在Common Lisp中等效的有序对的循环

结合这三个功能中的任何一个 - 每个功能都带有一个

map

-version，a

loop

-version和尾调用递归版本。 - 所以你可以选择创建一个

• 
  纯粹
  
map

  解
  


• 
  纯粹
  
loop

  解决方案或
  


• 
  纯粹的递归解决方案。
  

或者您

• 
  故意混淆他们;）
  

功能是：

    
      ;;;;;;;;;;;;;;;;;;;;
;; function collecting all cdrs of a list:
;; (tails ‘(a b c))
;; ;; returns: ((A B C) (B C) (C))
;;;;;;;;;;;;;;;;;;;;
;; with maps
(defun tails (l)
  (maplist #’identity l))
;; with loop
(defun tails (l)
  (loop for x on l collect x))
;; tail-call-recursion
(defun tails (l &optional (acc ‘()))
  (cond ((null l) (nreverse acc))
        (t (tails (cdr l) (cons l acc)))))
;;;;;;;;;;;;;;;;;;;;
;; function collecting car of a list consed with each list element
;; (1st-conses ‘(a b c))
;; ;; returns: ((A . A) (A . B) (A . C))
;;;;;;;;;;;;;;;;;;;;
;; with maps
(defun 1st-conses (l)
  (mapcar #’(lambda (x) (cons (car l) x)) l))
;; with loop
(defun 1st-conses (l)
  (loop for x in l collect (cons (car l) x)))
;; tail-call-recursion
(defun 1st-conses (l)
  (labels ((.1st-conses (l fst acc)
             (cond ((null l) (nreverse acc))
                   (t (.1st-conses (cdr l) fst (cons (cons fst (car l))
                                                                acc))))))
    (.1st-conses l (car l) ‘())))
;;;;;;;;;;;;;;;;;;;;
;; applying the second function on the first functions’ results
;; (combine-down ‘(a b c))
;; ;; returning: ((A . A) (A . B) (A . C) (B . B) (B . C) (C . C))
;;;;;;;;;;;;;;;;;;;;
;; with maps
(defun combine-down (l)
  (mapcan #’1st-conses (tails l)))
;; with loop
(defun combine-down (l)
  (loop for x in (tails l)
        nconc (1st-conses x)))
;; with tail-call-recursion
(defun combine-down (l)
  (labels ((.combine-down (l acc)
            (cond ((null l) acc)
                  (t (.combine-down (cdr l)
                                    (nconc acc
                                           (1st-conses (car l))))))))
    (.combine-down (tails l) ‘())))
</code>
  

    
      (combine-down ‘(1 3 5 7 9))
;; ((1 . 1) (1 . 3) (1 . 5) (1 . 7) (1 . 9) (3 . 3) (3 . 5) (3 . 7) (3 . 9)
;;  (5 . 5) (5 . 7) (5 . 9) (7 . 7) (7 . 9) (9 . 9))
</code>
  

    
      (let ((arr ‘(1 3 5 7 9))
      (res ‘()))
  (loop for i from 0 below 5 by 1
        do (loop for j from i below 5 by 1
                 do (setq res (cons (cons (elt arr i)
                                          (elt arr j))
                                          res))))
  (nreverse res))
</code>
  

    
      ((1 . 1) (1 . 3) (1 . 5) (1 . 7) (1 . 9) (3 . 3) (3 . 5) (3 . 7) (3 . 9)
 (5 . 5) (5 . 7) (5 . 9) (7 . 7) (7 . 9) (9 . 9))
</code>