适当使用reduce，嵌套循环

Question

下面是Dijkstra最短路径算法的一个实现。它的输入graph表示为源节点到目标节点和弧权重的关联列表。

我的问题是关于以下几个方面:我是否遗漏了一种简化min-path函数的方法？也许伟大而强大的loop可以做reduce的S的工作？

(defun dijkstra (source target graph)
  (labels ((min-path (&optional arc-1 arc-2)
             (let ((price-1 (cdr arc-1))
                   (price-2 (cdr arc-2)))
               (cond 
                 ((and (not (null arc-1)) (not (null arc-2))) (if (< price-1 price-2)
                                                                arc-1 arc-2) )
                 ((and (not (null arc-1)) (null (arc-2))) arc-1) 
                 ((and (not (null arc-2)) (null (arc-1))) arc-2))))
           (get-best-path (visited)
             (reduce #'min-path (loop for pick in visited 
                                      for start = (car pick) 
                                      for cost = (cdr pick) 
                                      append (loop for arc in (cdr (assoc start graph))
                                                   when (null (assoc (car arc) visited))
                                                   collect (cons (car arc) (+ cost (cdr arc))))))))
    (loop named main-loop
          with visited = (list (cons source 0))
          for best-arc = (get-best-path visited)

          when (not (null best-arc)) 
          do (push best-arc visited)  
          until (or (null best-arc)
                  (eql (car best-arc) target)) 
          finally 
          (format t "Found shortest path from ~A to ~A: ~A~%" source target best-arc)
          (return-from main-loop (cdr best-arc)))))

我是Lisp的新手，我希望得到任何建议，尤其是代码中关于Lisp坏习惯和坏Lisp风格的建议，谢谢。

UPD:新版本，有评论应用-没有警告，不像原来的版本。

(defun dijkstra (source target graph)
  (labels ((min-path (&optional arc-1 arc-2)
             (cond ((and arc-1 arc-2) (if (< (cdr arc-1) (cdr arc-2)) arc-1 arc-2))
                   (arc-1 arc-1)
                   (arc-2 arc-2)
                   (t nil)))
           (get-best-path (visited)
             (reduce #'min-path 
                     (loop for pick in visited 
                           for start = (car pick) 
                           for cost = (cdr pick) 
                           append (loop for arc in (cdr (assoc start graph))
                                        unless (assoc (car arc) visited)
                                        collect (cons (car arc) (+ cost (cdr arc))))))))
    (loop with visited = (list (cons source 0))
          for best-arc = (get-best-path visited)
          when best-arc do
          (push best-arc visited)  
          until (or (null best-arc) (eql (car best-arc) target)) 
          finally (return (cdr best-arc)))))

Answer 1

首先，(not (null x))不是非常惯用的；只需使用x。

接下来，您没有处理cond中的所有大小写(显然您指的是(null arc-2)而不是(null (arc-2)))。如果是这样的话，我会放弃let，转而使用嵌套的ifs。

最后，回答您最初的问题:当然，loop可以做reduce所能做的一切，但是如果您对后者的意图更明确，那么就使用它吧！

编辑：unless比when (null ...)更惯用。