How do I check whether a file exists?

spence91 Source

How do I see if a file exists or not, without using the try statement?



answered 10 years ago Paul #1

import os.path

if os.path.isfile(filepath):

answered 10 years ago benefactual #2

import os
os.path.exists(path) # Returns whether the path (directory or file) exists or not
os.path.isfile(path) # Returns whether the file exists or not

answered 10 years ago PierreBdR #3

You have the os.path.exists function:

import os.path

This returns True for both files and directories but you can instead use


to test if it's a file specifically. It follows symlinks.

answered 10 years ago rslite #4

If the reason you're checking is so you can do something like if file_exists: open_it(), it's safer to use a try around the attempt to open it. Checking and then opening risks the file being deleted or moved or something between when you check and when you try to open it.

If you're not planning to open the file immediately, you can use os.path.isfile

Return True if path is an existing regular file. This follows symbolic links, so both islink() and isfile() can be true for the same path.

import os.path

if you need to be sure it's a file.

Starting with Python 3.4, the pathlib module offers an object-oriented approach (backported to pathlib2 in Python 2.7):

from pathlib import Path

my_file = Path("/path/to/file")
if my_file.is_file():
    # file exists

To check a directory, do:

if my_file.is_dir():
    # directory exists

To check whether a Path object exists independently of whether is it a file or directory, use exists():

if my_file.exists():
    # path exists

You can also use resolve() in a try block:

    my_abs_path = my_file.resolve()
except FileNotFoundError:
    # doesn't exist
    # exists

answered 10 years ago zgoda #5

Additionally, os.access():

if os.access("myfile", os.R_OK):
    with open("myfile") as fp:

Being R_OK, W_OK, and X_OK the flags to test for permissions (doc).

answered 10 years ago bortzmeyer #6

Unlike isfile(), exists() will return True for directories.
So depending on if you want only plain files or also directories, you'll use isfile() or exists(). Here is a simple REPL output.

>>> print os.path.isfile("/etc/password.txt")
>>> print os.path.isfile("/etc")
>>> print os.path.isfile("/does/not/exist")
>>> print os.path.exists("/etc/password.txt")
>>> print os.path.exists("/etc")
>>> print os.path.exists("/does/not/exist")

answered 9 years ago pkoch #7

Prefer the try statement. It's considered better style and avoids race conditions.

Don't take my word for it. There's plenty of support for this theory. Here's a couple:

answered 7 years ago philberndt #8

You could try this (safer):

    # 'with' is safer to open a file
    with open('whatever.txt') as fh:
        # Do something with 'fh'
except IOError as e:

The ouput would be:

([Errno 2] No such file or directory: 'whatever.txt')

Then, depending on the result, your program can just keep running from there or you can code to stop it if you want.

answered 7 years ago aitchnyu #9

import os
path = /path/to/dir

root,dirs,files = os.walk(path).next()
if myfile in files:
   print "yes it exists"

This is helpful when checking for several files. Or you want to do a set intersection/ subtraction with an existing list.

answered 7 years ago Yugal Jindle #10

Use os.path.isfile() with os.access():

import os
import os.path


if os.path.isfile(PATH) and os.access(PATH, os.R_OK):
    print "File exists and is readable"
    print "Either the file is missing or not readable"

answered 5 years ago user2154354 #11

You should definitely use this one.

from os.path import exists

if exists("file") == True:
    print "File exists."
elif exists("file") == False:
    print "File doesn't exist."

answered 5 years ago un33k #12

This is the simplest way to check if a file exists. Just because the file existed when you checked doesn't guarantee that it will be there when you need to open it.

import os
fname = "foo.txt"
if os.path.isfile(fname):
    print("file does exist at this time")
    print("no such file exists at this time")

answered 5 years ago chad #13

It doesn't seem like there's a meaningful functional difference between try/except and isfile(), so you should use which one makes sense.

If you want to read a file, if it exists, do

    f = open(filepath)
except IOError:
    print 'Oh dear.'

