How to Create folder structures using Python?

Question

I want to automate my folder creation using python, the desired hierarchy as picture below,

I created list of the first hierarchy

A.1 = [
    'A.1', 'A.2'   
] 

A.2 = [
    'A2.1'
]

This is the main code that i run to automate the folder creation, but i managed to automate it until 2nd hierarchy (A.1.1, A.1.2).

import os


main_dir = [A.1, A.2]           # Loading the list of sub-directories
root_dir = 'A'
main_dir_names = ['A.1', 'A.2'] # Name of the sub-directories
def main():
    # Create directory
    for i in range(0, len(main_dir)):
        for j in range(0,len(main_dir[i])):
                dirName = str(root_dir) + '/' + str(main_dir_names[i]) +'/' + str(main_dir[i][j])
                
                try:
                    # Create target Directory
                    os.makedirs(dirName)
                    print("Directory " , dirName ,  " Created ") 
                except FileExistsError:
                    print("Directory " , dirName ,  " already exists")        

                # Create target Directory if don't exist
                if not os.path.exists(dirName):
                    os.makedirs(dirName)
                    print("Directory " , dirName ,  " Created ")
                else:    
                    print("Directory " , dirName ,  " already exists")

if __name__ == '__main__':
    main()

Do you guys have any idea how can i automate my code so i can automate folder creation until 3rd hierarchy (A.1.2.1)?

if you can able to generate full path string till the last folder you can use this to create the directories https://stackoverflow.com/a/600612/9050514 — deadshot, Jun 25 '20 at 07:25

score 1 · Answer 1 · answered Jun 25 '20 at 07:53

if you provide a 'configuration' for your dir structure like

dir_dict={
    'name':'A',
    'sub': [{
        'name': 'A.1',
        'sub':[{
            'name': 'A.1.1',
            'sub':[]
        },{
            'name': 'A.1.2',
            'sub':[{
                'name': 'A.1.2.1',
                'sub': []
            }]
        }]
    },{
        'name': 'A.2',
        'sub':[]
    }]
}

then you can iterate through the configuration and create your directories. It's not what you would call 'fully automated' but may be useful in your case.

def create_dir(curr_dir, sub_conf):

    curr_dir = curr_dir+'/'+sub_conf['name']
    # create dir
    print(curr_dir)

    for sub in sub_conf['sub']:
        create_dir(curr_dir, sub)


create_dir('path_from_root', dir_dict)

score -1 · Answer 2 · answered Jun 25 '20 at 07:19

-1

If the hierarchy is fixed, you can use os.makedirs() as follows:

import os


path_1 = os.path.join(os.getcwd(),'A','A1','A11')
path_2 = os.path.join(os.getcwd(),'A','A1','A12','A121')
path_3 = os.path.join(os.getcwd(),'A','A2','A21')

os.makedirs(path_1)
os.makedirs(path_2)
os.makedirs(path_3)

answered Jun 25 '20 at 07:19

Vineet Bhat

148
2
12

do you call this auomated folder creation – deadshot Jun 25 '20 at 07:20
Elements of list of sub-directories will have to be used in `os.path.join`. Again, I'm assuming the hierarchy is always fixed. Let me know if I'm missing something. – Vineet Bhat Jun 25 '20 at 07:27

How to Create folder structures using Python?

2 Answers2