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I want to implement a predicate of the form:

zipwithindex(?List1, ?List2)

which is true when the elements of List2 are the same as in List1 but are paired with the index of the element.

This is what I have:

zipwithindex([],I,S,S).
zipwithindex([H|T],I,S0,S):-
  append(S0,[H/I],S1),
  I1 is I+1,
  zipwithindex(T,I1,S1,S).
zipwithindex(List1,List2):-
  zipwithindex(List1,0,[],List2).

Is there a better way to implement this?

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migrated from stackoverflow.com May 12 '14 at 13:53

This question came from our site for professional and enthusiast programmers.

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sure:

zipwithindex(List1, List2) :-
   findall(E/I, nth0(I, List1, E), List2).

Edit - after Tudor' comment

bagof/3 it's much better than findall/3, as it doesn't introduce unwanted new variables. This can be of utmost importance when working with constrained variables - i.e. CLP(FD) or CHR

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  • 3
    \$\begingroup\$ I would suggest bagof/3 instead of findall/3 in this case as, if there are unbound variables in List1, List2 will have new variables instead of the original ones. E.g. ?- zipwithindex([X,Y,Z],L2). will produce L2 = [_G41/0, _G35/1, _G29/2]. \$\endgroup\$ – Tudor Berariu Jun 9 '14 at 7:46

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