But if you just wanted to rename a file if it exists, and therefore don't need to open it, do

if os.path.isfile(filepath):
    os.rename(filepath, filepath + '.old')

If you want to write to a file, if it doesn't exist, do

# python 2
if not os.path.isfile(filepath):
    f = open(filepath, 'w')

# python 3, x opens for exclusive creation, failing if the file already exists
    f = open(filepath, 'wx')
except IOError:
    print 'file already exists'

If you need file locking, that's a different matter.

answered 5 years ago John Allsup #14

If you want to do that in Bash it would be:

if [ -e "$FILE" ]; then
    prog "$FILE"

Which I sometimes do when using Python to do more complicated manipulation of a list of names (as I sometimes need to use Python for), the try open(file): except: method isn't really what's wanted, as it is not the Python process that is intended to open the file.

In one case, the purpose is to filter a list of names according to whether they exist at present (and there are no processes likely to delete the file, nor security issues since this is on my Raspberry Pi which has no sensitive files on its SD card).

I'm wondering whether a 'Simple Patterns' site would be a good idea? So that, for example, you could illustrate both methods with links between them and links to discussions as to when to use which pattern.

answered 4 years ago Cody Piersall #15

Python 3.4+ has an object-oriented path module: pathlib. Using this new module, you can check whether a file exists like this:

import pathlib
p = pathlib.Path('path/to/file')
if p.is_file():  # or p.is_dir() to see if it is a directory
    # do stuff

You can (and usually should) still use a try/except block when opening files:

    with as f:
        # do awesome stuff
except OSError:
    print('Well darn.')

The pathlib module has lots of cool stuff in it: convenient globbing, checking file's owner, easier path joining, etc. It's worth checking out. If you're on an older Python (version 2.6 or later), you can still install pathlib with pip:

# installs pathlib2 on older Python versions
# the original third-party module, pathlib, is no longer maintained.
pip install pathlib2

Then import it as follows:

# Older Python versions
import pathlib2 as pathlib

answered 4 years ago Chris #16

You can write Brian's suggestion without the try:.

from contextlib import suppress

with suppress(IOError), open('filename'):

suppress is part of Python 3.4. In older releases you can quickly write your own suppress:

from contextlib import contextmanager

def suppress(*exceptions):
    except exceptions:

answered 4 years ago user3197473 #17

You can use the following open method to check if a file exists + readable:

open(inputFile, 'r')

answered 4 years ago bergercookie #18

If the file is for opening you could use one of the following techniques:

>>> with open('somefile', 'xt') as f: #Using the x-flag, Python3.3 and above
...     f.write('Hello\n')

>>> if not os.path.exists('somefile'): 
...     with open('somefile', 'wt') as f:
...         f.write("Hello\n")
... else:
...     print('File already exists!')

answered 4 years ago Hanson #19

To check if a file exists,

from sys import argv

from os.path import exists
script, filename = argv
target = open(filename)
print "file exists: %r" % exists(filename)

answered 4 years ago Pradip Das #20

You can use the "OS" library of Python:

>>> import os
>>> os.path.exists("C:\\Users\\####\\Desktop\\test.txt") 
>>> os.path.exists("C:\\Users\\####\\Desktop\\test.tx")

answered 4 years ago Zizouz212 #21

Although I always recommend using try and except statements, here are a few possibilities for you (my personal favourite is using os.access):

  1. Try opening the file:

    Opening the file will always verify the existence of the file. You can make a function just like so:

    def File_Existence(filepath):
        f = open(filepath)
        return True

    If it's False, it will stop execution with an unhanded IOError or OSError in later versions of Python. To catch the exception, you have to use a try except clause. Of course, you can always use a try except` statement like so (thanks to hsandt for making me think):

    def File_Existence(filepath):
            f = open(filepath)
        except IOError, OSError: # Note OSError is for later versions of Python
            return False
        return True
  2. Use os.path.exists(path):

    This will check the existence of what you specify. However, it checks for files and directories so beware about how you use it.

    import os.path
    >>> os.path.exists("this/is/a/directory")
    >>> os.path.exists("this/is/a/file.txt")
    >>> os.path.exists("not/a/directory")
  3. Use os.access(path, mode):

    This will check whether you have access to the file. It will check for permissions. Based on the documentation, typing in os.F_OK, it will check the existence of the path. However, using this will create a security hole, as someone can attack your file using the time between checking the permissions and opening the file. You should instead go directly to opening the file instead of checking its permissions. (EAFP vs LBYP). If you're not going to open the file afterwards, and only checking its existence, then you can use this.

