list - 普通口齿不清 : Removing elements with the same cdr

Question

(作业帮助，对我放轻松)(我必须在不使用破坏性函数(setf)的情况下执行此操作)

使用普通的 lisp，作为一些代码的一部分，我需要能够:

获取列表列表，比较2个元素的cdr，如果等于忽略第一个元素，如果不相等，则尝试将第一个元素与列表中下一个未选中的元素进行比较。

一些例子来说明:

((1 1 1) (2 1 1) (3 1 1)) -> ((3 1 1))

((2 2 0) (4 1 1) (1 2 0) (3 0 1) (8 1 1)) -> ((1 2 0) (3 0 1) (8 1 1))

(defun simplify2 (vari)
    ;If last term: stop
    (if (equal (cdr vari) nil) vari

        ;If cdr of first and second term are equal...
        (if (equal (cdar vari) (cdr (cadr vari)))

            ;Ignore the first term and continue with the rest of the list
            (simplify2 (cdr vari))

            ;Otherwise (this is the line which isn't working)
            (cons (car vari) (simplify2 (cdr vari))))))

目前，只有当列表中所有“类似”字词并排放置时，代码才能正常工作。

Answer 1

Le Petit Prince's suggestion在评论中使用 remove-duplicates 可能是您想要的。 remove-duplicates 是非破坏性的(参见 delete-duplicates 可能是破坏性的)，它被指定为返回一个新列表，其中除最后一个实异常(exception)的所有实例省略了一个元素(添加了强调):

remove-duplicates

remove-duplicates returns a modified copy of sequence from which any element that matches another element occurring in sequence has been removed. … The elements of sequence are compared pairwise, and if any two match, then the one occurring earlier in sequence is discarded, unless from-end is true, in which case the one later in sequence is discarded.

您需要指定一个 key 参数来表明实际比较的是元素的 cdr 和一个 test 参数来表明它们应该与 equal 进行比较。因此:

(remove-duplicates '((1 1 1) (2 1 1) (3 1 1))
                   :test 'equal 
                   :key 'cdr)
;=> ((3 1 1))

(remove-duplicates '((2 2 0) (4 1 1) (1 2 0) (3 0 1) (8 1 1))
                   :test 'equal
                   :key 'cdr)
;=> ((1 2 0) (3 0 1) (8 1 1))