On May 14, 3:42 am, Jernej Azarija wrote:
> This is what the guys on #python (IRC) told me to consider
>
> >>> d1 = {-1: 0, -2: 0}>>> d2 = {-2: 0, -1: 0}>>> d1{-2: 0, -1: 0}>>> d2{-1:
> >>> 0, -2: 0}>>> d1 == d2True>>> d1.keys()[-2, -1]>>> d2.keys()[-1, -2]>>>
That is an extremely sneaky example
d1 = {-1: 0, -2: 0}
d2 = {-2: 0, -1: 0}
d1
> {-2: 0, -1: 0}
d2
> {-1: 0, -2: 0}
d1 == d2
> True
d1.keys()
> [-2, -1]
d2.keys()
> [-1, -2]
NOooo
Ookokokokok. Then the code is good a
This is what the guys on #python (IRC) told me to consider
>>> d1 = {-1: 0, -2: 0}>>> d2 = {-2: 0, -1: 0}>>> d1{-2: 0, -1: 0}>>> d2{-1: 0,
>>> -2: 0}>>> d1 == d2True>>> d1.keys()[-2, -1]>>> d2.keys()[-1, -2]>>>
On Tue, May 14, 2013 at 12:12 PM, Nathann Cohen wrote:
> Hello everybody
No, you cannot. You should always think dictionaries as unordered
structures. Dictionary keys depend not only on the keys themselves but
on the operations performed on the dictionary. Take a simple case with
hash collision:
{{{
sage: A={}
sage: B={}
sage: A[hash('a')]=0
sage: A['a']=1
sage: B['a']
