-
In your
for
loop, you’re iterating through the elements of a lista
. But in the body of the loop, you’re using those items to index that list, when you actually want indexes.
Imagine if the lista
would contain 5 items, a number 100 would be among them and the for loop would reach it. You will essentially attempt to retrieve the 100th element of the lista
, which obviously is not there. This will give you anIndexError
.We can fix this issue by iterating over a range of indexes instead:
for i in range(len(a))
and access the
a
‘s items like that:a[i]
. This won’t give any errors. -
In the loop’s body, you’re indexing not only
a[i]
, but alsoa[i+1]
. This is also a place for a potential error. If your list contains 5 items and you’re iterating over it like I’ve shown in the point 1, you’ll get anIndexError
. Why? Becauserange(5)
is essentially0 1 2 3 4
, so when the loop reaches 4, you will attempt to get thea[5]
item. Since indexing in Python starts with 0 and your list contains 5 items, the last item would have an index 4, so getting thea[5]
would mean getting the sixth element which does not exist.To fix that, you should subtract 1 from
len(a)
in order to get a range sequence0 1 2 3
. Since you’re using an indexi+1
, you’ll still get the last element, but this way you will avoid the error. -
There are many different ways to accomplish what you’re trying to do here. Some of them are quite elegant and more “pythonic”, like list comprehensions:
b = [a[i] + a[i+1] for i in range(len(a) - 1)]
This does the job in only one line.