    Anyway, here:

    >>> import os
    >>> os.access("/is/a/file.txt", os.F_OK)

I should also mention that there are two ways that you will not be able to verify the existence of a file. Either the issue will be permission denied or no such file or directory. If you catch an IOError, set the IOError as e (like my first option), and then type in print(e.args) so that you can hopefully determine your issue. I hope it helps! :)

answered 3 years ago Pedro Lobito #22

if os.path.isfile(path_to_file):
    except IOError as e:
        print "Unable to open file"

Raising exceptions is considered to be an acceptable, and Pythonic, approach for flow control in your program. Consider handling missing files with IOErrors. In this situation, an IOError exception will be raised if the file exists but the user does not have read permissions.


answered 3 years ago Bishop #23

import os
#Your path here e.g. "C:\Program Files\text.txt"
#For access purposes: "C:\\Program Files\\text.txt"
if os.path.exists("C:\..."):   
    print "File found!"
    print "File not found!"

Importing os makes it easier to navigate and perform standard actions with your operating system.

For reference also see How to check whether a file exists using Python?

If you need high-level operations, use shutil.

answered 3 years ago Khaled.K #24

import os.path

def isReadableFile(file_path, file_name):
    full_path = file_path + "/" + file_name
        if not os.path.exists(file_path):
            print "File path is invalid."
            return False
        elif not os.path.isfile(full_path):
            print "File does not exist."
            return False
        elif not os.access(full_path, os.R_OK):
            print "File cannot be read."
            return False
            print "File can be read."
            return True
    except IOError as ex:
        print "I/O error({0}): {1}".format(ex.errno, ex.strerror)
    except Error as ex:
        print "Error({0}): {1}".format(ex.errno, ex.strerror)
    return False

path = "/usr/khaled/documents/puzzles"
fileName = "puzzle_1.txt"

isReadableFile(path, fileName)

answered 3 years ago Aaron Hall #25

How do I check whether a file exists, using Python, without using a try statement?

Now available since Python 3.4, import and instantiate a Path object with the file name, and check the is_file method (note that this returns True for symlinks pointing to regular files as well):

>>> from pathlib import Path
>>> Path('/').is_file()
>>> Path('/initrd.img').is_file()
>>> Path('/doesnotexist').is_file()

If you're on Python 2, you can backport the pathlib module from pypi, pathlib2, or otherwise check isfile from the os.path module:

>>> import os
>>> os.path.isfile('/')
>>> os.path.isfile('/initrd.img')
>>> os.path.isfile('/doesnotexist')

Now the above is probably the best pragmatic direct answer here, but there's the possibility of a race condition (depending on what you're trying to accomplish), and the fact that the underlying implementation uses a try, but Python uses try everywhere in its implementation.

Because Python uses try everywhere, there's really no reason to avoid an implementation that uses it.

But the rest of this answer attempts to consider these caveats.

Longer, much more pedantic answer

Available since Python 3.4, use the new Path object in pathlib. Note that .exists is not quite right, because directories are not files (except in the unix sense that everything is a file).

>>> from pathlib import Path
>>> root = Path('/')
>>> root.exists()

So we need to use is_file:

>>> root.is_file()

Here's the help on is_file:

    Whether this path is a regular file (also True for symlinks pointing
    to regular files).

So let's get a file that we know is a file:

>>> import tempfile
>>> file = tempfile.NamedTemporaryFile()
>>> filepathobj = Path(
>>> filepathobj.is_file()
>>> filepathobj.exists()

By default, NamedTemporaryFile deletes the file when closed (and will automatically close when no more references exist to it).

