How to create a directory in Python?

Question

How can I use Python to create a certain directory?

For example:

app/
   main.py

How could I create a directory called temp inside app through Python main.py?

Answer 1

Reply using Python 3.*

It is possible to use os.makedirs or os.mkdir.

os.makedirs

Creates all directories that are specified in the parameter, recursively.
Ex.: os.makedir('./temp/2016/12/09') will create the folders temp, 2016, 12 and 09.

os.mkdir

Only creates the last directory. If the previous directories do not exist it will cause an error.
Ex.: os.mkdir('./temp/2016/12/09') - will only create the directory 09 and only if the previous ones exist, otherwise it will cause the following error

Filenotfounderror: [Winerror 3] The system cannot find the path specified: './1/2/3/4'

---

Example:

import os

dir = './temp'       
os.makedirs(dir)
# ou 
os.mkdir(dir)

Answer 2

There is more than one function for this:

• os.path.mkdir creates a folder (os.mkdir if it’s Python3)
• os.makedirs creates folder(s) recursively

The archive main.py is inside ./app then you can just use:

os.path.mkdir('./temp') #Python 2
os.mkdir('./temp') #Python 3

If you want to create date-based subfolders (this helps you "navigate" faster later):

os.makedirs('./temp/2016/12/9')

Answer 3

The right way to avoid running condition would be:

import os

try:
    os.makedirs("./temp")
except OSError:
    #faz o que acha que deve se não for possível criar

I put in the Github for future reference.

Answer 4

Although the previous answers are correct, from Python 3.4 give preference to the use of pathlib. In documentation and in the PEP 428 the reasons for.

Example: Little of the Functionality from os.path is reused. Many os.path functions are tied by Backwards Compatibility to confusing or Plain Wrong behaviour (for example, the Fact that os.path.abspath() simplifies ".." path Components without resolving symlinks first).

With pathlib, the solution to the question is as follows::

from pathlib import Path

Path('./temp').mkdir(exist_ok=True)

The exist_ok causes exceptions FileExistsError are ignored.

It is also possible to use the operator / to the Join directories. For example:

(Path('.') / 'temp' / 'subdir1' / 'sub2').mkdir(parents=True, exist_ok=True)

In this case it was necessary to add the option parents so that the entire directory chain could be created. Otherwise, the following error is launched: FileNotFoundError: [Errno 2] No such file or directory: 'temp/subdir1/sub2'

Answer 5

Eventually you may be interested in creating from the './temp' folder even if it already exists in this case you can do so:

import os

dirTemp = './temp'

try:
    os.mkdir(dirTemp)
except OSError:
    os.rmdir(dirTemp)
    os.mkdir(dirTemp)

Avoid using dir as variable name because dir is a Python function