在包含元素的嵌套列表中打印一个列表 [英] Print a List inside a nested List that contains an element

问题描述

我有以下问题.

我得到了一个 listOfLists，一个值 (row,col)，我需要获取包含该特定值的列表中的列表，直到该列表中我的值的索引.

I'm given a listOfLists, a value (row,col) and I need to get the list inside a list that contains that certain value, up to my value's index inside that list.

例如

?- find_list([[(1,2),(1,3),(1,4)], [(2,2),(2,3),(2,4)]], (1,3), List2).
List2 = [(1,2),(1,3)].

我的问题是，如果我使用 member/2，我只会得到 true 或 false，因为我的值是否在 listOfList 内，而不是我需要使用的列表.

My problem is that if I use member/2 I will only get true or false for if my value is inside listOfList or not, and not the list that I will need to be working with.

如何获得包含我价值的列表?

How can I get that list that has my value inside it?

推荐答案

值是二维坐标有关系吗?您是否必须尊重它们的顺序，或者仅仅是列表中元素的顺序?我会假设后者.

Does it matter that the values are two-dimensional coordinates? Is there an ordering on them that you must respect, or is it simply the ordering of the elements in the list? I will assume the latter.

如果你想在某个时候拆分一个列表，标准的 append/3 谓词通常是要走的路.例如，假设我们想将列表 [a, b, c, d, e] 切割成包含 c 之前元素的前缀和包含之后元素的后缀c.这是如何完成的:

If you want to split a list at some point, the standard append/3 predicate is usually the way to go. For example, assume we want to cut the list [a, b, c, d, e] into a prefix containing the elements before c and a suffix containing the elements after c. Here is how that is done:

?- append(Prefix, [c | Suffix], [a, b, c, d, e]).
Prefix = [a, b],
Suffix = [d, e] ;
false.

这里 c 被排除在前缀之外，但这很容易解决:

Here c is excluded from the prefix, but that's easy to fix:

?- append(Prefix, [c | Suffix], [a, b, c, d, e]), append(Prefix, [c], UpToAndIncludingC).
Prefix = [a, b],
Suffix = [d, e],
UpToAndIncludingC = [a, b, c] ;
false.

我们可以给这个谓词起一个好听的名字:

We can give this predicate a nice name:

list_pivot_prefix(List, Pivot, Prefix) :-
    append(Prefix0, [Pivot | _Suffix], List),
    append(Prefix0, [Pivot], Prefix).

然后，您的 find_list/3 谓词只需在给定列表中找到此关系成立的所有列表:

And your find_list/3 predicate then simply finds all the lists in the given list of lists for which this relation holds:

find_list(Lists, Element, Prefix) :-
    member(List, Lists),
    list_pivot_prefix(List, Element, Prefix).

这是你的测试用例:

?- find_list([[(1,2),(1,3),(1,4)],[(2,2),(2,3),(2,4)]],(1,3),List2).
List2 = [ (1, 2), (1, 3)] ;
false.

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