Python - Understanding error: IndexError: list index out of range
Solution 1
Using range
for iteration is nearly always not the best way. In Python you can iterate directly over a list, dict, set etc.:
for item in d.entries:
updates.append([{"url": feed_a_enviar["linktoourpage"], "msg": item.title + ", "}])
Obviously d.entries[i]
triggers the error because that list contains less than 8 items (feeds_updates
may contain 8, but you are not iterating over that list).
Solution 2
d.entries
has less than 8 elements. Iterate over d.entries
directly instead of some disconnected range.
André
Updated on November 20, 2022Comments
-
André over 1 year
I'm fairly new to python. I have an error that I need to understand.
The code:
config.py:
# Vou definir os feeds feeds_updates = [{"feedurl": "http://aaa1.com/rss/punch.rss", "linktoourpage": "http://www.ha.com/fun.htm"}, {"feedurl": "http://aaa2.com/rss.xml", "linktoourpage": "http://www.ha.com/fun.htm"}, {"feedurl": "http://aaa3.com/Heaven", "linktoourpage": "http://www.ha.com/fun.htm"}, {"feedurl": "http://aaa4.com/feed.php", "linktoourpage": "http://www.ha.com/fun.htm"}, {"feedurl": "http://aaa5.com/index.php?format=feed&type=rss", "linktoourpage": "http://www.ha.com/fun.htm"}, {"feedurl": "http://aaa6.com/rss.xml", "linktoourpage": "http://www.ha.com/fun.htm"}, {"feedurl": "http://aaa7.com/?format=xml", "linktoourpage": "http://www.ha.com/fun.htm"}, {"feedurl": "http://aaa8/site/component/rsssyndicator/?feed_id=1", "linktoourpage": "http://www.ha.com/fun.htm"}]
twitterC.py
# -*- coding: utf-8 -*- import config # Ficheiro de configuracao import twitter import random import sqlite3 import time import bitly_api #https://github.com/bitly/bitly-api-python import feedparser ... # Vou escolher um feed ao acaso feed_a_enviar = random.choice(config.feeds_updates) # Vou apanhar o conteudo do feed d = feedparser.parse(feed_a_enviar["feedurl"]) # Vou definir quantos feeds quero ter no i i = range(8) print i # Vou meter para "updates" 10 entradas do feed updates = [] for i in range(8): updates.append([{"url": feed_a_enviar["linktoourpage"], "msg": d.entries[i].title + ", "}]) # Vou escolher ums entrada ao acaso print updates # p debug so update_to_send = random.choice(updates) print update_to_send # Para efeitos de debug
And the error that appears sometimes because of the nature of the random:
Traceback (most recent call last): File "C:\Users\anlopes\workspace\redes_sociais\src\twitterC.py", line 77, in <module> updates.append([{"url": feed_a_enviar["linktoourpage"], "msg": d.entries[i].title + ", "}]) IndexError: list index out of range
I'am not getting to the error, the list "feeds_updates" is a list with 8 elements, I think is well declareted and the RANDOM will choose one out of the 8...
Can someone give me a clue on what is happenning here?
PS: Sorry for my bad english.
Best Regards,
-
user1066101 about 13 yearsYou are aware that indexes start at zero? 8 elements have indexes 0 to 7?
-
Tim Pietzcker about 13 years@S. Lott: And
range(8)
is[0,1,2,3,4,5,6,7]
, so that's not the problem here. -
user1066101 about 13 years@Tim Pietzcker: Of course, we're both assuming that the code which was presented actually has 8 items in the
feed_updates
list, aren't we? Yes, the evidence is good, but there has to be something going on that's not shown in the code snippet provided. -
Tim Pietzcker about 13 years@S. Lott: The number of items in
feed_updates
is irrelevant because he's iterating over a completely different list. See my answer.
-