exception - How can I safely create a nested directory in Python?

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 (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

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

Starting from Python 3.5, pathlib.Path.mkdir has an exist_ok flag:

from pathlib import Path
path = Path('/my/directory/filename.txt')
path.parent.mkdir(parents=True, exist_ok=True) 
# path.parent ~ os.path.dirname(path)

This recursively creates the directory and does not raise an exception if the directory already exists.

(just as os.makedirs got an exist_ok flag starting from python 3.2 e.g os.makedirs(path, exist_ok=True))

---

Note: when i posted this answer none of the other answers mentioned exist_ok...

Answer 6

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

Answer 7

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 8

Try the os.path.exists function

if not os.path.exists(dir):
    os.mkdir(dir)

Answer 9

I have put the following down. It's not totally foolproof though.

import os

dirname = 'create/me'

try:
    os.makedirs(dirname)
except OSError:
    if os.path.exists(dirname):
        # We are nearly safe
        pass
    else:
        # There was an error on creation, so make sure we know about it
        raise

Now as I say, this is not really foolproof, because we have the possiblity of failing to create the directory, and another process creating it during that period.

Answer 10

Check if a directory exists and create it if necessary?

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 the POSIX 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.

Answer 11

In Python 3.4 you can also use the brand new pathlib module:

from pathlib import Path
path = Path("/my/directory/filename.txt")
try:
    if not path.parent.exists():
        path.parent.mkdir(parents=True)
except OSError:
    # handle error; you can also catch specific errors like
    # FileExistsError and so on.

Answer 12

In Python3, os.makedirs supports setting exist_ok. The default setting is False, which means an OSError will be raised if the target directory already exists. By setting exist_ok to True, OSError (directory exists) will be ignored and the directory will not be created.

os.makedirs(path,exist_ok=True)

In Python2, os.makedirs doesn't support setting exist_ok. You can use the approach in heikki-toivonen's answer:

import os
import errno

def make_sure_path_exists(path):
    try:
        os.makedirs(path)
    except OSError as exception:
        if exception.errno != errno.EEXIST:
            raise

Answer 13

For a one-liner solution, you can use IPython.utils.path.ensure_dir_exists():

from IPython.utils.path import ensure_dir_exists
ensure_dir_exists(dir)

From the documentation: Ensure that a directory exists. If it doesn’t exist, try to create it and protect against a race condition if another process is doing the same.

Answer 14

The relevant Python documentation suggests the use of the EAFP coding style (Easier to Ask for Forgiveness than Permission). This means that the code

try:
    os.makedirs(path)
except OSError as exception:
    if exception.errno != errno.EEXIST:
        raise
    else:
        print "\nBE CAREFUL! Directory %s already exists." % path

is better than the alternative

if not os.path.exists(path):
    os.makedirs(path)
else:
    print "\nBE CAREFUL! Directory %s already exists." % path

The documentation suggests this exactly because of the race condition discussed in this question. In addition, as others mention here, there is a performance advantage in querying once instead of twice the OS. Finally, the argument placed forward, potentially, in favour of the second code in some cases --when the developer knows the environment the application is running-- can only be advocated in the special case that the program has set up a private environment for itself (and other instances of the same program).

Even in that case, this is a bad practice and can lead to long useless debugging. For example, the fact we set the permissions for a directory should not leave us with the impression permissions are set appropriately for our purposes. A parent directory could be mounted with other permissions. In general, a program should always work correctly and the programmer should not expect one specific environment.

Answer 15

You can use mkpath

# Create a directory and any missing ancestor directories. 
# If the directory already exists, do nothing.

from distutils.dir_util import mkpath
mkpath("test")    

Note that it will create the ancestor directories as well.

It works for Python 2 and 3.

Answer 16

I use os.path.exists(), here is a Python 3 script that can be used to check if a directory exists, create one if it does not exist, and delete it if it does exist (if desired).

It prompts users for input of the directory and can be easily modified.

Answer 17

I found this Q/A and I was initially puzzled by some of the failures and errors I was getting. I am working in Python 3 (v.3.5 in an Anaconda virtual environment on an Arch Linux x86_64 system).

Consider this directory structure:

└── output/         ## dir
   ├── corpus       ## file
   ├── corpus2/     ## dir
   └── subdir/      ## dir

Here are my experiments/notes, which clarifies things:

# ----------------------------------------------------------------------------
# [1] https://stackguides.com/questions/273192/how-can-i-create-a-directory-if-it-does-not-exist

import pathlib

""" Notes:
        1.  Include a trailing slash at the end of the directory path
            ("Method 1," below).
        2.  If a subdirectory in your intended path matches an existing file
            with same name, you will get the following error:
            "NotADirectoryError: [Errno 20] Not a directory:" ...
"""
# Uncomment and try each of these "out_dir" paths, singly:

# ----------------------------------------------------------------------------
# METHOD 1:
# Re-running does not overwrite existing directories and files; no errors.

