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From: =?UTF-8?Q?Nagy_L=c3=a1szl=c3=b3_Zsolt?= <gandalf@shopzeus.com>
Newsgroups: comp.lang.python
Subject: Re: I'm wrong or Will we fix the ducks limp?
Date: Fri, 3 Jun 2016 18:11:12 +0200
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Message-ID: <mailman.132.1464970271.1839.python-list@python.org>
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Xref: csiph.com comp.lang.python:109428


> def getsMeet(files=file_list):
>      """Get a File or List of Files.
>
>      From the list of files determine
>      what meetings exist and prepare them
>      to be parsed
>      """
>      pyqFiles = []
>      for filename in sorted(file_list):
>          pyqFiles = pyqFiles.append(pq(filename=my_dir + filename))
>
>      return pyqFiles
This may not be the answer you are seeking, but how about:

def getsMeet(files=file_list):
     return [pq(filename=my_dir + filename) for filename in sorted(file_list)]

If you can use any iterable instead of a list, then you can also 
consider using a generator:

def getsMeet(files=file_list):
     for filename in sorted(file_list):
         yield pq(filename=my_dir + filename)

for pg_obj in getsMeet(my_file_list):
     process(pg_obj)

This way you do not need to construct a list at all, and it won't create all pq objects in advance.