«

(please don't top-post)

> > It does exist in Python 2.3, and it does take one argument (and optionally
> > others):
> >
> > http://python.org/doc/lib/os-file-dir.html#l2h-1473

> It does not exist in Python 2.2, AFAIK.
> It is a good idea to write code that at least works for
> one previous Python version, hence the comment.

You should have said so in the first place -- explicit is better than
implicit

-Andrew.

I see. I understand now. Thank you for the advice.

Bengt Richter wrote:
> On Thu, 20 Nov 2003 22:52:40 -0500, hokiegal99 <> wrote:
>
>
>>When I use 'file' instead of 'open' on the 4th line of this script (the
>>line that begins with "outputFile") I get this error:
>>"UnboundLocalError: local variable 'file' referenced before assignment"
>>'open' works w/o problem and 'file' works in some other scripts that are
>>almost identical to this one... any ideas? I can post a script where
>>'file' works if anyone is interested.
>
>
> No, UIAM this is a good lesson on why it's not a good idea to use names of builtins as
> variable names.
>
>>import os, string
>>setpath = raw_input("Enter the path to the Mac files and folders: ")
>>def clean_spaces(setpath):
>> outputFile = open('fix-spaces.txt', 'w')
>
> ^^^^-- file would be fine here if you didn't shadow the name below
>
>> for root, dirs, files in os.walk(setpath):
>> for dir in dirs:
>> old_dname = dir
>> new_dname = old_dname.strip()
>> if new_dname != old_dname:
>> newpath = os.path.join(root,new_dname)
>> oldpath = os.path.join(root,old_dname)
>> print >> outputFile, "Replaced ", old_dname, "\nWith
>> ", new_dname
>> os.rename(oldpath,newpath)
>> for file in files:
>
> ^^^^ this is effectively an assignment in the local scope, so the compiler
> thinks 'file' is a local variable (which it actually becomes because of this).
> That means that 'file' in the place of 'open' above can't be referring
> to the intended builtin, and so it looks like it's referring to the 'file'
> variable you assign here, but before it's assigned.
> Try using a different name, e.g.,
> for a_file in files:
>
>> old_fname = file
>
> old_fname = a_file
>
>> new_fname = old_fname.strip()
>> if new_fname != old_fname:
>> newpath = os.path.join(root,new_fname)
>> oldpath = os.path.join(root,old_fname)
>> print >> outputFile, "Replaced ", old_fname, "\nWith
>> ", new_fname
>> os.rename(oldpath,newpath)
>> outputFile.close()
>>clean_spaces(setpath)
>>
>
>
> Regards,
> Bengt Richter

I tend to be brief while posting to newsgroups.
Most of the time, you get lambasted for verbosity here.
Hence I tend to stick to stuff and trust pythonistas
for their knowledge and intuition

-Anand

Andrew Bennetts <andrew-> wrote in message news:<mailman.972.1069429474.702.python->...
> (please don't top-post)
>
> > > It does exist in Python 2.3, and it does take one argument (and optionally
> > > others):
> > >
> > > http://python.org/doc/lib/os-file-dir.html#l2h-1473
>
> > It does not exist in Python 2.2, AFAIK.
> > It is a good idea to write code that at least works for
> > one previous Python version, hence the comment.
>
> You should have said so in the first place -- explicit is better than
> implicit
>
> -Andrew.