如何从一个点表中把其中的点都提取出来?
比如有(x1 y1 z1 x2 y2 z2)怎么提取为(x1 y1 z1) (x2 y2 z2)
求曲线交点的时候,有时候会有两个以上交点,但有的点并不需要,就想提取有用的点。
(setq jd'(x1 y1 z1 x2 y2 z2) i 0 a'())
(repeat (/ (length jd)3)
(setq a(append a(list(list(nth i jd)(nth(1+ i) jd)(nth(+ 2 i)jd))))
i(+ 3 i))
) http://bbs.xdcad.net/thread-668406-1-1.html llsheng_73 发表于 2013-11-19 09:04 static/image/common/back.gif
(setq jd'(x1 y1 z1 x2 y2 z2) i 0 a'())
(repeat (/ (length jd)3)
(setq a(append a(list(list(nth...
非常感谢! ll_j 发表于 2013-11-19 09:05 static/image/common/back.gif
http://bbs.xdcad.net/thread-668406-1-1.html
谢谢帮助! (setq PtLst nil)
(while Pts (setq PtLst (append PtLst (list (list (car Pts) (cadr Pts) (caddr Pts)))) Pts (cdddr pts))) NJZX05 发表于 2013-11-19 11:28 static/image/common/back.gif
(setq PtLst nil)
(while Pts (setq PtLst (append PtLst (list (list (car Pts) (cadr Pts) (caddr Pts)) ...
谢谢! NJZX05 发表于 2013-11-19 11:28 static/image/common/back.gif
(setq PtLst nil)
(while Pts (setq PtLst (append PtLst (list (list (car Pts) (cadr Pts) (caddr Pts)) ...
这个更简洁,效率也应该相当高学习了
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