On 06/17/10 20:21, candide 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) >>>>
Yep, you're right. They share the same subarray if you uses multiplication to build the array. > 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)] Right again. That's the way to go. Although if the elements are immutable, you can create the innermost array by multiplication: t=[[0]*2 for _ in range(3)] -- http://mail.python.org/mailman/listinfo/python-list
