Re: Iterative union



phantastic@xxxxxxxxxx wrote:

Hi,

I have a list like this: [[[a,b],[b,c]],[[c,d],[d,v]]] where the
number of elements (sublists) is two but number of elements inside
sublist can be random.
I want a list like this: [[a,b,c], [c,d,v]] (in this case)

I have considered union/3 predicate but I don't know how I can iterate
on every member of the list.
Any help?

thanks

Maybe like that :

map(_,[],[]).
map(F,[X|XS],[Y|YS]) :- apply(F,[X,Y]),map(F,XS,YS).

Regards
Thorsten
.



Relevant Pages

  • Re: Iterative union
    ... number of elements (sublists) is two but number of elements inside ... I have considered union/3 predicate but I don't know how I can iterate ... on every member of the list. ...
    (comp.lang.prolog)
  • Iterative union
    ... number of elements (sublists) is two but number of elements inside ... I have considered union/3 predicate but I don't know how I can iterate ... on every member of the list. ...
    (comp.lang.prolog)
  • Re: Iterative union
    ... number of elements (sublists) is two but number of elements inside ... Your list contains 2 sublists, which contain 2 subsublists each, which ... I have considered union/3 predicate but I don't know how I can iterate ... on every member of the list. ...
    (comp.lang.prolog)
  • Re: Iterative union
    ... number of elements (sublists) is two but number of elements inside ... I have considered union/3 predicate but I don't know how I can iterate ... on every member of the list. ...
    (comp.lang.prolog)
  • Re: mErge
    ... student wrote: ... every member of A is a member of B but there are at least two members of A occur in different order in B, ... These answers satisfy the "has each of the given lists as ... sublists" part of my definition but not the ...
    (comp.lang.prolog)