请根据输入的原始列表，输出目标列表：
原始列表：[1,2,None,None,None,None,None,None,3,4,5,6,None,None,None,None,7,8,9,10]
目标列表： [1,2] [3,4,5,6] [7,8,9,10]
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This topic created in 1615 days ago, the information mentioned may be changed or developed.
 | 1 princelai Dec 10, 2021  2 from itertools import groupby [list(g[1]) for g in groupby(ll,key=lambda x:x is not None) if g[0]] |
 | 3 learningman Dec 10, 2021 3 建议自己的作业自己写，楼上那哥们 pythonic 的写法你能理解吗 |
 | 4 ma6254 Dec 10, 2021 自己的作业自己写（滑稽 |
 | 5 deplivesb Dec 10, 2021 v2 已经沦陷为百度知道了吗 |
 | 6 stimw Dec 10, 2021 via Android 为什么大半年前就在问作业了...大半年后还停留在这种问题 |
 | 7 bytesfold Dec 10, 2021 1 @learningman 并不觉得 pythonic ，如果是公司代码不加备注基本无法维护 |
 | 8 JasonEWNL Dec 10, 2021 @deplivesb 高级点，Way to Overflow 。（ @going 话说回来鉴于是作业，或可回归不依赖任何库的一般思路，日后有机会亦能写出不一定最具效率但较易维护的代码。 ``` il = [1, 2, None, None, None, None, None, None, 3, 4, 5, 6, None, None, None, None, 7, 8, 9, 10] ol = [[]] for i in il: if i: ol[-1].append(i) elif ol[-1]: ol.append([]) print(ol) # [[1, 2], [3, 4, 5, 6], [7, 8, 9, 10]] ``` |
 | 9 keepeye Dec 10, 2021 遍历一遍不就可以了吗？这题是送分题 |
 | 10 cnrting Dec 10, 2021 via iPhone 1 print(il[0:2],il[8:12],il[16:20]) |
 | 12 fml87 Dec 10, 2021 lst = [1,2,None,None,None,None,None,None,3,4,5,6,None,None,None,None,7,8,9,10] [*map(eval,re.split(",,+",re.sub("[^0-9,]", "",str(lst) )))] |
 | 13 28Sv0ngQfIE7Yloe Dec 10, 2021 1 为啥我感觉 op 没有提问的礼貌 |
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