How do I check whether a file exists or not, without using the try
statement?
-
8Note: your program will not be 100% robust if it cannot handle the case where a file already exists or doesn't exist at the time you actually try to open or create it respectively. The filesystem is concurrently accessible to multiple programs, so the existance-check you did prior to these actions might already be outdated by the time your program acts on it.– masterxiloCommented Nov 18, 2023 at 17:00
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@masterxilo You can just check again– KaiLandoCommented Feb 13 at 22:32
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@KaiLando Not realistically. It would require you to check if the file exists every time you read/write to it, and even then there is the edge case that it disappears between the if and the access.– Ted Klein BergmanCommented Feb 14 at 13:39
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Sorry @masterxilo, I put in the wrong one.– KaiLandoCommented Feb 15 at 2:19
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@KaiLando It's not enough. If it exists it might no longer exist after the check is completed.– Ted Klein BergmanCommented Feb 15 at 8:38
41 Answers
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
if you need to be sure it's a file.
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)
pathlib
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
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87concerning 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.– makapufCommented Jun 20, 2018 at 7:58
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20Note that
FileNotFoundError
was introduced in Python 3. If you also need to support Python 2.7 as well as Python 3, you can useIOError
instead (whichFileNotFoundError
subclasses) stackoverflow.com/a/21368457/1960959 Commented Mar 29, 2019 at 13:44 -
20@makapuf You can open it for "updating" (
open('file', 'r+')
) and then seek to the end.– kyrillCommented Apr 30, 2019 at 17:45 -
2Wait, so
pathlib2
<pathlib
?pathlib
is for python3, right? I've been usingpathlib2
thinking it was superior.– theXCommented Jul 2, 2020 at 21:26 -
7@kyrill: Opening a file for appending is not the same as opening it for writing and seeking to the end: When you have concurrent writers, they will overwrite each other without
'a'
.– hagelloCommented Nov 20, 2020 at 14:48
Use os.path.exists
to check both files and directories:
import os.path
os.path.exists(file_path)
Use os.path.isfile
to check only files (note: follows symbolic links):
os.path.isfile(file_path)
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
>>> os.path.exists("/etc/password.txt")
True
>>> os.path.exists("/etc")
True
>>> os.path.exists("/does/not/exist")
False
Use os.path.isfile()
with os.access()
:
import os
PATH = './file.txt'
if os.path.isfile(PATH) and os.access(PATH, os.R_OK):
print("File exists and is readable")
else:
print("Either the file is missing or not readable")
-
75having multiple conditions, some of which are superfluous, is less clear and explicit.– wimCommented Apr 9, 2013 at 5:45
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19It is also redundant. If the file doesn't exist,
os.access()
will return false. Commented Mar 13, 2018 at 0:01 -
12
-
13since you
import os
, you do not need toimport os.path
again as it is already part ofos
. You just need to importos.path
if you are only going to use functions fromos.path
and not fromos
itself, to import a smaller thing, but as you useos.access
andos.R_OK
, the second import is not needed.– JesterCommented Aug 24, 2018 at 13:10 -
7Checking 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. Commented Jul 2, 2020 at 7:04
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
-
10Generally, not good practise to name variables the same as method names. Commented Apr 3, 2020 at 17:10
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:
Check file (arguable: also folder ("special" file) ?) existence
Don't use try / except / else / finally blocks
Possible solutions:
1. [Python.Docs]: 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. ReturnsFalse
for broken symbolic links. On some platforms, this function may returnFalse
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.Docs]: 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 functions (including os.path.isfile).
1.1. [Python.Docs]: pathlib - Path.is_file()
It's a fancier (and more [Wiktionary]: 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
2. [Python.Docs]: With Statement Context Managers
Either:
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) # Only swallow FileNotFoundError (not e.g. TypeError - if the user passes a wrong argument like None or float or ...) return exc_type in Swallow.swallowed_exceptions
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.Docs]: 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.Docs]: Compound statements - The with statement states:
This allows common try...except...finally usage patterns to be encapsulated for convenient reuse.
3. Filesystem traversal functions
Search the results for matching item(s):
[Python.Docs]: os.listdir(path='.') (or [Python.Docs]: os.scandir(path='.') on Python v3.5+, backport: [PyPI]: scandir)
Under the hood, both use:
Nix: [Man7]: OPENDIR(3) / [Man7]: READDIR(3) / [Man7]: CLOSEDIR(3)
Win: [MS.Learn]: FindFirstFileW function (fileapi.h) / [MS.Learn]: FindNextFileW function (fileapi.h) / [MS.Learn]: FindClose function (fileapi.h)
via [GitHub]: python/cpython - (main) cpython/Modules/posixmodule.c
Using scandir() instead of listdir() can significantly increase the performance of code that also needs file type or file attribute information, because os.DirEntry objects expose this information if the operating system provides it when scanning a directory. All os.DirEntry methods may perform a system call, but is_dir() and is_file() usually only require a system call for symbolic links; os.DirEntry.stat() always requires a system call on Unix, but only requires one for symbolic links on Windows.
