Reputation: 658
So I'm sure this is a stupid question, but I've looked through Python's documentation and attempted a couple of Google codes and none of them has worked.
It seems like the following should work, but it returns "False" for In my directory /foo/bar I have 3 items: 1 Folder "[Folder]", 1 file "test" (no extension), and 1 file "test.py".
I'm look to have a script that can distinguish folders from files for a bunch of functions, but I can't figure out anything that works.
#!/usr/bin/python
import os, re
for f in os.listdir('/foo/bar'):
print f, os.path.isdir(f)
Currently returns false for everything.
Upvotes: 3
Views: 1692
Reputation: 184200
This is because listdir()
returns the names of the files in /foo/bar
. When you later do os.path.isdir()
on one of these, the OS interprets it relative to the current working directory which is probably the directory your script is in, not /foo/bar
, and it probably does not contain a directory of the specified name. A path that doesn't exist is not a directory and so isdir()
returns False.
.
Use the complete pathname. Best way is to use os.path.join
, e.g., os.path.isdir(os.path.join('/foo/bar', f))
.
Upvotes: 6
Reputation: 50517
You might want to use os.walk
instead: http://docs.python.org/library/os.html#os.walk
When it returns the contents of the directory, it returns files and directories in separate lists, negating the need for checking.
So you could do:
import os
root, dirs, files = next(os.walk('/foo/bar'))
print 'directories:', dirs
print 'files:', files
Upvotes: 2
Reputation: 29727
I suppose that os.path.isdir(os.path.join('/foo/bar', f))
should work.
Upvotes: 0