Re: [Maxima-discuss] convert list access to function

[Pankaj S. <pan...@gm...>]

Hi,
Using Barton's suggestion I fixed it as,

ex(ll,x,f):=apply(substinpart,cons(apply(f,piece),cons(ll,x)))$
ex(ll,[3,3,4],fun) => [2,3,[a,b,[t,y,[u,i],fun(o,p),6,7]],8,8]

With this now one can get functionality similar to Mathematica's "MapAt"
feature.
I am thankful to provided suggestions and help.
​>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>​

On Mon, Jul 21, 2014 at 4:59 PM, Barton Willis <wi...@un...> wrote:

> Try using either inpart or substinpart; example
>
>     (%i12) substinpart(apply(f,piece), ll, 3,3,3);
>     (%o12) [2,3,[a,b,[t,y,f(u,i),[o,p],6,7]],8,8]
>
> --bw
>
> ________________________________________
>
> I want to access list elements based on their position but I don't
> intend to extract the elements out of list but instead keep my curosr
> there to apply some function on it, say,
>
> ll:[2,3,[a,b,[t,y,[u,i],[o,p],6,7]],8,8]
>
> I can reachout to various elements using,
>
> a) Using substitution,
> subst(apply(f,ll[3][3][3]),ll[3][3][3],ll)
>
> b) Or using positions directly as,
> ll[3][3][4]:apply(f,ll[3][3][4]);
>
> Showing the combined result of above two expressions,
> ll now, [2,3,[a,b,[t,y,f(u,i),f(o,p),6,7]],8,8]
>
> I want to create a function say,
> map_at_position(ll,[3,3,4],f) that shall give me
> [2,3,[a,b,[t,y,[u,i],f(o,p),6,7]],8,8]
>
> Function definition needs to be like,
>
>
> map_at_position(ll,a,f):=block([],ll[a1][a2]a[3]...[an]:apply(f,ll[a1],[a2],[a3]...[an]),ll)
> where a = [a1,a2,a3...an] representing the positions
>
> I am not able to figure out how to convert [3,3,3] to ll[3][3][3] kind
> of expression. I mean there can't be consed/joined/appended etc
> without comma.
>
> --
>


-- 
Regards,
Pankaj Sejwal
_______________________________________________
"The more I read, the more I acquire, the more certain I am that I know
nothing.” - Voltaire <http://www.goodreads.com/author/show/5754446.Voltaire>