> (map (apply +) '((0 1 2) (3 4 5)))
((0 1 2) (3 4 5))
> (map (fn(x) (apply + x)) '((0 1 2) (3 4 5)))
(3 12)
Why doesn't the first example work?
With:
we see the expected error:
ERR: missing argument in function max
I don't know why we don't see this with the + function, but the solution to both cases is to use curry:
Like in Scneme (+) or (apply +) evaluates to 0. So your map expression maps 0 onto two lists.
Using a number as operator is implicit resting. 0 gives you back the whole list.
Try also (map (apply *) '((0 1 2) (3 4 5))) to prove the point. Like in Scheme (*) gives 1.
So using curry for what you want is the right thing.
Ah, that makes sense. Thanks, Lutz.
Thanks a lot bairui and Luts! Hadn't used curry before.
Another question i just encountered is how to change all items in a list in one go, for example:
(set-ref-all "2" '(("1" "2" "3") ("4")) (int $it))
(("1" 2 "3") ("4"))
Works but i want to change all the strings to integers in one go but couldn't find an expression which matches
all the strings, see below some examples i tried but didn't work
> (set-ref-all '((*)) '(("1" "2" "3") ("4")) (int $it))
nil
> (set-ref-all '(?) '(("1" "2" "3") ("4")) (int $it))
nil
> (set-ref-all '(*) '(("1" "2" "3") ("4")) (int $it))
nil
> (set-ref-all '(+) '(("1" "2" "3") ("4")) (int $it))
nil
> (set-ref-all ? '(("1" "2" "3") ("4")) (int $it))
nil
> (set-ref-all * '(("1" "2" "3") ("4")) (int $it))
nil
How about:
;-> ((1 2 3) (4))
I admit that
Not as veratile as cormullion's match solution, but works for the given example:
Thx!
I understood the function from bairui directly (why didn't i think of that..) but i had to look in the documentation to understand cormullion's solution and then i saw in the documentation this example:
(set-ref-all '(oranges *) data (list (first $it) (apply + (rest $it))) match) ....
Great docs, just have to read them more carefully next time.