Avoiding use of .encode() in rss2html

Question

My concern with this code is the excessive use of .encode('utf-8'). Any advice on refining these functions would be very helpful.

rss2html GitHub repo

#!/usr/bin/env python
""" Simple rss to html converter """
__version__ = "0.0.1"
__author__ = "Ricky L Wilson"
from feedparser import parse as parsefeed
import StringIO
def entry2html(**kwargs):
 """ Format feedparser entry """
 title = kwargs['title'].encode('utf-8')
 link = kwargs['link'].encode('utf-8')
 description = kwargs['description'].encode('utf-8')
 template = """
 <h2 class='title'>{title}</h2>
 <a class='link' href='{link}'>{title}</a>
 <span class='description'>{description}</span>
 """
 return template.format(title=title, link=link, description=description)
def convert_feed(**kwargs):
 """ Main loop """
 out = StringIO.StringIO("")
 for entry in parsefeed(kwargs['url']).entries:
 title = entry['title']
 link = entry['link']
 description = entry['description']
 print >>out, entry2html(title=title, link=link,
 description=description)
 return out.getvalue()
print convert_feed(url='http://stackoverflow.com/feeds')

• 200_success: I've removed your new version of the code, since it adds confusion to the question. If you would like further advice, feel free to ask another question.

Answer 1

Work in Unicode and encode at the last moment:

def entry2html(**kwargs):
 """ Format feedparser entry """
 template = u"""
 <h2 class='title'>{title}</h2>
 <a class='link' href='{link}'>{title}</a>
 <span class='description'>{description}</span>
 """
 return template.format(**kwargs)
def convert_feed(**kwargs):
 """ Main loop """
 out = u'\n'.join(entry2html(**entry) 
 for entry in parsefeed(kwargs['url']).entries)
 return out.encode('utf-8')

You can pass entry directly as format does not mind extra keyword arguments. StringIO is not necessary for simple concatenation of strings.