[Tutor] Re: glob or filter help

[Javier Ruere]

Jay Loden wrote:
> I have the following code in my updates script (gets the five most recent 
> updated files on my site)
> 
> def get_fles(exts, upd_dir):
>  '''return list of all the files matching any extensions in list exts'''
>  fle_list = [] 
>  for each in exts:
>   cmd = upd_dir + "*." + each
>   ext_ls = glob.glob(cmd)
>   fle_list = fle_list + ext_ls
>  return filter(notlink, fle_list)
> 
> I wanted to just get one list, of all the .htm and .exe files in my upd_dir.  
> I was trying to make a far more elegant solution that what's above, that 
> could generate a list through a filter.  Is there a way to trim the code down 
> to something that does ONE sort through the directory and picks up the .htm 
> and .exe files? (note, it is not necessary for this to recurse through 
> subdirectories in the upd_dir).  I have cmd defined above because calling
> "glob.glob(upd_dir + "*." + each) returned the error "cannot concatenate 
> string and list objects" - is this the only way around that, or is there a 
> better way?

  How about:

>>> import os
>>> exts = ('gz', 'conf')
>>> upd_dir = '.'
>>> filter(lambda fn : fn.split('.')[-1] in exts, os.listdir(upd_dir))

  Do you like this better. It doesn't use glob and is terse.

> Also in the above code, "notlink" is just a function that returns True if 
> "islink()" returns true....there has to be a better way to use this with 
> filter(), how can i make filter use "if islink()!=true" as its condition?

  If the filter criteria is complex, it deserves it's own function:

>>> import os
>>> import os.path
>>> def get_files(exts, upd_dir):
...     def criteria(filename):
...             return filename.split('.')[-1] in exts \
...                     and not os.path.islink(filename)
...     return filter(criteria, os.listdir(upd_dir))
...
>>> get_files(('gz', 'conf'), '.')
['dynfun.pdf.gz', 'twander-3.160.tar.gz', 'PyCon2004DocTestUnit.pdf.gz',
'arg23.txt.gz', '.fonts.conf']


Javier