Python open() gives FileNotFoundError/IOError: Errno 2 No such file or directory

Santiago picture Santiago · Aug 30, 2012 · Viewed 401.3k times · Source

For some reason my code is having trouble opening a simple file:

This is the code:

file1 = open('recentlyUpdated.yaml')

And the error is:

IOError: [Errno 2] No such file or directory: 'recentlyUpdated.yaml'
  • Naturally I checked that this is the correct name of the file.
  • I have tried moving around the file, giving open() the full path to the file and none of it seems to work.

Answer

Lanaru picture Lanaru · Aug 30, 2012
  • Make sure the file exists: use os.listdir() to see the list of files in the current working directory
  • Make sure you're in the directory you think you're in with os.getcwd() (if you launch your code from an IDE, you may well be in a different directory)
  • You can then either:
    • Call os.chdir(dir), dir being the folder where the file is located, then open the file with just its name like you were doing.
    • Specify an absolute path to the file in your open call.
  • Remember to use a raw string if your path uses backslashes, like so: dir = r'C:\Python32'
    • If you don't use raw-string, you have to escape every backslash: 'C:\\User\\Bob\\...'
    • Forward-slashes also work on Windows 'C:/Python32' and do not need to be escaped.

Let me clarify how Python finds files:

  • An absolute path is a path that starts with your computer's root directory, for example 'C:\Python\scripts..' if you're on Windows.
  • A relative path is a path that does not start with your computer's root directory, and is instead relative to something called the working directory. You can view Python's current working directory by calling os.getcwd().

If you try to do open('sortedLists.yaml'), Python will see that you are passing it a relative path, so it will search for the file inside the current working directory. Calling os.chdir will change the current working directory.

Example: Let's say file.txt is found in C:\Folder.

To open it, you can do:

os.chdir(r'C:\Folder')
open('file.txt') #relative path, looks inside the current working directory

or

open(r'C:\Folder\file.txt') #full path