#### [SOLVED] How do I check whether a file exists without exceptions?

By spence91

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

#### @Devbrat Shukla 2020-05-12 14:35:54

use os.path.exists() to get file exist or not.

path = '/home/ie/SachinSaga/scripts/subscription_unit_reader_file/'
def file_check_at_location(filename):
return os.path.exists(path + str(filename).replace(' ',''))

file_name="dummy.txt"
responce = file_check_at_location(file_name)

if responce:
print('file found at location')
else:


#### @Yugal Jindle 2012-01-16 05:57:12

import os

PATH = './file.txt'
if os.path.isfile(PATH) and os.access(PATH, os.R_OK):
else:
print("Either the file is missing or not readable")


#### @wim 2013-04-09 05:45:27

having multiple conditions, some of which are superfluous, is less clear and explicit.

#### @Marquis of Lorne 2018-03-13 00:01:19

It is also redundant. If the file doesn't exist, os.access() will return false.

#### @e-info128 2018-07-16 21:30:18

@EJP In linux files can exist but not accesible.

#### @Jester 2018-08-24 13:10:26

since you import os, you do not need to import os.path again as it is already part of os. You just need to import os.path if you are only going to use functions from os.path and not from os itself, to import a smaller thing, but as you use os.access and os.R_OK, the second import is not needed.

#### @Martin Meeser 2020-07-02 07:04:45

Checking if the user has access rights to read the file is very professional. Often data is on local drive during dev, and on network share in prod. Then this might lead to such a situation. Also, the code is perfectly clear and readable and explicit.

#### @bortzmeyer 2008-09-17 15:01:14

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 some simple REPL output:

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


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:


UPDATE

Just to avoid confusion and based on the answers I got, current answer finds either a file or a directory with the given name.

#### @Chris Johnson 2015-08-01 13:55:01

This answer is wrong. os.path.exists returns true for things that aren't files, such as directories. This gives false positives. See the other answers that recommend os.path.isfile.

#### @Zorglub29 2018-05-19 21:33:55

got the false positive problem also.

#### @JayRizzo 2018-08-31 23:24:22

docs.python.org/3/library/os.path.html#os.path.exists To the above statement from chris >>os.path.exists(path) > 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. Changed in version 3.3: path can now be an integer: True is returned if it is an open file descriptor, False otherwise. Changed in version 3.6: Accepts a path-like object.

#### @user3197473 2014-05-23 10:01:20

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

file = open(inputFile, 'r')
file.close()


#### @Sam Dolan 2020-02-19 22:06:50

This definitely throws an exception if the file isn't there....

#### @Gopinath 2020-01-26 23:28:23

exists() and is_file() methods of 'Path' object can be used for checking if a given path exists and is a file.

Python 3 program to check if a file exists:

# File name:  check-if-file-exists.py

from pathlib import Path

filePath = Path(input("Enter path of the file to be found: "))

if filePath.exists() and filePath.is_file():
print("Success: File exists")
else:
print("Error: File does not exist")


Output:

$python3 check-if-file-exists.py Enter path of the file to be found: /Users/macuser1/stack-overflow/index.html Success: File exists$ python3 check-if-file-exists.py

Enter path of the file to be found: hghjg jghj

Error: File does not exist

#### @rslite 2008-09-17 12:57:51

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
os.path.isfile(fname)


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(strict=True) in a try block:

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


#### @makapuf 2018-06-20 07:58:09

concerning the first remark (use "try" if check before open) unfortunately this will not work if you want to open for appending being sure it exists before since 'a' mode will create if not exists.

#### @scottclowe 2019-03-29 13:44:15

Note that FileNotFoundError was introduced in Python 3. If you also need to support Python 2.7 as well as Python 3, you can use IOError instead (which FileNotFoundError subclasses) stackoverflow.com/a/21368457/1960959

#### @kyrill 2019-04-30 17:45:44

@makapuf You can open it for "updating" (open('file', 'r+')) and then seek to the end.

#### @theX 2020-07-02 21:26:18

Wait, so pathlib2 < pathlib? pathlib is for python3, right? I've been using pathlib2 thinking it was superior.

