[Tutor] update a list

[dn]

On 11/10/2020 21:44, Manprit Singh wrote:
> Dear sir ,
> 
> Consider a problem where I have to update a list , at even places with
> increment of 1 and at odd places with increment of 2 . Just need to check
> if this can be done in a better way. My solution is given below:
> 
> l = [2, 3, 5, 7, 6, 4]
> for i, j in enumerate(l):
>      if i % 2 == 0:
>          l[i] = j+1
>      else:
>          l[i] = j+2
> 
> the updated list is
> 
> [3, 5, 6, 9, 7, 6]


 From where are you finding these problems, or what do you intend to do 
with the results?


I am not in-favor of updating lists/list-elements in-place because 
altering iterables inside their iteration can cause big problems in 
certain other scenarios!

Accordingly, I prefer to create a new list (and if necessary, delete the 
old one, afterwards).

Remember that you can slice a sequence. Thus, a faster solution is:

 >>> l = [2, 3, 5, 7, 6, 4]
 >>> new_l = list()
 >>> for index in range( 0, len( l ), 2 ):
...     new_l.append( l[ index ] + 1 )
...     new_l.append( l[ index + 1 ] + 2 )
...
 >>> new_l
[3, 5, 6, 9, 7, 6]
 >>>

If you insist upon updating-in-place, replace the append()s with:

l[ index ] += 1


Performance comparison:
No integer-division/remainder comparison, no if, and with half(ish) the 
number of loops!
-- 
Regards =dn