Python create directory failing - python

I am using some pretty standard code:
1 if not os.path.exists(args.outputDirectory):
2 if not os.makedirs(args.outputDirectory, 0o666):
3 sys.exit('Fatal: output directory "' + args.outputDirectory + '" does not exist and cannot be created')
I remove the directory and the check at 1 drops through to 2. I one-step beyond that and hit the error message at 3.
However, when I check, the directory was created successfully.
drwxrwsr-x 2 userId userGroup 4096 Jun 25 16:07 output/
What am I missing??

os.makedirs does not indicate whether it succeeded through its return value: it always returns None.
None is False-y, therefore, not os.makedirs(args.outputDirectory, 0o666) is always True, which triggers your sys.exit code path.
Fortunately, you don't need any of that. If os.makedirs fails, it'll throw an OSError.
You should catch the exception, not check the return value:
try:
if not os.path.exists(args.outputDirectory):
os.makedirs(args.outputDirectory, 0o666):
except OSError:
sys.exit('Fatal: output directory "' + args.outputDirectory + '" does not exist and cannot be created')
If no OSError is thrown, that means the directory was successfully created.

You don't need to call os.path.exists() (or os.path.isdir()); os.makedirs() has exist_ok parameter.
And as #Thomas Orozco mentioned, you shouldn't check os.makedirs()' return value because os.makedirs() indicates errors by raising an exception instead:
try:
os.makedirs(args.output_dir, mode=0o666, exist_ok=True)
except OSError as e:
sys.exit("Can't create {dir}: {err}".format(dir=output_dir, err=e))
Note: Unlike os.path.exist()-based solution; it raises an error if the path exists but it is not a directory (or a symlink to a directory).
There could be issues with the mode parameter, see the note for versions of Python before 3.4.1

Related

Checking whether a sys.argv path exists

I am currently facing a problem in which I run a python script via cmd in which I set some of the arguments required for functions on that script also in cmd. So, I have been using sys.argv. To better explain, I would type the following:
python script.py arg.yaml C:\Users\JohnDoe\Documents
So, the sys.argv are the a yaml file and the second one a path. Please note that the second one is a optional argument, so the user might not choose to add that. My goal is to raise an error if the given path doesn't exist.
I have been try the following:
try:
export_path = sys.argv[2]
if len(sys.argv) == 3:
try:
Path('export_path').is_dir()
except FileNotFoundError:
print(custom_message_error)
except IndexError:
export_path = None
But the program is still proceeding and I am not been able to catch the error. So, how would be possible to check whether or not the sys.argv path folder exists.
I think you confused some concepts here. pathlib.Path.is_dir does not throw an exception if the directory was not found but simply returns False. Moreover, you should not put your variable export_path in quotes when passing it to Path:
from pathlib import Path
export_path = "C:\Users\JohnDoe\Documents"
if Path(export_path).is_dir():
print("Path exists!")
else:
print("Path does not exist")

How would you catch python Exception inside an except clause

How would you catch exception within exception in python ?
Consider the following case
bak_filename = filename + ".bak"
#This try is needed because if bak_filename existed, rename will fail
try:
os.rename(filename, bak_filename)
except WindowsError:
#However, os.remove may still fail e.g. file in use. How would you handle this exception within except clause ?
#try and except around this os.remove?
os.remove(bak_filename)
os.rename(filename, bak_filename)
Any thought of :
Rewriting to avoid double try?
Not necessarily this example, but in some case we cannot rewrite, how would you handle double try ?
A solution to avoid all this exception wrapping would be to use shutil.move
shutil.move(filename, bak_filename)
If the destination already exists but is not a directory, it may be overwritten depending on os.rename() semantics.
If the destination is on the current filesystem, then os.rename() is used. Otherwise, src is copied to dst using copy_function and then removed. In case of symlinks, a new symlink pointing to the target of src will be created in or as dst and src will be removed.
So it basically does what you're trying to do, but in a library which is available on all python distros.
Note that the performance may be bad if the file is big and the target file exists and os.rename refuses to overwrite it (completely depends on the operating system but for instance Windows will refuse to rename over an existing file), because the fallback when os.rename throws OSError is copying the source then deleting. The implementation doesn't try to remove the file then rename again, because if rename fails, Python assumes that we're trying to rename across filesystems, and a copy+delete is in order in that case (this is exactly how Unix mv works).
try:
os.rename(src, real_dst)
except OSError:
if os.path.islink(src):
...
else:
copy_function(src, real_dst)
os.unlink(src)
To workaround that possible file existence, a prior os.remove call wrapped in a try/except OSError statement could be done.
try:
os.remove(bak_filename)
except OSError:
pass
shutil.move(filename, bak_filename) # or os.rename(filename, bak_filename)
Of course if bak_filename is locked/not deletable, an exception can still be raised by shutil.mode. Also note that if we tried to delete the target file, os.rename will be as good as shutil.move. If the target file could not be deleted, the operation can't succeed anyway.
You can read more on Exception Chaining. But, that's for Python 3
In Python 2, you can store your exception in a variable, then raise it again explicitly like this:
try:
do something
except Exception as e:
try:
do something again
except:
pass
raise e

