How can I safely create a nested directory

Question

What is the most elegant way to check if the directory a file is going to be written to exists, and if not, create the directory using Python? Here is what I tried:

import os

file_path = "/my/directory/filename.txt"
directory = os.path.dirname(file_path)

try:
    os.stat(directory)
except:
    os.mkdir(directory)       

f = file(filename)

Somehow, I missed os.path.exists (thanks kanja, Blair, and Douglas). This is what I have now:

def ensure_dir(file_path):
    directory = os.path.dirname(file_path)
    if not os.path.exists(directory):
        os.makedirs(directory)

Is there a flag for "open", that makes this happen automatically?

Answer 1

On Python ≥ 3.5, use pathlib.Path.mkdir:

from pathlib import Path
Path("/my/directory").mkdir(parents=True, exist_ok=True)

For older versions of Python, I see two answers with good qualities, each with a small flaw, so I will give my take on it:

Try os.path.exists, and consider os.makedirs for the creation.

import os
if not os.path.exists(directory):
    os.makedirs(directory)

As noted in comments and elsewhere, there's a race condition – if the directory is created between the os.path.exists and the os.makedirs calls, the os.makedirs will fail with an OSError. Unfortunately, blanket-catching OSError and continuing is not foolproof, as it will ignore a failure to create the directory due to other factors, such as insufficient permissions, full disk, etc.

One option would be to trap the OSError and examine the embedded error code 

import os, errno try: os.makedirs(directory) except OSError as e: if e.errno != errno.EEXIST: raise

Alternatively, there could be a second os.path.exists, but suppose another created the directory after the first check, then removed it before the second one – we could still be fooled.

Depending on the application, the danger of concurrent operations may be more or less than the danger posed by other factors such as file permissions. The developer would have to know more about the particular application being developed and its expected environment before choosing an implementation.

Modern versions of Python improve this code quite a bit, both by exposing FileExistsError (in 3.3+)...

try:
    os.makedirs("path/to/directory")
except FileExistsError:
    # directory already exists
    pass

...and by allowing a keyword argument to os.makedirs called exist_ok (in 3.2+).

os.makedirs("path/to/directory", exist_ok=True)  # succeeds even if directory exists.

Answer 2

No need to put filename.txt

For directory check

path = '/my/directory_name'

if not os.path.exists(path):

os.mkdir(path)