# out_dir = 'output/corpus3'                ## no error but no dir created (missing tailing /)
# out_dir = 'output/corpus3/'               ## works
# out_dir = 'output/corpus3/doc1'           ## no error but no dir created (missing tailing /)
# out_dir = 'output/corpus3/doc1/'          ## works
# out_dir = 'output/corpus3/doc1/doc.txt'   ## no error but no file created (os.makedirs creates dir, not files!  ;-)
# out_dir = 'output/corpus2/tfidf/'         ## fails with "Errno 20" (existing file named "corpus2")
# out_dir = 'output/corpus3/tfidf/'         ## works
# out_dir = 'output/corpus3/a/b/c/d/'       ## works

# [2] https://docs.python.org/3/library/os.html#os.makedirs

# Uncomment these to run "Method 1":

#directory = os.path.dirname(out_dir)
#os.makedirs(directory, mode=0o777, exist_ok=True)

# ----------------------------------------------------------------------------
# METHOD 2:
# Re-running does not overwrite existing directories and files; no errors.

# out_dir = 'output/corpus3'                ## works
# out_dir = 'output/corpus3/'               ## works
# out_dir = 'output/corpus3/doc1'           ## works
# out_dir = 'output/corpus3/doc1/'          ## works
# out_dir = 'output/corpus3/doc1/doc.txt'   ## no error but creates a .../doc.txt./ dir
# out_dir = 'output/corpus2/tfidf/'         ## fails with "Errno 20" (existing file named "corpus2")
# out_dir = 'output/corpus3/tfidf/'         ## works
# out_dir = 'output/corpus3/a/b/c/d/'       ## works

# Uncomment these to run "Method 2":

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

Conclusion: in my opinion, "Method 2" is more robust.

[1] How can I create a directory if it does not exist?

[2] https://docs.python.org/3/library/os.html#os.makedirs

Answer 18

Use this command check and create dir

 if not os.path.isdir(test_img_dir):
     os.mkdir(test_img_dir)

Answer 19

Why not use subprocess module if running on a machine that supports command mkdir with -p option ? Works on python 2.7 and python 3.6

from subprocess import call
call(['mkdir', '-p', 'path1/path2/path3'])

Should do the trick on most systems.

In situations where portability doesn't matter (ex, using docker) the solution is a clean 2 lines. You also don't have to add logic to check if directories exist or not. Finally, it is safe to re-run without any side effects

If you need error handling:

from subprocess import check_call
try:
    check_call(['mkdir', '-p', 'path1/path2/path3'])
except:
    handle...

Answer 20

I saw Heikki Toivonen and A-B-B's answers and thought of this variation.

import os
import errno

def make_sure_path_exists(path):
    try:
        os.makedirs(path)
    except OSError as exception:
        if exception.errno != errno.EEXIST or not os.path.isdir(path):
            raise

Answer 21

You can use os.listdir for this:

import os
if 'dirName' in os.listdir('parentFolderPath')
    print('Directory Exists')

Answer 22

Call the function create_dir() at the entry point of your program/project.

import os

def create_dir(directory):
    if not os.path.exists(directory):
        print('Creating Directory '+directory)
        os.makedirs(directory)

create_dir('Project directory')

Answer 23

You have to set the full path before creating the directory:

import os,sys,inspect
import pathlib

currentdir = os.path.dirname(os.path.abspath(inspect.getfile(inspect.currentframe())))
your_folder = currentdir + "/" + "your_folder"

if not os.path.exists(your_folder):
   pathlib.Path(your_folder).mkdir(parents=True, exist_ok=True)

This works for me and hopefully, it will works for you as well

Answer 24

If you consider the following:

os.path.isdir('/tmp/dirname')

means a directory (path) exists AND is a directory. So for me this way does what I need. So I can make sure it is folder (not a file) and exists.

Answer 25

import os
if os.path.isfile(filename):
    print "file exists"
else:
    "Your code here"

Where your code here is use the (touch) command

This will check if the file is there if it is not then it will create it.

Answer 26

You can both create a file, and all its parent directories in 1 command with fastcore extension to pathlib: path.mk_write(data)

from fastcore.utils import Path
Path('/dir/to/file.txt').mk_write('Hello World')

Answer 27

Under Linux you can create directory in one line:

import os
os.system("mkdir -p {0}".format('mydir'))