candide <cand...@free.invalid> writes: > 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 ?
Not at all, actually. I'd be surprised if the multiplication operator was aware of object constructors. Even arrays are "objects" in Python. Should the multiplication operator know how to instantiate three arrays from a single array instance? What about an instance of a user-defined class? > So I suppose all the subarrays reférence > the same array : > >>>> id(t[0]), id(t[1]), id(t[2]) > (3077445996L, 3077445996L, 3077445996L) >>>> > As they should. > > 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 ? >>> 2d_zero_vector = lambda len: [[0, 0] for _ in range(len)] >>> t = 2d_zero_vector(3) >>> print t [[0, 0], [0, 0], [0, 0]] >>> t[0][0] = 1 >>> print t [[1, 0], [0, 0], [0, 0], [0, 0]] (Of course, if you're doing matrix math you'll probably want to work with numpy which has a function for doing just this) -- http://mail.python.org/mailman/listinfo/python-list
