Query explained:
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?
How to create a nested directory in Python? 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:
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.
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:
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:
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:
>>> dir = 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()
.
Answer #5:
Insights on the specifics of this situation
You give a particular file at a certain path and you pull the directory from the file path. Then after making sure you have the directory, you attempt to open a file for reading. To comment on this code:
filename = "/my/directory/filename.txt"
dir = os.path.dirname(filename)
We want to avoid overwriting the builtin function, dir
. Also, filepath
or perhaps fullfilepath
is probably a better semantic name than filename
so this would be better written:
import os
filepath = '/my/directory/filename.txt'
directory = os.path.dirname(filepath)
Your end goal is to open this file, you initially state, for writing, but you’re essentially approaching this goal (based on your code) like this, which opens the file for reading:
if not os.path.exists(directory):
os.makedirs(directory)
f = file(filename)
Assuming opening for reading
Why would you make a directory for a file that you expect to be there and be able to read?
Just attempt to open the file.
with open(filepath) as my_file:
do_stuff(my_file)
If the directory or file isn’t there, you’ll get an IOError
with an associated error number: errno.ENOENT
will point to the correct error number regardless of your platform. You can catch it if you want, for example:
import errno
try:
with open(filepath) as my_file:
do_stuff(my_file)
except IOError as error:
if error.errno == errno.ENOENT:
print 'ignoring error because directory or file is not there'
else:
raise
Assuming we’re opening for writing
This is probably what you’re wanting.
In this case, we probably aren’t facing any race conditions. So just do as you were, but note that for writing, you need to open with the w
mode (or a
to append). It’s also a Python best practice to use the context manager for opening files.
import os
if not os.path.exists(directory):
os.makedirs(directory)
with open(filepath, 'w') as my_file:
do_stuff(my_file)
However, say we have several Python processes that attempt to put all their data into the same directory. Then we may have contention over creation of the directory. In that case it’s best to wrap the makedirs
call in a try-except block.
import os
import errno
if not os.path.exists(directory):
try:
os.makedirs(directory)
except OSError as error:
if error.errno != errno.EEXIST:
raise
with open(filepath, 'w') as my_file:
do_stuff(my_file)
Answer #6:
The direct answer to this is, assuming a simple situation where you don’t expect other users or processes to be messing with your directory:
if not os.path.exists(d):
os.makedirs(d)
or if making the directory is subject to race conditions (i.e. if after checking the path exists, something else may have already made it) do this:
import errno
try:
os.makedirs(d)
except OSError as exception:
if exception.errno != errno.EEXIST:
raise
But perhaps an even better approach is to sidestep the resource contention issue, by using temporary directories via tempfile
:
import tempfile
d = tempfile.mkdtemp()
Here’s the essentials from the online doc:
mkdtemp(suffix='', prefix='tmp', dir=None)
User-callable function to create and return a unique temporary
directory. The return value is the pathname of the directory.
The directory is readable, writable, and searchable only by the
creating user.
Caller is responsible for deleting the directory when done with it.
New in Python 3.5: pathlib.Path
with exist_ok
There’s a new Path
object (as of 3.4) with lots of methods one would want to use with paths – one of which is mkdir
.
(For context, I’m tracking my weekly rep with a script. Here’s the relevant parts of code from the script that allow me to avoid hitting Stack Overflow more than once a day for the same data.)
First the relevant imports:
from pathlib import Path
import tempfile
We don’t have to deal with os.path.join
now – just join path parts with a /
:
directory = Path(tempfile.gettempdir()) / 'sodata'
Then I idempotently ensure the directory exists – the exist_ok
argument shows up in Python 3.5:
directory.mkdir(exist_ok=True)
Here’s the relevant part of the documentation:
If
exist_ok
is true,FileExistsError
exceptions will be ignored (same behavior as thePOSIX mkdir -p
command), but only if the last path component is not an existing non-directory file.
Here’s a little more of the script – in my case, I’m not subject to a race condition, I only have one process that expects the directory (or contained files) to be there, and I don’t have anything trying to remove the directory.
todays_file = directory / str(datetime.datetime.utcnow().date())
if todays_file.exists():
logger.info("todays_file exists: " + str(todays_file))
df = pd.read_json(str(todays_file))
Path
objects have to be coerced to str
before other APIs that expect str
paths can use them.
Perhaps Pandas should be updated to accept instances of the abstract base class, os.PathLike
.
Hope you learned something from this post.
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