Thanks everyone! The list-comprehension approach should work in my case
as I know a priori that my data structure is finite. Still, it'd be
awfully convenient for Python's copy module to offer a copy.deepercopy
function -- even if it comes with the caveat that it will explode if
given a data
On Tue, 27 Oct 2009 15:16:15 -0700, Gary Herron wrote:
Try this:
d = [1,2,3]
r = [[x for x in d] for i in range(3)]
r[1][2] = 999
r
[[1, 2, 3], [1, 2, 999], [1, 2, 3]]
[x for x in d] is better written as d[:]
--
Steven
--
http://mail.python.org/mailman/listinfo/python-list
copy.deepcopy apparently preserves multiple references to the same object:
$ python
Python 2.5.2 (r252:60911, Jan 4 2009, 17:40:26)
[GCC 4.3.2] on linux2
Type help, copyright, credits or license for more information.
import copy
d = [1,2,3]
r = copy.deepcopy([d]*3)
On 27Oct2009 14:01, Scott Pakin scott+...@pakin.org wrote:
| copy.deepcopy apparently preserves multiple references to the same object:
|
| $ python
| Python 2.5.2 (r252:60911, Jan 4 2009, 17:40:26)
| [GCC 4.3.2] on linux2
| Type help, copyright, credits or license for more
On 28Oct2009 08:23, I wrote:
| | I wanted to wind up with r being [[1, 2, 3], [1, 2, 999], [1, 2, 3]].
| | What's the right way to construct r as a list of *independent* d lists?
|
| Well, you would need to write your own. But consider this:
| x = [1, 2]
| x.append(x)
| Your deepercopy()
Scott Pakin wrote:
copy.deepcopy apparently preserves multiple references to the same object:
$ python
Python 2.5.2 (r252:60911, Jan 4 2009, 17:40:26)
[GCC 4.3.2] on linux2
Type help, copyright, credits or license for more information.
import copy
d = [1,2,3]
r =