>>> del file
>>> filepathobj.exists()
>>> filepathobj.is_file()

If you dig into the implementation, though, you'll see that is_file uses try:

def is_file(self):
    Whether this path is a regular file (also True for symlinks pointing
    to regular files).
        return S_ISREG(self.stat().st_mode)
    except OSError as e:
        if e.errno not in (ENOENT, ENOTDIR):
        # Path doesn't exist or is a broken symlink
        # (see
        return False

Race Conditions: Why we like try

We like try because it avoids race conditions. With try, you simply attempt to read your file, expecting it to be there, and if not, you catch the exception and perform whatever fallback behavior makes sense.

If you want to check that a file exists before you attempt to read it, and you might be deleting it and then you might be using multiple threads or processes, or another program knows about that file and could delete it - you risk the chance of a race condition if you check it exists, because you are then racing to open it before its condition (its existence) changes.

Race conditions are very hard to debug because there's a very small window in which they can cause your program to fail.

But if this is your motivation, you can get the value of a try statement by using the suppress context manager.

Avoiding race conditions without a try statement: suppress

Python 3.4 gives us the suppress context manager (previously the ignore context manager), which does semantically exactly the same thing in fewer lines, while also (at least superficially) meeting the original ask to avoid a try statement:

from contextlib import suppress
from pathlib import Path


>>> with suppress(OSError), Path('doesnotexist').open() as f:
...     for line in f:
...         print(line)
>>> with suppress(OSError):
...     Path('doesnotexist').unlink()

For earlier Pythons, you could roll your own suppress, but without a try will be more verbose than with. I do believe this actually is the only answer that doesn't use try at any level in the Python that can be applied to prior to Python 3.4 because it uses a context manager instead:

class suppress(object):
    def __init__(self, *exceptions):
        self.exceptions = exceptions
    def __enter__(self):
        return self
    def __exit__(self, exc_type, exc_value, traceback):
        if exc_type is not None:
            return issubclass(exc_type, self.exceptions)

Perhaps easier with a try:

from contextlib import contextmanager

def suppress(*exceptions):
    except exceptions:

Other options that don't meet the ask for "without try":


import os

from the docs:


Return True if path is an existing regular file. This follows symbolic links, so both islink() and isfile() can be true for the same path.

But if you examine the source of this function, you'll see it actually does use a try statement:

# This follows symbolic links, so both islink() and isdir() can be true
# for the same path on systems that support symlinks
def isfile(path):
    """Test whether a path is a regular file"""
        st = os.stat(path)
    except os.error:
        return False
    return stat.S_ISREG(st.st_mode)
>>> OSError is os.error

All it's doing is using the given path to see if it can get stats on it, catching OSError and then checking if it's a file if it didn't raise the exception.

If you intend to do something with the file, I would suggest directly attempting it with a try-except to avoid a race condition:

    with open(path) as f:
except OSError:


Available for Unix and Windows is os.access, but to use you must pass flags, and it does not differentiate between files and directories. This is more used to test if the real invoking user has access in an elevated privilege environment:

import os
os.access(path, os.F_OK)

It also suffers from the same race condition problems as isfile. From the docs:

Note: Using access() to check if a user is authorized to e.g. open a file before actually doing so using open() creates a security hole, because the user might exploit the short time interval between checking and opening the file to manipulate it. It’s preferable to use EAFP techniques. For example:

if os.access("myfile", os.R_OK):
    with open("myfile") as fp:
return "some default data"

is better written as:

    fp = open("myfile")
except IOError as e:
    if e.errno == errno.EACCES:
        return "some default data"
    # Not a permission error.
    with fp:

Avoid using os.access. It is a low level function that has more opportunities for user error than the higher level objects and functions discussed above.

Criticism of another answer:

Another answer says this about os.access:

Personally, I prefer this one because under the hood, it calls native APIs (via "${PYTHON_SRC_DIR}/Modules/posixmodule.c"), but it also opens a gate for possible user errors, and it's not as Pythonic as other variants:

This answer says it prefers a non-Pythonic, error-prone method, with no justification. It seems to encourage users to use low-level APIs without understanding them.