[Python.Docs]: os.walk(top, topdown=True, onerror=None, followlinks=False)
- Uses os.listdir (os.scandir when available)
[Python.Docs]: glob.iglob(pathname, *, root_dir=None, dir_fd=None, recursive=False, include_hidden=False) (or its predecessor: glob.glob)
- Doesn't seem a traversing function per se (at least in some cases), but it still uses os.listdir
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.Docs]: os.access(path, mode, *, dir_fd=None, effective_ids=False, follow_symlinks=True)
Its 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...
Security considerations:
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.
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, don't use it unless you know what you're doing:
Nix: [Man7]: ACCESS(2)
Warning: Using these calls to check if a user is authorized to, for example, open a file before actually doing so using open(2) creates a security hole, because the user might exploit the short time interval between checking and opening the file to manipulate it. For this reason, the use of this system call should be avoided.
As seen, this approach is highly discouraged (especially on Nix).
Note: calling native APIs is also possible via [Python.Docs]: ctypes - A foreign function library for Python, but in most cases it's more complicated. Before working with CTypes, check [SO]: C function called from Python via ctypes returns incorrect value (@CristiFati's answer) out.
(Win specific): since vcruntime###.dll (msvcr###.dll for older VStudio versions - I'm going to refer to it as UCRT) exports a [MS.Learn]: _access, _waccess function family as well, here's an example (note that the recommended [Python.Docs]: msvcrt - Useful routines from the MS VC++ runtime doesn't export them):
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 ctypes as cts, os >>> cts.CDLL("msvcrt")._waccess(u"C:\\Windows\\Temp", os.F_OK) 0 >>> cts.CDLL("msvcrt")._waccess(u"C:\\Windows\\Temp.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 Python 3 and Python 2 (in spite of [Wikipedia]: Unicode related differences between them - [SO]: Passing utf-16 string to a Windows function (@CristiFati's answer))
Although this targets a very specific area, it was not mentioned in any of the previous answers
The Linux (Ubuntu ([Wikipedia]: Ubuntu version history) 16 x86_64 (pc064)) 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 ctypes as cts, os >>> cts.CDLL("/lib/x86_64-linux-gnu/libc.so.6").access(b"/tmp", os.F_OK) 0 >>> cts.CDLL("/lib/x86_64-linux-gnu/libc.so.6").access(b"/tmp.notexist", os.F_OK) -1
Notes:
Instead hardcoding libc.so (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(3), 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 Windows (but that's not such a big deal, since UCRT is located in "%SystemRoot%\System32" which is in %PATH% by default). I wanted to take things further and replicate this behavior on Windows (and submit a patch), but as it turns out, [MS.Learn]: GetProcAddress function (libloaderapi.h) only "sees" exported symbols, so unless someone declares the functions in the main executable as
__declspec(dllexport)
(why on Earth the common person would do that?), the main program is loadable, but it is pretty much unusable
5. 3rd-party modules 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.
6. SysAdmin approach
I consider this a (lame) workaround (gainarie): use Python as a wrapper to execute shell commands:
Win:
(py35x64_test) [cfati@CFATI-5510-0:e:\Work\Dev\StackOverflow\q000082831]> "e:\Work\Dev\VEnvs\py35x64_test\Scripts\python.exe" -c "import os; print(os.system('dir /b \"C:\\Windows\\Temp\" > nul 2>&1'))" 0 (py35x64_test) [cfati@CFATI-5510-0:e:\Work\Dev\StackOverflow\q000082831]> "e:\Work\Dev\VEnvs\py35x64_test\Scripts\python.exe" -c "import os; print(os.system('dir /b \"C:\\Windows\\Temp.notexist\" > nul 2>&1'))" 1
Nix ([Wikipedia]: Unix-like) - Ubuntu:
[cfati@cfati-5510-0:/mnt/e/Work/Dev/StackOverflow/q000082831]> python3 -c "import os; print(os.system('ls \"/tmp\" > /dev/null 2>&1'))" 0 [cfati@cfati-5510-0:/mnt/e/Work/Dev/StackOverflow/q000082831]> 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 possible counterexample 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)
-
7@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). Commented Nov 19, 2017 at 23:54
-
5@CristiFati: As of 3.6,
glob.iglob
(andglob.glob
as well) are based onos.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 useglob
methods w/o any wildcards, the function is smart: It knows you can only have one hit, so it simplifies the globbing to justos.path.isdir
oros.path.lexists
(depending on whether path ends in/
). Commented Nov 29, 2017 at 18:29 -
5That 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
oros.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). Commented Nov 29, 2017 at 18:38
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
-
You can use
pathlib.Path.exists
, which covers more cases thanis_file
Commented Oct 11, 2020 at 21:13
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")
-
18As 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. Commented Nov 23, 2014 at 18:37
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1@IsaacSupeene Best practice is to make the window of (file) operation as small as possible followed by a proper exception handling Commented Jul 28, 2018 at 2:52
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The best would be to open the file, using try+catch and there is no time window. Commented Aug 31, 2021 at 8:50
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It is not unusual to have files that only exist for the benefit of one program, living in a file that noone else would reasonably open. In that case, I like this solution for its readability.– julaineCommented Mar 7 at 7:26
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()
andisfile()
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.
