How can I replace an element of a list using an in

问题:

I need do this:

replace(L,P,E,R), where L is the current list, P is the position where the element will be insert, E is the element to insert and R is the return, the new list.

Example:

L = [1,2,3,4,5],
replace(L,2,'@',R).
R = [1,2,'@',4,5]

I tried this, but nothing happens, the list continuous to be the same:

replaceInThePosition([_|T],1,E,[['1',E]|T]).
replaceInThePosition([H|T],P,E,[H|R]) :-
    P > 0, NP is P-1, replaceInThePosition(T,NP,E,R), !.

回答1:

You're not far.

You can try with

replaceInThePosition([_|T],0,E,[E|T]).
replaceInThePosition([H|T],P,E,[H|R]) :-
    P > 0, NP is P-1, replaceInThePosition(T,NP,E,R).

I mean:

1) 0 instead 1 in the first clause (or 1 in the first clause but P > 1 in the second one; but in this case from replaceInThePosition([1, 2, 3, 4, 5], 2, '@', R) you unify R = [1, @, 3, 4, 5], not R = [1, 2, @, 4, 5])

2) [E|T] for the last ergument in the first clause, not [['1',E]|T]

3) no cut (no !) at the end of the second clause (there is no need: the first clause i when P is zero; the secon one when P > 0; so they are mutually exclusive

回答2:

You could do this using append/3:

substitute(L, P, E, R) :-
    PreLen is P - 1,
    length(PreLen, P1),
    append(Pre, [_|T], L),
    append(Pre, [E|T], R).

回答3:

another possibility:

/*
L is the current list,
P is the position where the element will be insert,
E is the element to insert and
R is the return, the new list
*/
replaceInThePosition(L, P, E, R) :-
    findall(X, (nth0(I,L,Y), (I == P -> X=E ; X=Y)), R).