Common-LISP 调试（包含代码）

Question

我希望快速修复我的代码，该代码包含数字列表、数字列表和阈值以及 return 数字列表中大于阈值的项目数。我就是想不通哪里出了问题，也不熟悉调试。我对 Whosebug 和 LISP 还很陌生……欢迎任何 comments/criticism/advice。谢谢！

ex) (count-greater-than (list 1 2 3 4 5 6 6 7) 5) => 3

(defun count-greater-than (numberlist threshold)
  (if (null numberlist) 0
    (counter (numberlist threshold 0)))
  (defun counter (numberlist threshold count)
    (cond ((null numberlist) count)
          ((> (first numberlist) threshold) (counter (rest numberlist) threshold (+ 1 count)))
          (t (counter (rest numberlist) threshold count)))))

Answer 1

首先，请注意该标准实际上包含有助于解决此类问题的功能。有一个有用的 count-if 函数可以接受一个谓词并计算列表中有多少元素满足它。对于你的情况，你可以这样做：

CL-USER> (count-if #'(lambda (x)
                       (> x 5))
                   (list 1 2 3 4 5 6 6 7))
;=> 3

CL-USER> (defun count-greater-than (numbers threshold)
           (count-if (lambda (n) (> n threshold)) numbers))
COUNT-GREATER-THAN
CL-USER> (count-greater-than (list 1 2 3 4 5 6 6 7) 5)
3
CL-USER> (count-greater-than (list 1 2 3 4 5 6 6 7) 6)
1

在您的特定情况下，您似乎是手动执行此操作，但括号错误。看起来您正在尝试创建一个名为 counter 的本地辅助函数。您可以使用 defun 在函数外部定义它，例如：

(defun count-greater-than (numberlist threshold)
  (if (null numberlist) 0
    (counter (numberlist threshold 0))))

(defun counter (numberlist threshold count)
  (cond ((null numberlist) count)
        ((> (first numberlist) threshold) (counter (rest numberlist) threshold (+ 1 count)))
        (t (counter (rest numberlist) threshold count))))

或者您可以使用 labels:

使用本地定义来完成

(defun count-greater-than (numberlist threshold)
  (labels ((counter (numberlist threshold count)
             (cond ((null numberlist) count)
                   ((> (first numberlist) threshold) (counter (rest numberlist) threshold (+ 1 count)))
                   (t (counter (rest numberlist) threshold count)))))
    (if (null numberlist) 0
        (counter numberlist threshold 0))))

备注

正如 Xach 在评论中指出的那样，您实际上可以使用 count 的 :test 参数更简洁地执行此操作。我不知道它是否明确地捕捉了 "count things with this property" 的概念，但它提供了一个非常简短的解决方案：

CL-USER> (count 5 (list 1 2 3 4 5 6 6 7) :test #'<)
;=> 3

这会计算 5 在列表中出现的次数，但诀窍在于，而不是使用 eql 或 =[ 检查列表元素是否为 5 =46=]，它使用<。也就是说，count 将最终检查 (< 5 1)，然后 (< 5 2) , …, 直到 (< 5 6), (< 5 7)。它是指定将使用该顺序的参数调用测试。 glossary entry on satisfy the test 表示（强调）：

(for a two argument test) to be in a state such that the two-place predicate which is the sequence function's test argument returns true when given a first argument that is the object being considered, and when given a second argument that is the result of calling the sequence function's key argument on an element of the sequence function's sequence argument which is being tested for equality;

Common-LISP 调试（包含代码）

Common-LISP debugging (code included)

lisp

debugging

common-lisp

备注