How to index a Python list

Question

list=[1,3,0,3,0,3,0,3,0,3,0,3,0,3,0,3,0,3,0,3,0,3,0,3,1,0,1,3,0,3,0,3,1]

so now i want to know if the 5th object in this list is 3:

if list.index(3)==5:
  print("yes")
else:
  print("no")

-> answer is "no"... but it is, as you see. So how can i solve it?

I'm not familiar with python 3, but I have to feel that else: no is not valid. — Hoopdady
– Hoopdady, Commented Mar 6, 2013 at 14:28
@Hoopdady -- That depends on whether no is in the local namespace :P -- e.g. if no = "Hello World" before getting here, it would work on python2 or python3 — mgilson
– mgilson, Commented Mar 6, 2013 at 14:29
@mglison -- fair.... I guess I should have said useful instead of valid — Hoopdady
– Hoopdady, Commented Mar 6, 2013 at 14:30
" i want to know if the 5th object in this list is 3" : print list[5-1]==3 and the mass is over. The -1 is due to the fact that first element is indexed 0. — eyquem
– eyquem, Commented Mar 6, 2013 at 15:07
If you absolutely want to print YES or NO : print ('NO','YES')[list[4]==3]. Or print 'YES' if list[4]==3 else 'NO' — eyquem
– eyquem, Commented Mar 6, 2013 at 15:10

eumiro · Accepted Answer · 2013-03-06 14:23:40Z

5

don't use list as a variable name. It masks one very useful command.
index is a list's method returning the position of a given element within a list. You want the opposite: get the element for a given position.

This code will help you. It checks whether the 5th (position 4) element of a list is 3:

lst = [1,3,0,3,0,3,0,3,0,3,0,3,0,3,0,3,0,3,0,3,0,3,0,3,1,0,1,3,0,3,0,3,1]
if lst[4] == 3:
    print('yes')

Your code list.index(3) returns 1, because the first occurence of 3 is at the position 1.

answered Mar 6, 2013 at 14:23

eumiro

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1 Comment

I think you mean if lst[4] == 3