#### @The dark side of Gaming 2019-03-21 20:20:40

It isn't needed probably but if it is, here's some code

import os

def file_exists(path, filename):
for file_or_folder in os.listdir(path):
if file_or_folder == filename:
return True
return False


#### @moi 2019-10-08 12:46:25

This is overly complicated and unnecessarily wastes resources. Why? (Especially after there are already a bunch of useful answers...)

#### @CristiFati 2017-06-20 19:28:07

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.

Problem statement:

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

Possible solutions:

1. [Python 3]: os.path.exists(path) (also check other function family members like os.path.isfile, os.path.isdir, os.path.lexists for slightly different behaviors)

os.path.exists(path)


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 - posixpath.py (ntpath.py)

• genericpath.py, line ~#20+

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


it's just a try / except block around [Python 3]: 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).

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

def is_file(self):
"""
Whether this path is a regular file (also True for symlinks pointing
to regular files).
"""
try:
return S_ISREG(self.stat().st_mode)
except OSError as e:
if e.errno not in (ENOENT, ENOTDIR):
raise
# Path doesn't exist or is a broken symlink
# (see https://bitbucket.org/pitrou/pathlib/issue/12/)
return False

• Create one:

class Swallow:  # Dummy example
swallowed_exceptions = (FileNotFoundError,)

def __enter__(self):
print("Entering...")

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 os.path.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 3]: 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 3]: The with statement states:

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

2. 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.