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:
- Style: Section "Handling unusual conditions" of these course notes for Software Design (2007)
- Avoiding Race Conditions
-
4
-
12The cited Avoiding Race Conditions (apple dev support) link does not support your answer. It concerns only using temporary files that contain sensitive information on poorly designed operating systems that don't properly sandbox temporary files / directories via restricted permissions. Using
try...except
doesn't help resolve that problem anyway.– jstineCommented Sep 28, 2015 at 15:38 -
The problem with this method, is that if you have an important piece of code depending on the file not existing, putting it in the
except:
clause will make that an exception arising in this part of your code will raise a confusing message (second error raised during the processing of the first one.)– CamionCommented May 24, 2019 at 10:43 -
But one typically opens files with context managers, and plenty of packages don't seem to implement them with terribly good exceptions. E.g. xarray just throws a super generic "ValueError" with some confusing message about the backend if you try to open say a folder instead of a netcdf file. Catching a ValueError here could mask all sorts of other problems. Commented May 25, 2023 at 0:08
Use:
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:
print "File not found!"
Importing os
makes it easier to navigate and perform standard actions with your operating system.
For reference, also see How do I check whether a file exists without exceptions?.
If you need high-level operations, use shutil
.
-
15This 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 recommendos.path.isfile
. Commented Aug 1, 2015 at 13:56
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
TL;DR
The answer is: use the pathlib
module
Pathlib is probably the most modern and convenient way for almost all of the file operations. For the existence of a file or a folder a single line of code is enough. If file is not exists, it will not throw any exception.
from pathlib import Path
if Path("myfile.txt").exists(): # works for both file and folders
# do your cool stuff...
The pathlib
module was introduced in Python 3.4, so you need to have Python 3.4+. This library makes your life much easier while working with files and folders, and it is pretty to use. Here is more documentation about it: pathlib — Object-oriented filesystem paths.
BTW, if you are going to reuse the path, then it is better to assign it to a variable.
So it will become:
from pathlib import Path
p = Path("loc/of/myfile.txt")
if p.exists(): # works for both file and folders
# do stuffs...
#reuse 'p' if needed.
-
7Be aware that this returns True if the file is not present but path to file exists. If you are really interested in asking whether the file exists or not you should be using p.is_file()– bravhekCommented Jun 25, 2021 at 21:38
-
1@bravhek I think your comment is confusing. What do you mean by 'path to file'? If you mean 'path to dir', and if the directory exists, then of course it works as it should. The OP said
p.exists()
works for both files and folders. Commented Mar 13, 2023 at 2:25 -
@starriet, The question is about the existence of the file, however p.exists() will return a false positive when the folder containing the checked file exists but the file itself doesn't.– BravhekCommented Mar 14, 2023 at 10:50
-
@Bravhek, can you please give an example to better highlight the issue, I may also update my answer accordingly, to better inform the users. Thank you.– MeminCommented Mar 14, 2023 at 14:19
-
@Memin, suppose you want to check if the file "myfile.txt", exists in folder "c:\temp". if you declare fpath = Path(r"c:\temp\myfile.txt"), and then check for fpath.exists(), it will return True but the file does not exists! (false positive)... in contrast fpath.is_file() will return false, wich is the desired output.– BravhekCommented Mar 14, 2023 at 19:09
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():
...
-
4
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.– KaiBuxeCommented Feb 25, 2016 at 10:44
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
try:
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
try:
f = open(filepath, 'wx')
except IOError:
print 'file already exists'
If you need file locking, that's a different matter.