It also creates a context manager which, by unconditionally returning True, allows all Exceptions (including KeyboardInterrupt and SystemExit!) to pass silently, which is a good way to hide bugs.

This seems to encourage users to adopt poor practices.

answered 3 years ago Love and peace - Joe Codeswell #26

Here's a 1 line Python command for the Linux command line environment. I find this VERY HANDY since I'm not such a hot Bash guy.

python -c "import os.path; print os.path.isfile('/path_to/')"

I hope this is helpful.

answered 3 years ago Unai Sainz de la Maza #27

In Python 3.4 the language provides a new module to manage files:

import pathlib
path = pathlib.Path('path/to/file')
if path.is_file(): # If you want to check a directory: path.is_dir()
    # If it is true, return true on your code.

answered 2 years ago KaiBuxe #28

In 2016 the best way is still using os.path.isfile:

>>> os.path.isfile('/path/to/some/file.txt')

Or in Python 3 you can use pathlib:

import pathlib
path = pathlib.Path('/path/to/some/file.txt')
if path.is_file():

answered 2 years ago Mike McKerns #29

I'm the author of a package that's been around for about 10 years, and it has a function that addresses this question directly. Basically, if you are on a non-Windows system, it uses Popen to access find. However, if you are on Windows, it replicates find with an efficient filesystem walker.

The code itself does not use a try block… except in determining the operating system and thus steering you to the "Unix"-style find or the hand-buillt find. Timing tests showed that the try was faster in determining the OS, so I did use one there (but nowhere else).

>>> import pox
>>> pox.find('*python*', type='file', root=pox.homedir(), recurse=False)

And the doc…

>>> print pox.find.__doc__
find(patterns[,root,recurse,type]); Get path to a file or directory

    patterns: name or partial name string of items to search for
    root: path string of top-level directory to search
    recurse: if True, recurse down from root directory
    type: item filter; one of {None, file, dir, link, socket, block, char}
    verbose: if True, be a little verbose about the search

    On some OS, recursion can be specified by recursion depth (an integer).
    patterns can be specified with basic pattern matching. Additionally,
    multiple patterns can be specified by splitting patterns with a ';'
    For example:
        >>> find('pox*', root='..')
        ['/Users/foo/pox/pox', '/Users/foo/pox/scripts/']

        >>> find('*shutils*;*init*')
        ['/Users/foo/pox/pox/', '/Users/foo/pox/pox/']


The implementation, if you care to look, is here:

answered 2 years ago iPhynx #30

You can use os.listdir to check if a file is in a certain directory.

import os
if 'file.ext' in os.listdir('dirpath'):

answered 2 years ago Marcel Wilson #31

Adding one more slight variation which isn't exactly reflected in the other answers.

This will handle the case of the file_path being None or empty string.

def file_exists(file_path):
    if not file_path:
        return False
    elif not os.path.isfile(file_path):
        return False
        return True

Adding a variant based on suggestion from Shahbaz

def file_exists(file_path):
    if not file_path:
        return False
        return os.path.isfile(file_path)

Adding a variant based on suggestion from Peter Wood

def file_exists(file_path):
    return file_path and os.path.isfile(file_path):

answered 2 years ago Tom Fuller #32

Testing for files and folders with os.path.isfile(), os.path.isdir() and os.path.exists()

Assuming that the "path" is a valid path, this table shows what is returned by each function for files and folders:

enter image description here

You can also test if a file is a certain type of file using os.path.splitext() to get the extension (if you don't already know it)

>>> import os
>>> path = "path to a word document"
>>> os.path.isfile(path)
>>> os.path.splitext(path)[1] == ".docx" # test if the extension is .docx

answered 2 years ago Inconnu #33

How do I check whether a file exists, without using the try statement?

