如果有两个相等长度的列表a，b：
>>> a = [1,2,3]
>>> b = [4,5,6]
要得到他们各个元素的合并列表c：
>>> print c
[[1, 4], [2, 5], [3, 6]]

我想到的是：
c = []
for i in range(a):
    c.append([a[i], b[i]])

 但总觉得Python一定有更简单优雅的方法，谁能指点一下？
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zip()ÄÚÖú¯Êý¡£
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On 4/8/07, Bing Yuan <aceyuan在gmail.com> wrote:
>
> Èç¹ûÓÐÁ½¸öÏàµÈ³¤¶ÈµÄÁбía£¬b£º
> >>> a = [1,2,3]
> >>> b = [4,5,6]
> ÒªµÃµ½ËûÃǸ÷¸öÔªËصĺϲ¢Áбíc£º
> >>> print c
> [[1, 4], [2, 5], [3, 6]]
>
> ÎÒÏëµ½µÄÊÇ£º
> c = []
> for i in range(a):
>     c.append([a[i], b[i]])
>
>  µ«×ܾõµÃPythonÒ»¶¨Óиü¼òµ¥ÓÅÑŵķ½·¨£¬Ë­ÄÜÖ¸µãһϣ¿
>
>
> _______________________________________________
> python-chinese
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thanks!

On 4/8/07, jessinio smith <jessinio at gmail.com> wrote:
>
> zip()内置函数。
> list 的 extend函数
>
>  On 4/8/07, Bing Yuan <aceyuan at gmail.com> wrote:
>
> >  如果有两个相等长度的列表a，b：
> > >>> a = [1,2,3]
> > >>> b = [4,5,6]
> > 要得到他们各个元素的合并列表c：
> > >>> print c
> > [[1, 4], [2, 5], [3, 6]]
> >
> > 我想到的是：
> > c = []
> > for i in range(a):
> >     c.append([a[i], b[i]])
> >
> >  但总觉得Python一定有更简单优雅的方法，谁能指点一下？
> >
> >
> > _______________________________________________
> > python-chinese
> > Post: send python-chinese at lists.python.cn
> > Subscribe: send subscribe to python-chinese-request at lists.python.cn
> > Unsubscribe: send unsubscribe to
> > python-chinese-request at lists.python.cn
> > Detail Info: http://python.cn/mailman/listinfo/python-chinese
> >
>
>
>
> --
> 注意身体，身体是革命的本钱！！
> _______________________________________________
> python-chinese
> Post: send python-chinese at lists.python.cn
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>>> print zip(a,b)
[[1, 4], [2, 5], [3, 6]]

>>> print [ [a[i],b[i]] for i in range( len(a))]
[[1, 4], [2, 5], [3, 6]]

jessinio smith 写道:
> zip()内置函数。
> list 的 extend函数
>
> On 4/8/07, *Bing Yuan* <aceyuan at gmail.com aceyuan at gmail.com>>
> wrote:
>
>     如果有两个相等长度的列表a，b：
>     >>> a = [1,2,3]
>     >>> b = [4,5,6]
>     要得到他们各个元素的合并列表c：
>     >>> print c
>     [[1, 4], [2, 5], [3, 6]]
>     我想到的是：
>     c = []
>     for i in range(a):
>     c.append([a[i], b[i]])
>     但总觉得Python一定有更简单优雅的方法，谁能指点一下？
>