python if and - How do I check whether a file exists without exceptions?





15 Answers

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.

create php delete

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




import os.path

if os.path.isfile(filepath):



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



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")






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 to check whether a file exists using Python?

If you need high-level operations, use shutil.




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():
    ...



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.




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.



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!')

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.




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



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/file.xxx')"

I hope this is helpful.




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




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']}



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).




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