Get previous object without len(list)

Question

How to get previous or next object with this format of code?

alignment = [[a,b,c],[2,3,4],[q,w,e]]

for obj in alignment:
    some code here to get previous object

I know how to do that with:

for i in range(0,len(alignment)):
    alignment[i-1][objIndex]

Answer 1

You can use enumerate :

alignment = [[a,b,c],[2,3,4],[q,w,e]]

for index,obj in enumerate(alignment):
    alignment[index-1] # pre
    alignment[index+1] # next

Note that as a more efficient way for accessing to next items and refuse of multiple indexing you can use iter() function to create an iterator object from your list (from second element to end) and access to next elements in each iteration with next :

>>> l=[1,2,3,4]
>>> it=iter(l[1:])
>>> for i in l :
...   print i,next(it,None)
... 
1 2
2 3
3 4
4 None

Note that if you don't pass the None as the second argument to next() function it will raise a StopIteration error.You can also handle it with a try-except statement.

Also for short lists you can use zip function and for long lists itertools.izip() function (zip in python 3):

>>> for i,j in zip(l,l[1:]):
...   print i,j
... 
1 2
2 3
3 4

zip(l,l[1:]) will give you the following pairs of items :

[(1, 2), (2, 3), (3, 4)]

and in the loop you can use i as the current item then j will be the next item or use j as the current then i will be the previous!:)

Answer 2

There are many different options depending on what your use for the neighbour entry is. For instance, if you only want to process pairs, and not modify the list:

for left,right in zip(somelist[:-1], somelist[1:]):
  pass

Or if you need to keep the prior entry as a reference:

prev = None
for item in somelist:
  pass
  prev = item

None here is used in place of a prior item for the first, similar to how it's used for the next item after the last in Kasra's iter()-based example.