-
5This 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 recommendos.path.isfile
. Commented Aug 1, 2015 at 13:54 -
6On your third example, I create a link named
filepath
with the right timing, and BAM, you overwrite the target file. You should doopen(filepath, 'wx')
in atry...except
block to avoid the issue.– spectrasCommented Aug 24, 2015 at 14:05 -
1In your second example, at least in Windows, you will get an
OSError
iffilepath + '.old'
already exists: "On Windows, if dst already exists, OSError will be raised even if it is a file; there may be no way to implement an atomic rename when dst names an existing file." Commented May 24, 2016 at 14:14 -
@TomMyddeltyn: As of Python 3.3,
os.replace
portably performs silent replacement of the destination file (it's identical toos.rename
's Linux behavior) (it only errors if the destination name exists and is a directory). So you're stuck on 2.x, but Py3 users have had a good option for several years now. Commented Nov 29, 2017 at 18:14 -
On the
rename
example: It should still be done withtry
/except
.os.rename
(oros.replace
on modern Python) is atomic; making it check then rename introduces an unnecessary race and additional system calls. Just dotry: os.replace(filepath, filepath + '.old') except OSError: pass
Commented Nov 29, 2017 at 18:17
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.
-
22The original question asked for a solution that does not use
try
– rrsCommented Apr 23, 2014 at 13:10 -
8This 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. Commented Feb 17, 2016 at 18:52
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
# Additionally, check a directory
os.path.isdir('~/folder') # Returns True if the folder exists, else False
# Check either a directory or a file
os.path.exists('~/file')
I made an exhaustive cheat sheet for your reference:
# os.path methods in exhaustive cheat sheet
{'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']}
Although I always recommend using try
and except
statements, here are a few possibilities for you (my personal favourite is using os.access
):
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
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
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! :)
If the file is for opening you could use one of the following techniques:
with open('somefile', 'xt') as f: # Using the x-flag, Python 3.3 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!')
Note: This finds either a file or a directory with the given name.
-
10This 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 recommendos.path.isfile
. Commented Aug 1, 2015 at 13:55 -
-
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.– JayRizzoCommented Aug 31, 2018 at 23:24
Additionally, os.access()
:
if os.access("myfile", os.R_OK):
with open("myfile") as fp:
return fp.read()
Being R_OK
, W_OK
, and X_OK
the flags to test for permissions (doc).
if os.path.isfile(path_to_file):
try:
open(path_to_file)
pass
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.
-
6The OP asked how to check if a file exists. It's possible for a file to exist but for you to not be able to open it. Therefore using opening a file as a proxy for checking if the file exists is not correct: will have false negatives. Commented Feb 17, 2016 at 18:58
Check file or directory exists
You can follow these three ways:
1. Using isfile()
Note 1: 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
2. Using exists
Note 2: The os.path.exists
is 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
3. The pathlib.Path
method (included in Python 3+, installable with pip for Python 2)
from pathlib import Path
Path(filename).exists()
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.
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
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
Here's a one-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/file.xxx')"
-
10One-line check in bash:
[ -f "${file}" ] && echo "file found" || echo "file not found"
(which is the same asif [ ... ]; then ...; else ...; fi
). Commented Oct 1, 2015 at 7:48 -
2Python would be a completely wrong tool for this job on the command line.– ruoholaCommented Oct 25, 2021 at 10:54
-
-
1On one-liners in Python: "...which makes Python close to useless for command-line one-liner programs ... You can get away with a sequence of simple statements, separated by semi-colon ... as soon as you add a construct that introduces an indented block (like if), you need the line break..." Commented Oct 4, 2022 at 20:08
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):
-
4
if (x) return true; else return false;
is really justreturn x
. Your last four lines can becomereturn os.path.isfile(file_path)
. While we're at it, the whole function can be simplified asreturn file_path and os.path.isfile(file_path)
.– ShahbazCommented Jan 4, 2017 at 22:50 -
You have to be careful with
return x
in the case ofif (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. Commented Jan 5, 2017 at 17:08 -
1True. In this case however,
x
isos.path.isfile(..)
so it's already bool.– ShahbazCommented Jan 5, 2017 at 17:10 -
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. Commented Jan 5, 2017 at 17:13 -
4
return file_path and os.path.isfile(file_path)
Commented Apr 6, 2017 at 10:35
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).
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
-
6This 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 recommendos.path.isfile
. Commented Aug 1, 2015 at 13:55 -
@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 Commented Aug 2, 2015 at 14:51
-
So, this answer works. Great. Iff the path isn't that of a file. Is that what the question was about? No. Commented Apr 14, 2016 at 23:33
-
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). Commented Sep 5, 2017 at 11:24