List 序言：选择在列表递归期间不插入元素

我做了一个谓词，它将2个list作为参数，并返回一个与“recipesub”的乘积连接的列表，但是我需要做第三个规则，如果手头的产品是空列表，则禁止插入该产品

因此，第一个列表可以如下所示：

recipe([ingredient(rice,4),ingredient(salt,3),ingredient(water,5)]).

ingredients([ingredient(rice,3),ingredient(salt,4),ingredient(water,4),

need_to_buy([], _, []).

% case 1: we don't have the ingredient at all
need_to_buy([ingredient(Type, AmountNeeded) | Ingredients],
            OnHand,
            [ingredient(Type, AmountNeeded)|Needed]) :-
    \+ memberchk(ingredient(Type, _), OnHand),
    need_to_buy(Ingredients, OnHand, Needed).

% case 2: we have it, but not enough
need_to_buy([ingredient(Type, AmountNeeded) | Ingredients],
            OnHand,
            [ingredient(Type, AmountToBuy)|RestNeeded]) :-
    memberchk(ingredient(Type, AmountOnHand), OnHand),
    AmountNeeded > AmountOnHand,
    AmountToBuy is AmountNeeded - AmountOnHand,
    need_to_buy(Ingredients, OnHand, RestNeeded).

% case 3: we have enough
need_to_buy([ingredient(Type, AmountNeeded) | Ingredients],
            OnHand,
            RestNeeded) :-
    memberchk(ingredient(Type, AmountOnHand), OnHand),
    AmountNeeded =< AmountOnHand,
    need_to_buy(Ingredients, OnHand, RestNeeded).

第二个是这样的：

recipe([ingredient(rice,4),ingredient(salt,3),ingredient(water,5)]).

ingredients([ingredient(rice,3),ingredient(salt,4),ingredient(water,4),

need_to_buy([], _, []).

% case 1: we don't have the ingredient at all
need_to_buy([ingredient(Type, AmountNeeded) | Ingredients],
            OnHand,
            [ingredient(Type, AmountNeeded)|Needed]) :-
    \+ memberchk(ingredient(Type, _), OnHand),
    need_to_buy(Ingredients, OnHand, Needed).

% case 2: we have it, but not enough
need_to_buy([ingredient(Type, AmountNeeded) | Ingredients],
            OnHand,
            [ingredient(Type, AmountToBuy)|RestNeeded]) :-
    memberchk(ingredient(Type, AmountOnHand), OnHand),
    AmountNeeded > AmountOnHand,
    AmountToBuy is AmountNeeded - AmountOnHand,
    need_to_buy(Ingredients, OnHand, RestNeeded).

% case 3: we have enough
need_to_buy([ingredient(Type, AmountNeeded) | Ingredients],
            OnHand,
            RestNeeded) :-
    memberchk(ingredient(Type, AmountOnHand), OnHand),
    AmountNeeded =< AmountOnHand,
    need_to_buy(Ingredients, OnHand, RestNeeded).

当它返回时：

List=[配料（大米，1），[]，配料（水，1）]

我希望它返回：

List=[配料（大米，1），配料（水，1）]

下面是我在解决方案方面取得的进展

/*need_to_buy([H|Hs],[X|Xs],List):-
   H = ingredient(Type,Amount),
   recipesub(Type,Amount,[X|Xs],Difference),
   Difference = [],
   need_to_buy(Hs,[X|Xs],List).*/

这是支持谓词，recipesub

recipesub(Type,Amount,[],Difference):-
    Difference = ingredient(Type,Amount).
recipesub(Type,Amount,[Z|_],Difference):-
    Z = ingredient(Type,Stock),
    Amount>Stock,
    NewAmount is Amount-Stock,
    Difference = ingredient(Type,NewAmount).
recipesub(Type,Amount,[Z|_],Difference):-
    Z = ingredient(Type, Stock),
    Stock >= Amount,
    Difference = [].
recipesub(Type,Amount,[Z|Zs],Difference):-
    Z = ingredient(WrongType,_),
    WrongType \= Type,
    recipesub(Type,Amount,Zs,Difference).

---

我通常不会做很多嵌套条件，但这次“感觉不错”，这就是我找到的解决方案：

need_to_buy([], _, []).
need_to_buy([ingredient(Type, AmountNeeded)|Ingredients], OnHand, Needed) :-
    % Do we have any on-hand?
    member(ingredient(Type, AmountOnHand), OnHand) ->

        % If the amount on-hand is greater than the amount needed, 
        % just hand off the rest
        (AmountOnHand >= AmountNeeded ->
             need_to_buy(Ingredients, OnHand, Needed)

         % otherwise, calculate the amount needed and recur
         ; (AmountToBuy is AmountNeeded - AmountOnHand,
            need_to_buy(Ingredients, OnHand, RestNeeded),
            Needed = [ingredient(Type, AmountToBuy)|RestNeeded]))

    % If we have none on-hand, we can just use the amount needed
    % to form the request, and recur
    ; need_to_buy(Ingredients, OnHand, RestNeeded),
      Needed = [ingredient(Type, AmountNeeded)|RestNeeded].

