CODEWITHSUNDEEP
pythondictionarydirectory-structure
Ashik Mydeen
Ashik Mydeen

Reputation: 104

Parsing list of directories into nested dictionary

I have the following items in a list with the format of directories structure.

[
    'fold/2021/',
    'fold/2021/11/',
    'fold/2021/11/01/',
    'fold/2021/11/01/123.gz',
    'fold/2021/11/01/345.gz',
    'fold/2021/12/',
    'fold/2021/12/02/',
    'fold/2022/'
]

I need this in the following nested dictionary structure:

{
  "fold": {
    "2021": {
      "11": {
        "01": {
          "123.gz": None,
          "345.gz": None
        }
      },
      "12": {
        "02": {}
      }
    },
    "2022": {}
  }
}

I tried a lot with recursion and some other methods, but I am not getting this structure.

Here is what I tried:

def get_directory_structure(path):
    global main_dict

    local_dict = {}

    a = path.rstrip('/').split('/')
    local_dict.setdefault(a[0], {})

    if len(a) > 1:
        return_dict = get_directory_structure(path[path.find('/')+1:])
        
        local_dict[a[0]] = return_dict

        if a[0] == 'fold':
            main_dict.update(**local_dict)
        
    return local_dict

main_dict = {}
for path in paths:
    get_directory_structure(main_dict, path)

print(main_dict)

Please help me with this. Thank you

Note:- I don't have the folder on my PC. I just have the items in the list

Upvotes: 2

Views: 492

Answers (1)

tobias_k
tobias_k

Reputation: 82899

You can try like this, not using recursion but using *-unpacking to separate the items into the file (or '') and the path leading up to that, and using setdefault do "expand" deeper levels of the dict, if they do not exist yet, and finally add the file, if any.

res = {}
for item in lst:
    d = res
    *path, last = item.split("/")
    for p in path:
        d = d.setdefault(p, {})
    if last != "":
        d[last] = None

Afterwards, res should be your desired result:

{'fold': {'2021': {'11': {'01': {'123.gz': None, '345.gz': None}}, '12': {'02': {}}}, '2022': {}}}

Upvotes: 3

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