一个新挑战substr可以提取字符串，有人发现了可以对表操作的函数吗

nonsmall

(defun sublist (lst be end / new_list)
 (setq end (+ be end -3) be (- be 2));为了和substr一样
 (while (<= be end)
  (setq new_list (append new_list (list (nth (setq be (1+ be)) lst))))
 )
)

(sublist (list 1 2 3 4 5 6 7 8 9) 3 6)

(3 4 5 6 7 8)

nonsmall

命令: (ko-substr-<list (list 1 2 3 4) 2 7)
nil

命令: (ko-substr-<list (list 1 2 3 4) 2 4)
(2 3 4 nil)

命令: (substr "1234" 2 7)
"234"

有错误;看来还不是99.9％

nonsmall

我修改一下

命令: (sublist (list 1 2 3 4) 2 7)
(2 3 4)

caoyin

一个另类的写法：

(defun sublist (lst st len)
  (repeat (1- st) (setq lst (cdr lst)))
  (setq lst (reverse lst))
  (repeat (- (length lst) len) (setq lst (cdr lst)))
  (reverse lst)
)

nonsmall

局部变量都省下了 果然另类！