lambda function

[LibertyOne]

Just another thought building on what I added earlier. If you still wanted to keep the last element in the first list, you could add a 0 to the end of the second list to be added to the first list. =>

(setq lst (list 1 2 3 4 5 6 7 8 9))
;returns (1 2 3 4 5 6 7 8 9)
(mapcar '+ lst (reverse (cons 0 (reverse (cdr lst)))))
;returns (3 5 7 9 11 13 15 17 9)

you could even loop around to the beginning of the list and add the first element to the last =>

(mapcar '+ lst (reverse (cons (car lst) (reverse (cdr lst)))))
;returns (3 5 7 9 11 13 15 17 10)

[Lee Mac]

Apologies but this task appears to be made overcomplicated - refer to my post #17.

[LibertyOne]

Apologies but this task appears to be made overcomplicated - refer to my post #17.

 

true, but like you also said is that the result wasn't really clearly defined in the beginning of the thread. I was solving more on the lines of list manipulation by returning a modified list and not just that of an equation by adding up a list of numbers. What it all boils down to is what you want as a result and how generic can you build a function to suit different needs of many different programs.