Python: Getting a WindowsError instead of an IOError

I am trying to understand exceptions with Python 2.7.6, on Windows 8.
Here's the code I am testing, which aims to create a new directory at My_New_Dir. If the directory already exists, I want to delete the entire directory and its contents, and then create a fresh directory.
import os
dir = 'My_New_Dir'
try:
os.mkdir(dir)
except IOError as e:
print 'exception thrown'
shutil.rmtree(dir)
os.mkdir(dir)
The thing is, the exception is never thrown. The code works fine if the directory does not already exist, but if the directory does exist, then I get the error:
WindowsError: [Error 183] Cannot create a file when that file already exists: 'My_New_Dir'
But according to the Python documentation for os.mkdir(),
If the directory already exists, OSError is raised.
So why is the Windows error being thrown, rather than the Python exception?
WindowsError is a subclass of OSError. From the exceptions documentation:
Raised when a Windows-specific error occurs or when the error number does not correspond to an errno value. The winerror and strerror values are created from the return values of the GetLastError() and FormatMessage() functions from the Windows Platform API. The errno value maps the winerror value to corresponding errno.h values. This is a subclass of OSError.
You are trying to catch IOError however, which is not such a parent class of WindowsError; as a result it won't suffice to catch either OSError nor WindowsError.
Alter your code to use the correct exception here:
try:
os.mkdir(dir)
except OSError as e:
or use WindowsError; this would tie your code to the Windows platform.
You mis-read it. It is "OSError" not "IOError", and WindowsError is a subclasss of "OSError" for your specific working OS.
>>> issubclass(WindowsError, OSError)
True
Besides, for your propose, this API is better:
os.path.isdir(path): Return True if path is an existing directory. This follows symbolic links, so both islink() and isdir() can be true for the same path.
if os.path.isdir(dir):
shutil.rmtree(dir)
os.mkdir(dir)
If you search for "WindowsError" on the exceptions page, you will see that WindowsError is in fact a Python exception. It is a subclass of OSError, so the documentation is still correct. If you change to
try:
os.mkdir(dir)
except OSError as e:
print 'exception thrown'
shutil.rmtree(dir)
os.mkdir(dir)
then you will catch the exception.

