building a directory structure using Python with recursion

Question

I am trying to build following structure using Python

├───Baseframework
│   ├───0_src
│   │   ├───Appfun
│   │   │   ├───Cpugeneric
│   │   │   ├───i5t
│   │   │   ├───Tricore
│   │   │   └───vpprint
│   │   └───Base
│   │       ├───gnfru
│   │       ├───ills
│   │       └───service
│   ├───1_toolbar
│   │   └───0_build
│   │       ├───0_utilites
│   │       ├───1_config
│   │       └───9_make
│   └───2_out
│       ├───tasking
│       └───tricore

Here's my code

import os
import argparse


base_path = "c:/usr"

# level_zero = "Baseframework"
levels = {"Baseframework":{"0_src":{"Appfun":["Cpugeneric", "i5t", "Tricore", "vpprint"], "Base":["ills","gnfru","service"]},
                           "1_toolbar":{"0_build":["0_utilities", "1_config", "9_make"]},
                           "2_out": ["ist_tasking","tricore_gyne"]}}

def iterdict(d, bd, flag = 1):

    for k,v in d.items():
        if isinstance(v, dict):
            print(flag)
            if flag==1:
                bd = os.path.join(bd,k)
                print(bd)
            else:
                bd = os.path.join(os.path.abspath(bd+"/.."),k)
                print(bd)

            if not os.path.exists(bd):
                    os.mkdir(bd)
            iterdict(v, bd)
        elif isinstance(v, list):
            bd1 = os.path.join(bd, k)
            if not os.path.exists(bd1):
                os.mkdir(bd1)
            for i in v:
                os.mkdir(os.path.join(bd1, i))
            flag = 0

iterdict(levels, base_path)

The above code creates a structure:

├───Baseframework
│   └───0_src
│       ├───1_toolbar
│       │   ├───0_build
│       │   │   ├───0_utilities
│       │   │   ├───1_config
│       │   │   └───9_make
│       │   └───2_out
│       │       ├───ist_tasking
│       │       └───tricore_gyne
│       ├───Appfun
│       │   ├───Cpugeneric
│       │   ├───i5t
│       │   ├───Tricore
│       │   └───vpprint
│       └───Base
│           ├───gnfru
│           ├───ills
│           └───service

Logic and Explanation:

Whenever I encounter list, I set flag=False so that I can go 1 step back for next iteration when value is dict.

But unfortunately value of flag variable is not changing. I can't get my head around it why value is not changing it is always 1.

Note: Kindly ignore if there's mismatch between folder name in Tree structure and code. It's just an example.

sahasrara62 · Accepted Answer · 2020-04-05 07:46:27Z

1

whenever you are calling function , then you need to pass the new path to the function not the base path

import os
import argparse


base_path = r"<base directory path>"

# level_zero = "Baseframework"
levels = {"Baseframework":{"0_src":{"Appfun":["Cpugeneric", "i5t", "Tricore", "vpprint"], "Base":["ills","gnfru","service"]},
                           "1_toolbar":{"0_build":["0_utilities", "1_config", "9_make"]},
                           "2_out": ["ist_tasking","tricore_gyne"]}}

def func(data, path):
    new_path = ''
    if isinstance(data, dict):
        for k, v in data.items():
            new_path = os.path.join(path, k)
            os.mkdir(new_path)
            func(v, new_path)
    elif isinstance(data, list):
        for _dir in data:
            new_path = os.path.join(path, _dir)
            os.mkdir(new_path)
    else:
        exit(0)

func(levels, base_path)

answered Apr 5, 2020 at 7:46

sahasrara62

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1 Comment

Sociopath Over a year ago

Thanks, it worked as expected. I guess using for loop before if was causing the issue.

Back2Basics · Accepted Answer · 2020-04-05 08:01:08Z

1

I would suggest using pathlib for this for anything python 3.6+

from pathlib import Path

base_path = Path("c:/usr")

levels = {"Baseframework":{"0_src":{"Appfun":["Cpugeneric", "i5t", "Tricore", "vpprint"], "Base":["ills","gnfru","service"]},
                           "1_toolbar":{"0_build":["0_utilities", "1_config", "9_make"]},
                           "2_out": ["ist_tasking","tricore_gyne"]}}

def iterdict(d:dict, bd: Path):
    """recursively creates directory structure from dictionaries"""
    for k,v in d.items():
        if isinstance(v, dict):
            bd.pathjoin(k).mkdir(exist_ok=True)
            iterdict(v, bd.pathjoin(k))
        else:
            for file in v:
                bd.pathjoin(k).mkdir(exist_ok=True)


iterdict(levels, base_path)

edited Apr 5, 2020 at 8:01

answered Apr 5, 2020 at 7:54

Back2Basics

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2 Comments

Sociopath Over a year ago

throws an error AttributeError: 'str' object has no attribute 'joinpath'

Back2Basics Over a year ago

oops backwards.

Philippe · Accepted Answer · 2020-04-05 08:08:36Z

This is a version which allows list to contains other dicts:

import os

base_path = "c:/usr"

levels = {"Baseframework":{"0_src":{"Appfun":["Cpugeneric", "i5t", "Tricore", "vpprint"], "Base":["ills","gnfru","service"]},
                           "1_toolbar":{"0_build":["0_utilities", "1_config", "9_make"]},
                           "2_out": ["ist_tasking","tricore_gyne"]}}

def iterdict(levels, base_path):
    if  isinstance(levels, dict):
        for k, v in levels.items():
            iterdict(v, os.path.join(base_path, k))
    elif isinstance(levels, list):
        [iterdict(e, base_path) for e in levels]
    else:
        os.makedirs(os.path.join(base_path, levels), exist_ok=True)

iterdict(levels, base_path)

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