否则我认为你会有很多相当低效的测试和重新测试。我在代码中看到的主要错误是第二个参数的模式匹配。依靠

member/2

或

memberchk/2

可以更轻松地从手头的东西中找到合适的成分

如果我用一堆子句来代替，它可能会是这样的：

recipe([ingredient(rice,4),ingredient(salt,3),ingredient(water,5)]).

ingredients([ingredient(rice,3),ingredient(salt,4),ingredient(water,4),

need_to_buy([], _, []).

% case 1: we don't have the ingredient at all
need_to_buy([ingredient(Type, AmountNeeded) | Ingredients],
            OnHand,
            [ingredient(Type, AmountNeeded)|Needed]) :-
    \+ memberchk(ingredient(Type, _), OnHand),
    need_to_buy(Ingredients, OnHand, Needed).

% case 2: we have it, but not enough
need_to_buy([ingredient(Type, AmountNeeded) | Ingredients],
            OnHand,
            [ingredient(Type, AmountToBuy)|RestNeeded]) :-
    memberchk(ingredient(Type, AmountOnHand), OnHand),
    AmountNeeded > AmountOnHand,
    AmountToBuy is AmountNeeded - AmountOnHand,
    need_to_buy(Ingredients, OnHand, RestNeeded).

% case 3: we have enough
need_to_buy([ingredient(Type, AmountNeeded) | Ingredients],
            OnHand,
            RestNeeded) :-
    memberchk(ingredient(Type, AmountOnHand), OnHand),
    AmountNeeded =< AmountOnHand,
    need_to_buy(Ingredients, OnHand, RestNeeded).

需要购买（[]，[]）。
%案例1：我们根本没有原料
需要购买（[配料（类型、所需数量）|配料]，
现在，
[成分（类型、所需数量）|所需]：-
\+memberchk（成分（类型），现有），
需要购买（原料、现货、需要）。
%案例2：我们有，但还不够
需要购买（[配料（类型、所需数量）|配料]，
现在，
[成分（类型、购买量）|需要时]：-
memberchk（成分（类型、数量和数量），现有），
需要的数量>需要的数量，
AmountToBuy是所需数量-AmountOnHand，
需要购买（原料、现货、需要时）。
%案例3：我们已经足够了
需要购买（[配料（类型、所需数量）|配料]，
现在，
（需要时）：-
memberchk（成分（类型、数量和数量），现有），
AmountNeeded=

---

这在堆栈上留下了一个选择点，通常看起来像是为了我的口味重新测试相同的条件和遍历。但是，如果你觉得它更好的话，它也应该起同样的作用。

我最终解决了这个问题，将第二条“需要购买”规则分为两条，一条处理差异是空列表的情况，另一条处理差异不是空列表的情况

一开始我遇到了一些麻烦，但事实证明规则的“方向”给我带来了麻烦，所以我不得不将处理差异\=[]情况的规则置于差异=[]的规则之上。 现在看起来是这样的：

need_to_buy([],_,List):- List = [].
need_to_buy([H|Hs],[X|Xs],[Difference|List]):-
   H = ingredient(Type,Amount),
   recipesub(Type,Amount,[X|Xs],Difference),
   Difference \= [],
   need_to_buy(Hs,[X|Xs],List).
need_to_buy([H|Hs],[X|Xs],List):-
   H = ingredient(Type,Amount),
   recipesub(Type,Amount,[X|Xs],Difference),
   Difference = [],
   need_to_buy(Hs,[X|Xs],List).

---

请这是什么？

recipesub/4

？recipesub是一个谓词，它递归地将第一个列表的头与第二个列表中的每个元素进行比较，如果“H”中的变量“Amount”小于X中的“Amount”，则返回差=成分（type，requiredAmount），但如果X中的可变金额大于H中的可变金额，则返回一个空列表。好的，对不起，我现在已尽我所能进行了更正。您可以在

需要购买

谓词中检查

[]

（从

配方返回

），或者你可以让

recipesub

失败，而不是提供

[]

作为解决方案，当

recipesub

失败时，让

need\u\u购买

“做正确的事情”。谢谢你的回复，但当我看到你的代码时，我已经想出了一个解决方案。