In 2016, this is still arguably the easiest way to check if both a file exists and if it is a file:

import os
os.path.isfile('./file.txt')    # Returns True if exists, else False

isfile is actually just a helper method that internally uses os.stat and stat.S_ISREG(mode) underneath. This os.stat is a lower-level method that will provide you with detailed information about files, directories, sockets, buffers, and more. More about os.stat here

Note: However, this approach will not lock the file in any way and therefore your code can become vulnerable to "time of check to time of use" (TOCTTOU) bugs.

So raising exceptions is considered to be an acceptable, and Pythonic, approach for flow control in your program. And one should consider handling missing files with IOErrors, rather than if statements (just an advice).

answered 1 year ago CristiFati #34

2017 / 12 / 22:

Although almost every possible way has been listed in (at least one of) the existing answers (e.g. Python 3.4 specific stuff was added), I'll try to group everything together.

Note: every piece of Python standard library code that I'm going to post, belongs to version 3.5.3 (doc quotes are version 3 specific).

Problem statement:

  1. Check file (arguable: also folder ("special" file) ?) existence
  2. Don't use try / except / else / finally blocks

Possible solutions:

  1. [Python]: os.path.exists(path) (also check other function family members like os.path.isfile, os.path.isdir, os.path.lexists for slightly different behaviors)


    Return True if path refers to an existing path or an open file descriptor. Returns False for broken symbolic links. On some platforms, this function may return False if permission is not granted to execute os.stat() on the requested file, even if the path physically exists.

    All good, but if following the import tree:

    • os.path - (

      •, line ~#20+

        def exists(path):
            """Test whether a path exists.  Returns False for broken symbolic links"""
                st = os.stat(path)
            except os.error:
                return False
            return True

    it's just a try/except block around [Python]: os.stat(path, *, dir_fd=None, follow_symlinks=True). So, your code is try/except free, but lower in the framestack there's (at least) one such block. This also applies to other funcs (including os.path.isfile).

    1.1. [Python]: pathlib.Path.is_file()

    • It's a fancier (and more pythonic) way of handling paths, but
    • Under the hood, it does exactly the same thing (, line ~#1330):

      def is_file(self):
          Whether this path is a regular file (also True for symlinks pointing
          to regular files).
              return S_ISREG(self.stat().st_mode)
          except OSError as e:
              if e.errno not in (ENOENT, ENOTDIR):
              # Path doesn't exist or is a broken symlink
              # (see
              return False
  2. [Python]: With Statement Context Managers. Either:

    • Create one:

      class Swallow:  # Dummy example
          swallowed_exceptions = (FileNotFoundError,)
          def __enter__(self):
          def __exit__(self, exc_type, exc_value, exc_traceback):
              print("Exiting:", exc_type, exc_value, exc_traceback)
              return exc_type in Swallow.swallowed_exceptions  # only swallow FileNotFoundError (not e.g. TypeError - if the user passes a wrong argument like None or float or ...)
      • And its usage - I'll replicate the isfile behavior (note that this is just for demonstrating purposes, do not attempt to write such code for production):

        import os
        import stat
        def isfile_seaman(path):  # Dummy func
            result = False
            with Swallow():
                result = stat.S_ISREG(os.stat(path).st_mode)
            return result
    • Use [Python]: contextlib.suppress(*exceptions) - which was specifically designed for selectively suppressing exceptions

    But, they seem to be wrappers over try/except/else/finally blocks, as [Python]: The with statement states:

    This allows common try...except...finally usage patterns to be encapsulated for convenient reuse.

  3. Filesystem traversal functions (and search the results for matching item(s))

    Since these iterate over folders, (in most of the cases) they are inefficient for our problem (there are exceptions, like non wildcarded globbing - as @ShadowRanger pointed out), so I'm not going to insist on them. Not to mention that in some cases, filename processing might be required.

