> Do you know a better way to do this?
>
> if i < A_Len-1:
>
> a = A_List[i-1]
> b = A_List[i]
> c = A_List[i+1]
>
> else:
>
> a = A_List[i-1]
> b = A_List[i]
> c = A_List[0]
Hi Carlos,
Here's one way to handle it. If you're familiar with modulo "clock"
arithmetic, you might want to apply it above.
Here's an example:
########################################
>>> for i in [1, 2, 3, 11, 12, 13, 14]:
... print i, i % 12
...
1 1
2 2
3 3
11 11
12 0
13 1
14 2
########################################
The idea is that the second columns results are the result of doing:
i % 12
and that number "rolls" around twelve. Your index can "roll" around
A_Len.
There are other alternative approaches to handle the boundary case. We
can talk about them if you'd like.
Good luck!
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