use of index (beginner's question)

Wed Apr 27 21:23:51 EDT 2011

Chris Angelico wrote:
> Rusty Scalf wrote:
>> list1 = ['pig', 'horse', 'moose']
>> list2 = ['62327', '49123', '79115']
>> n = 2
>> s2 = "list" + `n`

I would prefer the clearer
 s2 = "list" + str(n)
or
 s2 = "list%s" % n
>> a = s2[list1.index('horse')]
>> print a
>> s2 is a string with the value "list2"; this is not the same as the
> variable list2. You could use eval to convert it, but you'd do better
> to have a list of lists:
>> lists = [
> ['pig', 'horse', 'moose']
> ['62327', '49123', '79115']
> ]

You forgot a comma after the first `]', to separate the list elements.
A way to reuse the existing code is
 lists = [list1, list2]

> Then you could use:
> n = 2
> a = lists[n][list1.index('horse')]

That would throw an IndexError exception, though, since list indexes are
0-based, and this list has only two items (so the highest index is 1, not 
2).
But even if this was fixed, this could still throw a ValueError exception
if there was no 'horse' in `list1'. While you could catch that –
 needle = 'horse'
 try:
 a = lists[n][list1.index(needle)]
 except ValueError:
 a = 'N/A'
– perhaps a better way to store this data is a dictionary:
 data = {
 'pig': '62327',
 'horse': '49123',
 'moose': '79115'
 }
 print data.get('horse')
 print data.get('cow')
 print data.get('cow', 'N/A')
Such a dictionary can be built from the existing lists as well:
 data = dict(zip(list1, list2))
 print data
 print data.get('horse', 'N/A')
HTH
-- 
PointedEars