Error opening file - Python 3.3

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, Jan 26 '13 at 19:31
@MartijnPieters [Full code](http://pastebin.com/qEnHgQHP) - it's only few lines, so I'm not sure what could be wrong with it, really — P.K., Jan 26 '13 at 19:34

Rodrigue · Accepted Answer · 2013-01-26T22:05:29.430

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')

edited Jan 26 '13 at 22:05

answered Jan 26 '13 at 19:30

Rodrigue

I already stated that in a comment, but the OP is listing the *current directory*. – Martijn Pieters Jan 26 '13 at 19:30
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. Jan 26 '13 at 19:31
@P.K.: relative paths *should* work just fine. – Martijn Pieters Jan 26 '13 at 19:32
Oh well, that worked, no errors, thanks! – P.K. Jan 26 '13 at 19:43

1 Answers1