  4. [Python]: os.access(path, mode, *, dir_fd=None, effective_ids=False, follow_symlinks=True) whose behavior is close to os.path.exists (actually it's wider, mainly because of the 2nd argument)

    • user permissions might restrict the file "visibility" as the doc states:

      ...test if the invoking user has the specified access to path. mode should be F_OK to test the existence of path...

    os.access("/tmp", os.F_OK)

    Since I also work in C, I use this method as well because under the hood, it calls native APIs (again, via "${PYTHON_SRC_DIR}/Modules/posixmodule.c"), but it also opens a gate for possible user errors, and it's not as Pythonic as other variants. So, as @AaronHall rightly pointed out, don't use it unless you know what you're doing:

    Note: calling native APIs is also possible via [Python]: ctypes — A foreign function library for Python, but in most cases it's more complicated.

    (Win specific): Since msvcr*(vcruntime*) exports a [MSDN]: _access, _waccess function family as well, here's an example:

    Python 3.5.3 (v3.5.3:1880cb95a742, Jan 16 2017, 16:02:32) [MSC v.1900 64 bit (AMD64)] on win32
    Type "help", "copyright", "credits" or "license" for more information.
    >>> import os, ctypes
    >>> ctypes.CDLL("msvcrt")._waccess(u"C:\\Windows\\System32\\cmd.exe", os.F_OK)
    >>> ctypes.CDLL("msvcrt")._waccess(u"C:\\Windows\\System32\\___cmd.exe", os.F_OK)


    • Although it's not a good practice, I'm using os.F_OK in the call, but that's just for clarity (its value is 0)
    • I'm using _waccess so that the same code works on Python3 and Python2 (in spite of unicode related differences between them)
    • Although this targets a very specific area, it was not mentioned in any of the previous answers

    The Lnx (Ubtu (16 x64)) counterpart as well:

    Python 3.5.2 (default, Nov 17 2016, 17:05:23)
    [GCC 5.4.0 20160609] on linux
    Type "help", "copyright", "credits" or "license" for more information.
    >>> import os, ctypes
    >>> ctypes.CDLL("/lib/x86_64-linux-gnu/").access(b"/tmp", os.F_OK)
    >>> ctypes.CDLL("/lib/x86_64-linux-gnu/").access(b"/tmp1", os.F_OK)


    • Instead hardcoding libc's path ("/lib/x86_64-linux-gnu/") which may (and most likely, will) vary across systems, None (or the empty string) can be passed to CDLL constructor (ctypes.CDLL(None).access(b"/tmp", os.F_OK)). According to [man]: DLOPEN(3):

      If filename is NULL, then the returned handle is for the main program. When given to dlsym(), this handle causes a search for a symbol in the main program, followed by all shared objects loaded at program startup, and then all shared objects loaded by dlopen() with the flag RTLD_GLOBAL.

      • Main (current) program (python) is linked against libc, so its symbols (including access) will be loaded
      • This has to be handled with care, since functions like main, Py_Main and (all the) others are available; calling them could have disastrous effects (on the current program)
      • This doesn't also apply to Win (but that's not such a big deal, since msvcrt.dll is located in "%SystemRoot%\System32" which is in %PATH% by default). I wanted to take things further and replicate this behavior on Win (and submit a patch), but as it turns out, [MSDN]: GetProcAddress function only "sees" exported symbols, so unless someone declares the functions in the main executable as __declspec(dllexport) (why on Earth the regular person would do that?), the main program is loadable but pretty much unusable
  5. Install some 3rd Party module with filesystem capabilities

    Most likely, will rely on one of the ways above (maybe with slight customizations).
    One example would be (again, Win specific) [GitHub]: Python for Windows (pywin32) Extensions, which is a Python wrapper over WINAPIs.

    But, since this is more like a workaround, I'm stopping here.

