How can I safely create a nested directory?
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Chapters
00:00 How Can I Safely Create A Nested Directory?
00:12 Accepted Answer Score 7229
01:53 Answer 2 Score 133
02:24 Answer 3 Score 662
02:59 Answer 4 Score 1574
06:26 Thank you
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Full question
https://stackoverflow.com/questions/2731...
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Content licensed under CC BY-SA
https://meta.stackexchange.com/help/lice...
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Tags
#python #exception #path #directory #operatingsystem
#avk47
ACCEPTED ANSWER
Score 7229
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 (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 os.makedirs called exist_ok (in 3.2+).
os.makedirs("path/to/directory", exist_ok=True) # succeeds even if directory exists.
ANSWER 2
Score 1574
Python 3.5+:
import pathlib
pathlib.Path('/my/directory').mkdir(parents=True, exist_ok=True)
pathlib.Path.mkdir as used above recursively creates the directory and does not raise an exception if the directory already exists. If you don't need or want the parents to be created, skip the parents argument.
Python 3.2+:
Using pathlib:
If you can, install the current pathlib backport named pathlib2. Do not install the older unmaintained backport named pathlib. Next, refer to the Python 3.5+ section above and use it the same.
If using Python 3.4, even though it comes with pathlib, it is missing the useful exist_ok option. The backport is intended to offer a newer and superior implementation of mkdir which includes this missing option.
Using os:
import os
os.makedirs(path, exist_ok=True)
os.makedirs as used above recursively creates the directory and does not raise an exception if the directory already exists. It has the optional exist_ok argument only if using Python 3.2+, with a default value of False. This argument does not exist in Python 2.x up to 2.7. As such, there is no need for manual exception handling as with Python 2.7.
Python 2.7+:
Using pathlib:
If you can, install the current pathlib backport named pathlib2. Do not install the older unmaintained backport named pathlib. Next, refer to the Python 3.5+ section above and use it the same.
Using os:
import os
try:
os.makedirs(path)
except OSError:
if not os.path.isdir(path):
raise
While a naive solution may first use os.path.isdir followed by os.makedirs, the solution above reverses the order of the two operations. In doing so, it prevents a common race condition having to do with a duplicated attempt at creating the directory, and also disambiguates files from directories.
Note that capturing the exception and using errno is of limited usefulness because OSError: [Errno 17] File exists, i.e. errno.EEXIST, is raised for both files and directories. It is more reliable simply to check if the directory exists.
Alternative:
mkpath creates the nested directory, and does nothing if the directory already exists. This works in both Python 2 and 3. Note however that distutils has been deprecated, and is scheduled for removal in Python 3.12.
import distutils.dir_util
distutils.dir_util.mkpath(path)
Per Bug 10948, a severe limitation of this alternative is that it works only once per python process for a given path. In other words, if you use it to create a directory, then delete the directory from inside or outside Python, then use mkpath again to recreate the same directory, mkpath will simply silently use its invalid cached info of having previously created the directory, and will not actually make the directory again. In contrast, os.makedirs doesn't rely on any such cache. This limitation may be okay for some applications.
With regard to the directory's mode, please refer to the documentation if you care about it.
ANSWER 3
Score 662
Using try except and the right error code from errno module gets rid of the race condition and is cross-platform:
import os
import errno
def make_sure_path_exists(path):
try:
os.makedirs(path)
except OSError as exception:
if exception.errno != errno.EEXIST:
raise
In other words, we try to create the directories, but if they already exist we ignore the error. On the other hand, any other error gets reported. For example, if you create dir 'a' beforehand and remove all permissions from it, you will get an OSError raised with errno.EACCES (Permission denied, error 13).
ANSWER 4
Score 133
I would personally recommend that you use os.path.isdir() to test instead of os.path.exists().
>>> os.path.exists('/tmp/dirname')
True
>>> os.path.exists('/tmp/dirname/filename.etc')
True
>>> os.path.isdir('/tmp/dirname/filename.etc')
False
>>> os.path.isdir('/tmp/fakedirname')
False
If you have:
>>> directory = raw_input(":: ")
And a foolish user input:
:: /tmp/dirname/filename.etc
... You're going to end up with a directory named filename.etc when you pass that argument to os.makedirs() if you test with os.path.exists().