Error opening file - Python 3.3 [duplicate]

Question

Possible Duplicate:
IOError when trying to open existing files

I'm having problems opening a file with open() in python 3.3, any idea why?
I'm trying

import os

filelist = [ f for f in os.listdir( os.curdir )]
singleFile = filelist[a]
hppfile = open(singleFile, 'r')

And I get

FileNotFoundError: [Errno 2] No such file or directory: '-file that is actually inside the directory-'

Ideas?
On Windows, I just started this to learn this to write few quick scripts

os.listdir() already returns a list, and the default argument is already os.curdir btw. But I cannot reproduce this problem with files in the current directory, so I suspect your code sample does not match your real code. — Martijn Pieters
– Martijn Pieters, Commented Jan 26, 2013 at 19:31
@MartijnPieters Full code - it's only few lines, so I'm not sure what could be wrong with it, really — P.K.
– P.K., Commented Jan 26, 2013 at 19:34

Rodrigue · Accepted Answer · 2013-01-26 22:05:29Z

1

If you read the documentation for listdir you will see that it returns filenames and not full path.

You will need something like

current_dir_path = os.getcwd()
open(os.path.join(curren_dir_path, file), 'r')

answered Jan 26, 2013 at 19:30

Rodrigue

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4 Comments

I already stated that in a comment, but the OP is listing the current directory.

So in python you need whole path to open a file? I thought it could be a relative path like in C++ is in default

@P.K.: relative paths should work just fine.

Oh well, that worked, no errors, thanks!