3. [Python 3]: 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 3]: ctypes - A foreign function library for Python, but in most cases it's more complicated. (Win specific): Since vcruntime* (msvcr*) .dll exports a [MS.Docs]: _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) 0 >>> ctypes.CDLL("msvcrt")._waccess(u"C:\\Windows\\System32\\cmd.exe.notexist", os.F_OK) -1  Notes: • 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/libc.so.6").access(b"/tmp", os.F_OK) 0 >>> ctypes.CDLL("/lib/x86_64-linux-gnu/libc.so.6").access(b"/tmp.notexist", os.F_OK) -1  Notes: • Instead hardcoding libc's path ("/lib/x86_64-linux-gnu/libc.so.6") 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 [man7]: 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, [MS.Docs]: 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 4. Install some third-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]: mhammond/pywin32 - Python for Windows (pywin32) Extensions, which is a Python wrapper over WINAPIs. But, since this is more like a workaround, I'm stopping here. 5. 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'))" 0 (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'))" 1  • Nix (Lnx (Ubtu)): [[email protected]:~]> python3 -c "import os; print(os.system('ls \"/tmp\" > /dev/null 2>&1'))" 0 [[email protected]:~]> python3 -c "import os; print(os.system('ls \"/tmp.notexist\" > /dev/null 2>&1'))" 512  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 #### @sk8asd123 2017-11-19 01:46:40 Can you elaborate on this statement? "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)" #### @CristiFati 2017-11-19 23:54:34 @sk8asd123: Kind of hard to doo it in a comment: generally, it's best to use constants with functions that they come together with. That applies when working with multiple modules that define the same constant, because some might not be up to date, and it's best for the functions and constants to be in sync. When working with ctypes (calling the functions directly) I should have defined the constant (from MSDN), or not use a constant at all. It's just a guideline that I use, in 99.9% it probably makes no difference (functionally). #### @ShadowRanger 2017-11-29 18:29:02 @CristiFati: As of 3.6, glob.iglob (and glob.glob as well) are based on os.scandir, so it's lazy now; to get the first hit in a directory of 10M files, you only scan until you reach the first hit. And even pre-3.6, if you use glob methods w/o any wildcards, the function is smart: It knows you can only have one hit, so it simplifies the globbing to just os.path.isdir or os.path.lexists (depending on whether path ends in /). #### @ShadowRanger 2017-11-29 18:38:49 That second part of my comment (non-wildcarded globbing doesn't actually iterate the folder, and never has) does mean it's a perfectly efficient solution to the problem (slower than directly calling os.path.isdir or os.path.lexist since it's a bunch of Python level function calls and string operations before it decides the efficient path is viable, but no additional system call or I/O work, which is orders of magnitude slower). #### @Calculus 2017-12-04 08:51:49 Date:2017-12-04 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('~/file.md') # 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 os.path.exists('~/file')  I made an exhaustive cheatsheet for your reference: #os.path methods in exhaustive cheatsheet {'definition': ['dirname', 'basename', 'abspath', 'relpath', 'commonpath', 'normpath', 'realpath'], 'operation': ['split', 'splitdrive', 'splitext', 'join', 'normcase'], 'compare': ['samefile', 'sameopenfile', 'samestat'], 'condition': ['isdir', 'isfile', 'exists', 'lexists' 'islink', 'isabs', 'ismount',], 'expand': ['expanduser', 'expandvars'], 'stat': ['getatime', 'getctime', 'getmtime', 'getsize']}  #### @Vimal Maheedharan 2018-10-31 13:22:09 import os # for testing purpose args defaulted to current folder & file. # returns True if file found def file_exists(FOLDER_PATH='../', FILE_NAME=__file__): return os.path.isdir(FOLDER_PATH) \ and os.path.isfile(os.path.join(FOLDER_PATH, FILE_NAME))  Basically a folder check, then a file check with proper directory separator using os.path.join. #### @PierreBdR 2008-09-17 12:57:08 You have the os.path.exists function: import os.path os.path.exists(file_path)  This returns True for both files and directories but you can instead use os.path.isfile(file_path)  to test if it's a file specifically. It follows symlinks. #### @durjoy 2017-08-10 05:50:16 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")  This will return true or false based on its existence. #### @benefactual 2008-09-17 12:56:22 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  #### @Homunculus Reticulli 2020-04-03 17:10:44 Generally, not good practise to name variables the same as method names. #### @Ali Hallaji 2018-03-04 06:24:37 # 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 Path(filename).exists()  #### @Aaron Hall 2015-08-11 03:54:25 ## 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() False >>> Path('/initrd.img').is_file() True >>> Path('/doesnotexist').is_file() False  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('/') False >>> os.path.isfile('/initrd.img') True >>> os.path.isfile('/doesnotexist') False  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() True  So we need to use is_file: >>> root.is_file() False  Here's the help on is_file: is_file(self) 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(file.name) >>> filepathobj.is_file() True >>> filepathobj.exists() True  By default, NamedTemporaryFile deletes the file when closed (and will automatically close when no more references exist to it). >>> del file >>> filepathobj.exists() False >>> filepathobj.is_file() False  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). """ try: return S_ISREG(self.stat().st_mode) except OSError as e: if e.errno not in (ENOENT, ENOTDIR): raise # Path doesn't exist or is a broken symlink # (see https://bitbucket.org/pitrou/pathlib/issue/12/) 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  Usage: >>> 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 @contextmanager def suppress(*exceptions): try: yield except exceptions: pass  ## Other options that don't meet the ask for "without try": isfile import os os.path.isfile(path)  from the docs: os.path.isfile(path) 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""" try: st = os.stat(path) except os.error: return False return stat.S_ISREG(st.st_mode)  >>> OSError is os.error True  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: try: with open(path) as f: f.read() except OSError: pass  os.access 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 fp.read() return "some default data"  is better written as: try: fp = open("myfile") except IOError as e: if e.errno == errno.EACCES: return "some default data" # Not a permission error. raise else: with fp: return fp.read()  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.

#### @un33k 2013-06-27 13:38:04

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")
else:
print("no such file exists at this time")


#### @Isaac Supeene 2014-11-23 18:37:59

As long as you intend to access the file, the race condition does exist, regardless of how your program is constructed. Your program cannot guarantee that another process on the computer has not modified the file. It's what Eric Lippert refers to as an exogenous exception. You cannot avoid it by checking for the file's existence beforehand.

#### @un33k 2018-07-28 02:52:48

@IsaacSupeene Best practice is to make the window of (file) operation as small as possible followed by a proper exception handling

#### @Cody Piersall 2014-02-08 02:38:42

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:

try:
with p.open() 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


#### @Paul 2008-09-17 12:55:48

import os.path

if os.path.isfile(filepath):


#### @KaiBuxe 2016-02-24 12:44:10

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():
...


#### @Joko 2016-02-25 08:55:33

May I ask: What's the advantage of using the module 'pathlib' instead of the module 'os' in python3 for this checking?

#### @KaiBuxe 2016-02-25 10:44:49

pathlib is python's OOP solution for paths. You can do a lot more with it. If you just need to check existance, the advantage is not so big.

