On 6/18/10 8:40 AM, bart.c wrote:
> I suppose there are pros and cons to both approaches; copying all the time
> at least avoids some of the odd effects and inconsistencies you get using
> Python:
What inconsistencies? All your examples are perfectly consistent. Its
just consistent to different id
"Steven D'Aprano" wrote in message
news:4c1b8ac6$0$14148$c3e8...@news.astraweb.com...
On Fri, 18 Jun 2010 12:07:38 +0100, bart.c wrote:
(Although I have an issue with the way that that append works. I tried
it in another, simpler language (which always does deep copies):
L:=(1,2,3)
L append:
On Fri, 18 Jun 2010 12:07:38 +0100, bart.c wrote:
> (Although I have an issue with the way that that append works. I tried
> it in another, simpler language (which always does deep copies):
>
> L:=(1,2,3)
> L append:= L
> print L
>
> output: (1,2,3,(1,2,3))
>
> which is exactly what I'd expect
Lie Ryan wrote:
On 06/18/10 20:00, bart.c wrote:
(I
don't know if Python allows circular references, but that would give
problems anyway: how would you even print out such a list?)
Python uses ellipsis to indicate recursive list:
a = [1, 2, 3]
a.append(a)
a
[1, 2, 3, [...]]
Ok, perhaps w
On 06/18/10 20:00, bart.c wrote:
> (I
> don't know if Python allows circular references, but that would give
> problems anyway: how would you even print out such a list?)
Python uses ellipsis to indicate recursive list:
>>> a = [1, 2, 3]
>>> a.append(a)
>>> a
[1, 2, 3, [...]]
--
http://mail.pyt
Benjamin Kaplan wrote:
On Thu, Jun 17, 2010 at 4:20 PM, bart.c wrote:
I don't know how Python does things, but an object should either
specify a special way of duplicating itself, or lend itself to some
standard way of doing so. (So for a list, it's just a question of
copying the data in the
On Fri, 18 Jun 2010 00:20:30 +0100, bart.c wrote:
> The code is clearly trying to set only t[0][0] to 1, not t[1][0] and
> t[2][0] as well.
Trying to guess the motivation of the person writing code is tricky, but
in this case, that's a reasonable assumption. I can't think of any reason
why some
On 06/18/10 09:20, bart.c wrote:
>
> "J Kenneth King" wrote in message
> news:87wrtxh0dq@agentultra.com...
>> candide 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 s
On Jun 17, 6:44 pm, Benjamin Kaplan wrote:
> It's the recursively duplicating each element that's the problem. How
> do you know when to stop?
Thats easy, stack overflow! ;-)
--
http://mail.python.org/mailman/listinfo/python-list
On Thu, Jun 17, 2010 at 4:20 PM, bart.c wrote:
>
> "J Kenneth King" wrote in message
> news:87wrtxh0dq@agentultra.com...
>>
>> candide writes:
>>
>>> Let's the following code :
>>>
>> t=[[0]*2]*3
>> t
>>>
>>> [[0, 0], [0, 0], [0, 0]]
>>
>> t[0][0]=1
>> t
>>>
>>> [[1, 0],
"J Kenneth King" wrote in message
news:87wrtxh0dq@agentultra.com...
candide 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.
The code is clearly trying to set only
candide 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 arr
candide wrote:
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)]
That's overkill :) You can skip the inner loop by using a list display, eg
t=[[0,0] for _ in range(3)]
It seems there is no more trouble now :
Yes you are. List comprehension makes you create list of lists without
reference-sharing. You should also find a recipe about that on the
python cookbook.
On Thu, Jun 17, 2010 at 12:21 PM, candide wrote:
> Let's the following code :
>
t=[[0]*2]*3
t
> [[0, 0], [0, 0], [0, 0]]
t[0][0
On 06/17/10 20:21, candide wrote:
> 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 ? So I suppose all the subarrays reférence
> the same array :
>
id(t[0]), id(t[1]), id(t[2])
>
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