How do I check if the directory that a file is going to be written to exists, and if it does not, create the directory using Python?
Now we will see Solution for issue: How can I safely create a nested directory?
On Python ≥ 3.5, use
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:
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 (see Is there a cross-platform way of getting information from Python’s OSError):
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
exist_ok (in 3.2+).
os.makedirs("path/to/directory", exist_ok=True) # succeeds even if directory exists.
This question is answered By – Blair Conrad
This answer is collected from stackoverflow and reviewed by FixPython community admins, is licensed under cc by-sa 2.5 , cc by-sa 3.0 and cc by-sa 4.0