Set multiple similar variables in Python

Question

How would I clean up the following code and make it look better?

I wonder if there is a way to set all these variables and values in an easier and shorter version. What would be an example of doing so? :-)

import feedparser
import memcache
import sys

mc = memcache.Client(["127.0.0.1:11211"])
rss = feedparser.parse('http://example.com/example.rss')
song0 = rss.entries[0].title.encode('latin-1', 'replace').replace("?" , "-")
song1 = rss.entries[1].title.encode('latin-1', 'replace').replace("?" , "-")
song2 = rss.entries[2].title.encode('latin-1', 'replace').replace("?" , "-")
song3 = rss.entries[3].title.encode('latin-1', 'replace').replace("?" , "-")
song4 = rss.entries[4].title.encode('latin-1', 'replace').replace("?" , "-")
song5 = rss.entries[5].title.encode('latin-1', 'replace').replace("?" , "-")
song6 = rss.entries[6].title.encode('latin-1', 'replace').replace("?" , "-")
song7 = rss.entries[7].title.encode('latin-1', 'replace').replace("?" , "-")
song8 = rss.entries[8].title.encode('latin-1', 'replace').replace("?" , "-")
song9 = rss.entries[9].title.encode('latin-1', 'replace').replace("?" , "-")

mc.set("track0", song0);
mc.set("track1", song1);
mc.set("track2", song2);
mc.set("track3", song3);
mc.set("track4", song4);
mc.set("track5", song5);
mc.set("track6", song6);
mc.set("track7", song7);
mc.set("track8", song8);
mc.set("track9", song9);

This is not really deserving of a question so don't be too surprised if it is closed. However, rss must be an iterable so for item in iterable: do_something it could be done in one line with a list comprehension — PyNEwbie, Mar 05 '14 at 18:59
What I mainly was looking for and failed was the function list. — toby, Mar 05 '14 at 19:48

score 2 · Answer 1 · answered Mar 05 '14 at 19:00

2

You have 10 variables with similar names only differing by a number. It's a sure sign that what you really want to use is a list:

songs = [i.title.encode("latin-1", "replace").replace("?" , "-") for i in rss.entries]

answered Mar 05 '14 at 19:00

Max Noel

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Valentin Lorentz · Answer 2 · 2014-03-05T19:49:10.717

2

The cleanest (thus best) way:

import feedparser
import memcache
import sys

mc = memcache.Client(["127.0.0.1:11211"])
rss = feedparser.parse('http://example.com/example.rss')
songs = [] # Initialize “songs” to the empty list
for i in range(0, 10):
    # Add items in “songs”
    songs.append(rss.entries[i].title.encode('latin-1', 'replace').replace("?" , "-"))

for (i, song) in enumerate(songs): # This is equivalent to “for i in range(0, len(songs)+1):” and “song = songs[i]”
    mc.set("track%i" % i, song);

If you really want to keep songs in different variables (I'm almost sure you don't actually want to; more over it is not recommended to use it (see the comments on this question)):

import feedparser
import memcache
import sys

mc = memcache.Client(["127.0.0.1:11211"])
rss = feedparser.parse('http://example.com/example.rss')
for i in range(0, 10):
    locals()['song%i' % i] = rss.entries[0].title.encode('latin-1', 'replace').replace("?" , "-")

for i in range(0, 10):
    mc.set("track%i" % i, locals()['song%i' % i]);

edited Mar 05 '14 at 19:49

answered Mar 05 '14 at 19:00

Valentin Lorentz

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have you actually tried the second solution (using `locals()`)? i'm pretty sure that doesn't work. – Corley Brigman Mar 05 '14 at 19:09
It works for me on a small example (`locals()['spam'] = 'egg'` and `print(locals()['spam'])`) – Valentin Lorentz Mar 05 '14 at 19:12
please look at this: http://stackoverflow.com/questions/10488296/how-to-dynamically-modify-a-functions-local-namespace?lq=1 even if it works, the doc explicitly warns against it as unsafe. – Corley Brigman Mar 05 '14 at 19:16
note that the way you have used `locals()`, it could be any dictionary - wouldn't specifically have to be the locals, since you get the value out also by dereferencing `locals()` in your example. in fact, that's how it works - if you try to access the variable `song1` directly, it won't work - it's not really added to the local namespace. but `locals()` returns the same dictionary every time, so you see the addition later. – Corley Brigman Mar 05 '14 at 19:23
I just tested, accessing `spam` after doing `locals()['spam'] = 'egg'`) works. (But I agree the doc says we should not rely on it) – Valentin Lorentz Mar 05 '14 at 19:30
When I try this I get IndexError: list index out of range – toby Mar 05 '14 at 19:47

score 1 · Answer 3 · edited Mar 05 '14 at 19:52

1

import feedparser
import memcache
import sys

mc = memcache.Client(["127.0.0.1:11211"])
rss = feedparser.parse('http://example.com/example.rss')
song=[]
for i in range(0,10):
    song[i] = rss.entries[i].title.encode('latin-1', 'replace').replace("?" , "-")
    mc.set("track"+str(i), song[i]);

edited Mar 05 '14 at 19:52

Valentin Lorentz

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answered Mar 05 '14 at 19:01

Ashoka Lella

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score 0 · Accepted Answer · edited Feb 02 '15 at 11:53

0

Using enumerate() with rss.entries would free you from the burden of manually maintaining an index. Best of all, it'd work regardless of the length of rss.entries.

import feedparser
import memcache
import sys

mc = memcache.Client(["127.0.0.1:11211"])
rss = feedparser.parse('http://example.com/example.rss')

song = []
for i, entry in enumerate(rss.entries):
    title = entry.title.encode('latin-1', 'replace').replace("?" , "-")
    song.append(title)
    mc.set("track%d" % i, title)

edited Feb 02 '15 at 11:53

Peter Mortensen

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answered Mar 05 '14 at 19:14

Imran

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I had to initialize the list with song = {} otherwise I get IndexError: list index out of range – toby Mar 05 '14 at 19:44
You ended up using dictionary, not list (`{}` stands for dictionary literal). Updated the code to work correctly for list. – Imran Mar 05 '14 at 22:35

Set multiple similar variables in Python

4 Answers4