Python: Represent directory structure tree as list of lists

Question

I have a list of directories extracted from os.walk . I removed the files because I don't need them.

.
|____A
     |____G
     |____H
          |____K
          |____L
|____B
     |____I
     |____J
|____C
|____D
|____E
|____F
     |____M

So it looks like this:

['.', ['A', 'B', 'C', 'D', 'E', 'F']], ['A', ['G', 'H']], ['A\\G', []], ['A\\H', ['K', 'L']], ['A\\G\\K', []], ['A\\G\\L', []], ['B', ['I', 'J']], ['B\\I', []], ['B\\J', []], ['C', []], ['D', []], ['E', []], ['F', ['M']], ['F\\M', []]

What I actually need is a real representation of the tree structure in a list like so:

['.', ['A' ['G', 'H' ['K', 'L']], ['B' ['I', 'J']], 'C', 'D', 'E', 'F' ['M']] 

ty ;)

Answer 1

You can construct a dictionary from the flattened values, and then use recursion:

import re
d = ['.', ['A', 'B', 'C', 'D', 'E', 'F']], ['A', ['G', 'H']], ['A\\G', []], ['A\\H', ['K', 'L']], ['A\\G\\K', []], ['A\\G\\L', []], ['B', ['I', 'J']], ['B\\I', []], ['B\\J', []], ['C', []], ['D', []], ['E', []], ['F', ['M']], ['F\\M', []]
new_d = {re.findall('.$', a)[0]:b for a, b in d}
def _tree(_start):
  if not new_d[_start]:
    return _start
  _c = [_tree(i) for i in new_d[_start]]
  return [_start, *(_c if any(not isinstance(i, str) for i in _c) else [_c])]

print(_tree('.'))

Output:

['.', ['A', 'G', ['H', ['K', 'L']]], ['B', ['I', 'J']], 'C', 'D', 'E', ['F', ['M']]]

Edit: Python2 version:

import re
d = ['.', ['A', 'B', 'C', 'D', 'E', 'F']], ['A', ['G', 'H']], ['A\\G', []], ['A\\H', ['K', 'L']], ['A\\G\\K', []], ['A\\G\\L', []], ['B', ['I', 'J']], ['B\\I', []], ['B\\J', []], ['C', []], ['D', []], ['E', []], ['F', ['M']], ['F\\M', []]
new_d = {re.findall('.$', a)[0]:b for a, b in d}
def _tree(_start):
  if not new_d[_start]:
    return _start
  _c = [_tree(i) for i in new_d[_start]]
  return [_start]+(_c if any(not isinstance(i, str) for i in _c) else [_c])

print(_tree('.'))

Answer 2

this does not return the datatype you are looking for but a nested dictionary (as this feels more natural to me as a tree structure):

from collections import defaultdict

lst = (['.', ['A', 'B', 'C', 'D', 'E', 'F']],
       ['A', ['G', 'H']], ['A\\G', []],
       ['A\\H', ['K', 'L']], ['A\\G\\K', []], ['A\\G\\L', []],
       ['B', ['I', 'J']], ['B\\I', []], ['B\\J', []], ['C', []],
       ['D', []], ['E', []], ['F', ['M']], ['F\\M', []])

def rec_dd():
    """"recursive default dict"""
    return defaultdict(rec_dd)

tree = rec_dd()
for here, dirs in lst:
    if not here.startswith('.'):
        cur_tree = tree['.']
    else:
        cur_tree = tree
    for key in here.split('\\'):
        cur_tree = cur_tree[key]

    for d in dirs:
        cur_tree[d] = rec_dd()

you can pretty print it this way:

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

and the result is:

{
    ".": {
        "A": {
            "G": {
                "K": {},
                "L": {}
            },
            "H": {
                "K": {},
                "L": {}
            }
        },
        "B": {
            "I": {},
            "J": {}
        },
        "C": {},
        "D": {},
        "E": {},
        "F": {
            "M": {}
        }
    }
}