CODEWITHSUNDEEP
pythonpython-2.7os.path
user7811364
user7811364

Reputation:

Why does os.path.abspath() return the path of cwd+file?

If I have a structure like such:

root/
-- group1/
---- names/
---- places/
------ foo.zip

Why is that when I call os.path.abspath('foo.zip') I get the file path of where the Python script is located plus the foo.zip?

Looks like: H:\Program\Scripts\foo.zip

Needs to be: H:\Progran\Groups\group1\names\places\foo.zip

This is the code I have for a function where the problem is resulting:

def unzip(in_dir):
    # in_dir is places passed to unzip()
    files = [f for f in os.listdir(os.path.abspath(in_dir)) if f.endswith('.zip')]
    for zip in files:
        # This prints the 'looks like' path above
        print os.path.abspath(zip)

Shouldn't print os.path.abspath(zip) give me the full path of each file that was found in the os.listdir(os.path.abspath(in_dir))?

Upvotes: 2

Views: 2175

Answers (2)

mkrieger1
mkrieger1

Reputation: 23139

Why does os.path.abspath() return the path of cwd+file?

Because that's quite literally what abspath is supposed to do:

os.path.abspath(path)

Return a normalized absolutized version of the pathname path. On most platforms, this is equivalent to calling the function normpath() as follows: normpath(join(os.getcwd(), path)).

(Emphasis mine)

Upvotes: 4

Barmar
Barmar

Reputation: 781068

os.path.abspath() has no idea where the name foo.zip came from -- it doesn't know it came from os.listdir() of some directory. So it doesn't know that's the correct directory to use as the prefix. A relative pathname is always interpreted relative to the current directory.

If you want to create the desired absolute pathname, use os.path.join:

print os.path.join(in_dir, zip)

Upvotes: 3

Related Questions