
# output:
# 1
#    2
#       5
#          3
#       6
#          7
#       4

Answer 2

Using a dictionary (if needed collections.OrderedDict) would make the looping easier. Using the recommendable anytree package it would give the wanted graphical output and the whole code would be just:

import anytree

# parent-child relations
table = [
    [1, 2],
    [5, 3],
    [2, 5],
    [2, 6],
    [2, 4],
    [6, 7],
]

def build_tree_recursively(p_num, p_node):
    for c_num in parent_children[p_num]:  # add children
        c_node = anytree.Node(str(c_num), parent=p_node)
        if c_num in parent_children:  # dive into
            build_tree_recursively(c_num, c_node)

# map parents to list of children
parent_children = {}
for p, c in table:  # numbers
    if p in parent_children:
        parent_children[p].append(c)
    else:
        parent_children[p] = [c]

p = 1  # assuming single root node (else add loop over elements not in column B)
tree = anytree.Node(str(p))
build_tree_recursively(p, tree)

# render
for pre, fill, node in anytree.RenderTree(tree):
    print("{}{}".format(pre, node.name))