Yes you are. List comprehension makes you create list of lists without reference-sharing. You should also find a recipe about that on the python cookbook.
On Thu, Jun 17, 2010 at 12:21 PM, candide <cand...@free.invalid> wrote: > Let's the following code : > >>>> t=[[0]*2]*3 >>>> t > [[0, 0], [0, 0], [0, 0]] >>>> t[0][0]=1 >>>> t > [[1, 0], [1, 0], [1, 0]] > > Rather surprising, isn't it ? So I suppose all the subarrays reférence the > same array : > >>>> id(t[0]), id(t[1]), id(t[2]) > (3077445996L, 3077445996L, 3077445996L) >>>> > > > So what is the right way to initialize to 0 a 2D array ? Is that way correct > : > > >>>> t=[[0 for _ in range(2)] for _ in range(3)] > > It seems there is no more trouble now : > >>>> t > [[0, 0], [0, 0], [0, 0]] >>>> t[0][0]=1 >>>> t > [[1, 0], [0, 0], [0, 0]] >>>> > > Correct ? > -- > http://mail.python.org/mailman/listinfo/python-list > -- Matteo Landi http://www.matteolandi.net/ -- http://mail.python.org/mailman/listinfo/python-list
