Asked : Nov 17
Viewed : 20 times
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 make this happen automatically?
Nov 17
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.
answered Jan 18
Check os.makedirs
: (It makes sure the complete path exists.)
To handle the fact the directory might exist, catch OSError
. (If exist_ok
is False
(the default), an OSError
is raised if the target directory already exists.)
import os
try:
os.makedirs('./path/to/somewhere')
except OSError:
pass
answered Jan 18
we want to bundle everything together into a tidy function we can do so as follows:
def makeDirectory(path_to_directory):
p = Path(path_to_directory)
p.mkdir(exists_ok=True, parents=True)
answered Jan 18
To safely create a nested directory in Python, we can use the Path
instance’s mkdir
method.
For instance, we write:
from pathlib import Path
Path("./my/directory").mkdir(parents=True, exist_ok=True)
We invoke the Path
constructor with the path of the directory we want to create.
Then we set parents
and exist_ok
to True
so that it’ll create the nested directory without the parent directory being present.
answered Jan 18