On Tue, Nov 6, 2012 at 1:21 AM, Andrew Robinson <andr...@r3dsolutions.com> wrote: > If you nest it another time; > [[[None]]]*4, the same would happen; all lists would be independent -- but > the objects which aren't lists would be refrenced-- not copied. > > a=[[["alpha","beta"]]]*4 would yield: > a=[[['alpha', 'beta']], [['alpha', 'beta']], [['alpha', 'beta']], [['alpha', > 'beta']]] > and a[0][0]=1 would give [[1],[['alpha', 'beta']], [['alpha', 'beta']], > [['alpha', 'beta']]]] > rather than a=[[1], [1], [1], [1]] > > Or at another level down: a[0][0][0]=1 would give: a=[[[1, 'beta']], > [['alpha', 'beta']], [['alpha', 'beta']], [['alpha', 'beta']] ] > rather than a=[[[1, 'beta']], [[1, 'beta']], [[1, 'beta']], [[1, 'beta']]]
You wrote "shallow copy". When the outer-level list is multiplied, the mid-level lists would be copied. Because the copies are shallow, although the mid-level lists are copied, their contents are not. Thus the inner-level lists would still be all referencing the same list. To demonstrate: >>> from copy import copy >>> class ShallowCopyList(list): ... def __mul__(self, number): ... new_list = ShallowCopyList() ... for _ in range(number): ... new_list.extend(map(copy, self)) ... return new_list ... >>> a = ShallowCopyList([[["alpha", "beta"]]]) >>> a [[['alpha', 'beta']]] >>> b = a * 4 >>> b [[['alpha', 'beta']], [['alpha', 'beta']], [['alpha', 'beta']], [['alpha', 'beta']]] >>> b[0][0][0] = 1 >>> b [[[1, 'beta']], [[1, 'beta']], [[1, 'beta']], [[1, 'beta']]] >>> b[0][0] = 1 >>> b [[1], [[1, 'beta']], [[1, 'beta']], [[1, 'beta']]] This shows that assignments at the middle level are independent with a shallow copy on multiplication, but assignments at the inner level are not. In order to achieve the behavior you describe, a deep copy would be needed. > That really is what people *generally* want. > If the entire list is meant to be read only -- the change would affect > *nothing* at all. The time and memory cost of the multiplication operation would become quadratic instead of linear. > See if you can find *any* python program where people desired the > multiplication to have the die effect that changing an object in one of the > sub lists -- changes all the objects in the other sub lists. > > I'm sure you're not going to find it -- and even if you do, it's going to be > 1 program in 1000's. Per the last thread where we discussed extremely rare scenarios, shouldn't you be rounding "1 in 1000s" up to 20%? ;-) -- http://mail.python.org/mailman/listinfo/python-list
