alex23 schreef op wo 21-12-2011 om 16:50 [-0800]:
> On Dec 22, 8:25 am, Eric <einazaki...@yahoo.com> wrote:
> > This surprises me, can someone tell me why it shouldn't? I figure if
> > I want to create and initialize three scalars the just do "a=b=c=7",
> > for example, so why not extend it to arrays.
>
> The thing to remember is that everything is an object, and that it's
> better to think of variables as labels on an object.
>
> So: a=b=c=7 means that _one_ integer object with the value of 7 can be
> referenced using any of the labels a, b or c. x=y=z=[] means that
> _one_ empty list can be referenced using x, y or z.
>
> The difference is that the value of a number object _cannot be
> changed_ ('immutable') while a list can be modified to add or remove
> items ('mutable'). a=10 just reassigns the label a to an integer
> object of value 10. x.append("foo") _modifies_ the list referred to by
> x, which is the same list known as y & z.
>
> > Also, is there a more pythonic way to do "x=[], y=[], z=[]"?
>
> I'd say that _is_ the most pythonic way, it's very obvious in its
> intent (or would be with appropriate names). If it bothers you that
> much:
>
> def listgen(count, default=[]):
> for _ in xrange(count):
> yield default[:]
>
> x, y, z = listgen(3)
>
I'm afraid it's dangerous to encourage the use of '[]' as assignment to
a parameter in a function definition. If you use the function several
times 'default' always points to the same list.
>>> def return_list(list_ = []):
>>> return list_
>>> a_list = return_list()
>>> a_list
[]
>>> a_list.append(3)
>>> a_list
[3]
>>> b_list = return_list()
>>> b_list
>>> [3] # !!??
>>> def return_list():
>>> return []
>>> a_list = return_list()
>>> a_list
[]
>>> a_list.append(3)
>>> a_list
[3]
>>> b_list = return_list()
>>> b_list
>>> [] # OK!
I only use python3 so I don't know how these things work in other
versions.
No problem in your function since you yield a copy, but I've already
seen long threads about this.
I would change your function to (Python3.x):
def empty_lists(count):
for _ in range(count):
yield []
Regards,
Rolf
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