[LISP程序]：可否有简单方法？

zhoujun74518

(setq pt1 '(0 0) pt2 '(1 1) pt3 '(2 2) pt4 '(3 3) pt5 '(4 4) loc_1 5)
(setq        pt1 (list (* loc_1 (car pt1)) (cadr pt1))
        pt2 (list (* loc_1 (car pt2)) (cadr pt2))
        pt3 (list (* loc_1 (car pt3)) (cadr pt3))
        pt4 (list (* loc_1 (car pt4)) (cadr pt4))
        pt5 (list (* loc_1 (car pt5)) (cadr pt5))
)
请问，可否有简单一点的方法？点有几百个呢！

LUCAS

2. 
   
3. (setq PT1   '(0 0)
   
4.       PT2   '(1 1)
   
5.       PT3   '(2 2)
   
6.       PT4   '(3 3)
   
7.       PT5   '(4 4)
   
8.       LOC_1 5
   
9. )
   
10. 
    
11. (setq N 1)
    
12. (setq LL NIL)
    
13. (repeat        5 ;;点数
    
14.   (setq LL (append LL (list (eval (read (strcat "PT" (rtos N)))))))
    
15.   (setq N (1+ N))
    
16. )
    
17. (setq N 1)
    
18. (foreach PT LL
    
19.   (set (read (eval (strcat "PT" (rtos N))))
    
20.        (list (* LOC_1 (car PT))
    
21.              (cadr PT)
    
22.        )
    
23.   )
    
24.   (setq N (1+ N))
    
25. )
    

<br /> (setq PT1   '(0 0)<br />       PT2   '(1 1)<br />       PT3   '(2 2)<br />       PT4   '(3 3)<br />       PT5   '(4 4)<br />       LOC_1 5<br /> )<br /> <br /> (setq N 1)<br /> (setq LL NIL)<br /> (repeat        5 ;;点数<br />   (setq LL (append LL (list (eval (read (strcat "PT" (rtos N)))))))<br />   (setq N (1+ N))<br /> )<br /> (setq N 1)<br /> (foreach PT LL<br />   (set (read (eval (strcat "PT" (rtos N))))<br />        (list (* LOC_1 (car PT))<br />              (cadr PT)<br />        )<br />   )<br />   (setq N (1+ N))<br /> )<br />

雄啸

(defun ptx ( pt loc_1 / )
(setq pt (* loc_1 (car pt)(cadr pt)))
)
调用的时候,如：   (ptx '(0 0) 5)

雄啸

2. 
   
3.   [FONT=courier new]
   
4. ...
   
5. (setq li (list '(0 0) '(1 1) '(2 2) '(3 3) '(4 4) ) ptl nil)
   
6. (while (setq pt (car li))
   
7.   (setq ptl (append ptl (ptx pt 5))
   
8.           li (cdr li)
   
9.   )
   
10. )
    
11. ...
    
12. (defun ptx ( pt loc_1 / )
    
13. (setq pt (* loc_1 (car pt)(cadr pt)))
    
14. )
    
15.   [/FONT]
    

<br />   [FONT=courier new]<br /> ...<br /> (setq li (list '(0 0) '(1 1) '(2 2) '(3 3) '(4 4) ) ptl nil)<br /> (while (setq pt (car li))<br />   (setq ptl (append ptl (ptx pt 5))<br />           li (cdr li)<br />   )<br /> )<br /> ...<br /> (defun ptx ( pt loc_1 / )<br /> (setq pt (* loc_1 (car pt)(cadr pt)))<br /> )<br />   [/FONT]<br />

雄啸

用mapcar更简单
(setq li (list '(0 0) '(1 1) '(2 2) '(3 3) '(4 4) ) loc_1 5)
(setq ptl (mapcar '(lambda (x) (* loc_1 (car x) (cadr x)) ) li))
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(0 5 20 45 80)