#### @Mike McKerns 2016-05-05 12:00:42

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)
['/Users/mmckerns/.python']


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/pox_launcher.py']

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

>>>


The implementation, if you care to look, is here: https://github.com/uqfoundation/pox/blob/89f90fb308f285ca7a62eabe2c38acb87e89dad9/pox/shutils.py#L190

#### @Inconnu 2016-12-02 06:39:08

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).

#### @Zizouz212 2014-12-26 20:05:32

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):
try:
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")
True
>>> os.path.exists("this/is/a/file.txt")
True
>>> os.path.exists("not/a/directory")
False

3. Use os.access(path, mode):

This will check whether you have access to the file. It will check for permissions. Based on the os.py 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)
True


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! :)

You can use the "OS" library of Python:

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


#### @Chris Johnson 2015-08-01 13:55:13

This answer is wrong. os.path.exists returns true for things that aren't files, such as directories. This gives false positives. See the other answers that recommend os.path.isfile.

@Chris Johnson , os.path.exists() function checks whether a path exists in system. PATH may be a DIRECTORY or FILE. It will work fine on both the cases. Please try with some example

#### @Debosmit Ray 2016-04-14 23:33:56

So, this answer works. Great. Iff the path isn't that of a file. Is that what the question was about? No.

#### @starturtle 2017-09-05 11:24:42

It depends. If the goal of determining the existence of a "file" is to find out whether the path already exists (and is therefore not a path where new data can be stored without deleting other information), then exists is fine. If the goal is to determine whether it's safe to open a presumably existing file, then the criticism is justified and exists is not precise enough. Sadly, the OP doesn't specify which is the desired goal (and probably won't do so any more).

#### @Chris 2014-02-10 21:30:20

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

from contextlib import suppress

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


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

from contextlib import contextmanager

@contextmanager
def suppress(*exceptions):
try:
yield
except exceptions:
pass


#### @philberndt 2011-01-25 23:00:01

You could try this (safer):

try:
# http://effbot.org/zone/python-with-statement.htm
# 'with' is safer to open a file
with open('whatever.txt') as fh:
# Do something with 'fh'
except IOError as e:
print("({})".format(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.

#### @rrs 2014-04-23 13:10:01

The original question asked for a solution that does not use try

#### @Chris Johnson 2016-02-17 18:52:12

This answer misses the point of the OP. Checking is a file exists is not the same as checking if you can open it. There will be cases where a file does exist but for a variety of reasons, you can't open it.

#### @loxsat 2015-05-25 18:29:22

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!"
else:


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.

#### @Chris Johnson 2015-08-01 13:56:22

This answer is wrong. os.path.exists returns true for things that aren't files, such as directories. This gives false positives. See the other answers that recommend os.path.isfile.

#### @Marcel Wilson 2016-08-05 15:54:37

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
else:
return True


Adding a variant based on suggestion from Shahbaz

def file_exists(file_path):
if not file_path:
return False
else:
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):


#### @Shahbaz 2017-01-04 22:50:16

if (x) return true; else return false; is really just return x. Your last four lines can become return os.path.isfile(file_path). While we're at it, the whole function can be simplified as return file_path and os.path.isfile(file_path).

#### @Marcel Wilson 2017-01-05 17:08:43

You have to be careful with return x in the case of if (x). Python will consider an empty string False in which case we would be returning an empty string instead of a bool. The purpose of this function is to always return bool.

#### @Shahbaz 2017-01-05 17:10:24

True. In this case however, x is os.path.isfile(..) so it's already bool.

#### @Marcel Wilson 2017-01-05 17:13:10

os.path.isfile(None) raises an exception which is why I added the if check. I could probably just wrap it in a try/except instead but I felt it was more explicit this way.

#### @Peter Wood 2017-04-06 10:35:01

return file_path and os.path.isfile(file_path)

#### @Tom Fuller 2016-10-08 12:43:02

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:

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)
True
>>> os.path.splitext(path)[1] == ".docx" # test if the extension is .docx
True


#### @iPhynx 2016-06-08 12:45:32

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

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

#### @Jean-François Fabre 2017-01-07 12:24:21

won't work in windows since filesystem isn't case sensitive. And very uneffective because it scans the whole directory.

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