  6. Another (lame) workaround (gainarie) is (as I like to call it,) the sysadmin approach: use Python as a wrapper to execute shell commands

    • Win:

      (py35x64_test) e:\Work\Dev\StackOverflow\q000082831>"e:\Work\Dev\VEnvs\py35x64_test\Scripts\python.exe" -c "import os; print(os.system('dir /b \"C:\\Windows\\System32\\cmd.exe\" > nul 2>&1'))"
      (py35x64_test) e:\Work\Dev\StackOverflow\q000082831>"e:\Work\Dev\VEnvs\py35x64_test\Scripts\python.exe" -c "import os; print(os.system('dir /b \"C:\\Windows\\System32\\cmd.exe.notexist\" > nul 2>&1'))"
    • Lnx (Ubtu):

      [[email protected]:~]> python3 -c "import os; print(os.system('ls \"/tmp\" > /dev/null 2>&1'))"
      [[email protected]:~]> python3 -c "import os; print(os.system('ls \"/tmp.notexist\" > /dev/null 2>&1'))"

Bottom line:

  • Do use try / except / else / finally blocks, because they can prevent you running into a series of nasty problems. A counter-example that I can think of, is performance: such blocks are costly, so try not to place them in code that it's supposed to run hundreds of thousands times per second (but since (in most cases) it involves disk access, it won't be the case).

Final note(s):

  • I will try to keep it up to date, any suggestions are welcome, I will incorporate anything useful that will come up into the answer

answered 11 months ago durjoy #35

If you imported NumPy already for other purposes then there is no need to import other libraries like pathlib, os, paths, etc.

import numpy as np np.DataSource().exists("path/to/your/file")

will return true or false based on its existence.

answered 8 months ago Prashant Godhani #36

In this case you can check if the file name exist in listdir or not. If not, then simply the file does not exist.

import os
filename = 'D:\\Python\\Python27\\prz\\passwords\\alpa.txt'
li = filename.split('\\')
st = ''
for i in range(0, len(li)-1):
    st = st + li[i] + '\\'
dirlist = os.listdir(st)
if(li[len(li)-1] in dirlist):
    fp = open(filename, 'r')
    print "FILE NOT FOUND"

This works perfect.

answered 8 months ago JawSaw #37


Every possible solution has been listed in other answers.

An intuitive and arguable way to check if a file exists is the following:

import os
os.path.isfile('~/')    # Returns True if exists, else False
additionaly check a dir
os.path.isdir('~/folder') # Returns True if the folder exists, else False
check either a dir or a file

I made an exhaustive cheatsheet for your reference:

#os.path methods in exhaustive cheatsheet
{'definition': ['dirname',
'operation': ['split', 'splitdrive', 'splitext',
               'join', 'normcase'],
'compare': ['samefile', 'sameopenfile', 'samestat'],
'condition': ['isdir',
 'expand': ['expanduser',
 'stat': ['getatime', 'getctime', 'getmtime',

answered 5 months ago Ali Hallaji #38

Check file or directory exists

You can follow these three ways:

Note1: The os.path.isfile used only for files

import os.path
os.path.isfile(filename) # True if file exists
os.path.isfile(dirname) # False if directory exists

Note2: The os.path.exists used for both files and directories

import os.path
os.path.exists(filename) # True if file exists
os.path.exists(dirname) #True if directory exists

The pathlib.Path method (included in Python 3+, installable with pip for Python 2)

from pathlib import Path

answered 2 months ago job #39

I found this string while trying to build the following functionality:

  1. Check if a file is present in static directories within a Django view
  2. If the file is present, serve it in template
  3. If the file is not present - serve a dialogue in template saying file not present.

Having worked through the various solutions I found that no single solution worked in both the development environment (with local static files) and production environment (using S3 storage).

Thus the following:

import os.path
import urllib2
from django.templatetags.static import static
from django.contrib.staticfiles import finders

## Determine if file present to serve
doc_title = 'document.pdf'
doc_available = False
    if site_settings.CLOUD_STATIC :
        doc_path = static('app/document/%s' % doc_title)
            ret = urllib2.urlopen(doc_path)
        except Exception as e:
            if ret.code == 200:
                doc_available = True
        doc_path = finders.find('app/document/%s' % doc_title)
        doc_available = os.path.isfile(doc_path)

except Exception as e:
    doc_path = None

The answer may not be very Pythonic so please improve.

site_settings.CLOUD_STATIC is a settings bool that determines if cloud storage should be used based on environment (could use site_settings.DEBUG).

I thus pass the doc_available bool to the template and process accordingly.

comments powered by Disqus