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
– PyNEwbie, Commented Mar 5, 2014 at 18:59
What I mainly was looking for and failed was the function list. — toby
– toby, Commented Mar 5, 2014 at 19:48

Max Noel · Accepted Answer · 2014-03-05 19:00:30Z

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 5, 2014 at 19:00

Max Noel

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Valentin Lorentz · Accepted Answer · 2014-03-05 19:49:10Z

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 5, 2014 at 19:49

answered Mar 5, 2014 at 19:00

Valentin Lorentz

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6 Comments

Corley Brigman Over a year ago

have you actually tried the second solution (using locals())? i'm pretty sure that doesn't work.

Valentin Lorentz Over a year ago

It works for me on a small example (locals()['spam'] = 'egg' and print(locals()['spam']))

Corley Brigman Over a year ago

please look at this: stackoverflow.com/questions/10488296/… even if it works, the doc explicitly warns against it as unsafe.

Corley Brigman Over a year ago

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.

Valentin Lorentz Over a year ago

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 · Accepted Answer · 2014-03-05 19:52:57Z

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 5, 2014 at 19:52

Valentin Lorentz

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answered Mar 5, 2014 at 19:01

Ashoka Lella

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Comments

Peter Mortensen · Accepted Answer · 2015-02-02 11:53:18Z

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 2, 2015 at 11:53

Peter Mortensen

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answered Mar 5, 2014 at 19:14

Imran

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2 Comments

toby Over a year ago

I had to initialize the list with song = {} otherwise I get IndexError: list index out of range

Imran Over a year ago

You ended up using dictionary, not list ({} stands for dictionary literal). Updated the code to work correctly for list.

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