Modifying a symlink in python

How do I change a symlink to point from one file to another in Python?
The os.symlink function only seems to work to create new symlinks.
If you need an atomic modification, unlinking won't work.
A better solution would be to create a new temporary symlink, and then rename it over the existing one:
os.symlink(target, tmpLink)
os.rename(tmpLink, linkName)
You can check to make sure it was updated correctly too:
if os.path.realpath(linkName) == target:
# Symlink was updated
According to the documentation for os.rename though, there may be no way to atomically change a symlink in Windows. In that case, you would just delete and re-create.
A little function for Python2 which tries to symlink and if it fails because of an existing file, it removes it and links again.
import os, errno
def symlink_force(target, link_name):
try:
os.symlink(target, link_name)
except OSError, e:
if e.errno == errno.EEXIST:
os.remove(link_name)
os.symlink(target, link_name)
else:
raise e
For python3 except condition should be except OSError as e:
You could os.unlink() it first, and then re-create using os.symlink() to point to the new target.
Given overwrite=True, this function will safely overwrite an existing file with a symlink.
It is cognisant of race conditions, which is why it is not short, but it is safe.
import os, tempfile
def symlink(target, link_name, overwrite=False):
'''
Create a symbolic link named link_name pointing to target.
If link_name exists then FileExistsError is raised, unless overwrite=True.
When trying to overwrite a directory, IsADirectoryError is raised.
'''
if not overwrite:
os.symlink(target, link_name)
return
# os.replace() may fail if files are on different filesystems
link_dir = os.path.dirname(link_name)
# Create link to target with temporary filename
while True:
temp_link_name = tempfile.mktemp(dir=link_dir)
# os.* functions mimic as closely as possible system functions
# The POSIX symlink() returns EEXIST if link_name already exists
# https://pubs.opengroup.org/onlinepubs/9699919799/functions/symlink.html
try:
os.symlink(target, temp_link_name)
break
except FileExistsError:
pass
# Replace link_name with temp_link_name
try:
# Pre-empt os.replace on a directory with a nicer message
if not os.path.islink(link_name) and os.path.isdir(link_name):
raise IsADirectoryError(f"Cannot symlink over existing directory: '{link_name}'")
os.replace(temp_link_name, link_name)
except:
if os.path.islink(temp_link_name):
os.remove(temp_link_name)
raise
Notes for pedants:
If the function fails (e.g. computer crashes), an additional random link to the target might exist.
An unlikely race condition still remains: the symlink created at the randomly-named temp_link_name could be modified by another process before replacing link_name.
I raised a python issue to highlight the issues of os.symlink() requiring the target to not exist, where I was advised to raise my suggestion on the python-ideas mailing list
Credit to Robert Siemer’s input.
I researched this question recently, and found out that the best way is indeed to unlink and then symlink. But if you need just to fix broken links, for example with auto-replace, then you can do os.readlink:
for f in os.listdir(dir):
path = os.path.join(dir, f)
old_link = os.readlink(path)
new_link = old_link.replace(before, after)
os.unlink(path)
os.symlink(new_link, path)
Don't forget to add a raise command in the case when e.errno != errno.EEXIST You don't want to hide an error then:
if e.errno == errno.EEXIST:
os.remove(link_name)
os.symlink(target, link_name)
else:
raise
A quick and easy solution:
while True:
try:
os.symlink(target, link_name)
break
except FileExistsError:
os.remove(link_name)
However this has a race condition when replacing a symlink which should
always exist, eg:
/lib/critical.so -> /lib/critical.so.1.2
When upgrading by:
my_symlink('/lib/critical.so.2.0', '/lib/critical.so')
There is a point in time when /lib/critical.so doesn't exist.
This answer avoids the race condition.

How can I safely create a directory (possibly including intermediate directories)?

I am writing a file using Python, and I want it to be placed in a specific path. How can I safely make sure that the path exists?
That is: how can I check whether the folder exists, along with its parents? If there are missing folders along the path, how can I create them?
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.
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.
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).
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...
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().
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
Try the os.path.exists function
if not os.path.exists(dir):
os.mkdir(dir)
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)
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.
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.
fastest safest way to do it is:
it will create if not exists and skip if exists:
from pathlib import Path
Path("path/with/childs/.../").mkdir(parents=True, exist_ok=True)
Best way to do this in python
#Devil
import os
directory = "./out_dir/subdir1/subdir2"
if not os.path.exists(directory):
os.makedirs(directory)
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.
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.
IPython is an extension package, not part of the standard library.
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
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.
I found this Q/A after I was puzzled by some of the failures and errors I was getting while working with directories in Python. 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 provides clarification:
# ----------------------------------------------------------------------------
# [1] https://stackoverflow.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 safely create a nested directory?
[2] https://docs.python.org/3/library/os.html#os.makedirs
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.
In case you're writing a file to a variable path, you can use this on the file's path to make sure that the parent directories are created.
from pathlib import Path
path_to_file = Path("zero/or/more/directories/file.ext")
parent_directory_of_file = path_to_file.parent
parent_directory_of_file.mkdir(parents=True, exist_ok=True)
Works even if path_to_file is file.ext (zero directories deep).
See pathlib.PurePath.parent and pathlib.Path.mkdir.
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...
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
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
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.
Use this command check and create dir
if not os.path.isdir(test_img_dir):
os.mkdir(test_img_dir)
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')
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.
You can use os.listdir for this:
import os
if 'dirName' in os.listdir('parentFolderPath')
print('Directory Exists')
This may not exactly answer the question. But I guess your real intention is to create a file and its parent directories, given its content all in 1 command.
You can do that with fastcore extension to pathlib: path.mk_write(data)
from fastcore.utils import Path
Path('/dir/to/file.txt').mk_write('Hello World')
See more in fastcore documentation

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