f

#### List of lists surprising behaviour

```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 ref�rence
the same array :

>>> id(t[0]), id(t[1]), id(t[2])
(3077445996L, 3077445996L, 3077445996L)
>>>

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)]

It seems there is no more trouble now :

>>> t
[[0, 0], [0, 0], [0, 0]]
>>> t[0][0]=1
>>> t
[[1, 0], [0, 0], [0, 0]]
>>>

Correct ?
```
 0
candide (110)
6/17/2010 10:21:31 AM
comp.lang.python 77058 articles. 6 followers.

15 Replies
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```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 ref�rence
> the same array :
>
>>>> id(t[0]), id(t[1]), id(t[2])
> (3077445996L, 3077445996L, 3077445996L)
>>>>

Yep, you're right. They share the same subarray if you uses
multiplication to build the array.

> 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)]

Right again. That's the way to go. Although if the elements are
immutable, you can create the innermost array by multiplication:

t=[[0]*2 for _ in range(3)]
```
 0
Lie
6/17/2010 10:59:52 AM
```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 <candide@free.invalid> wrote:
> Let's the following code :
>
>>>> t=3D[[0]*2]*3
>>>> t
> [[0, 0], [0, 0], [0, 0]]
>>>> t[0][0]=3D1
>>>> t
> [[1, 0], [1, 0], [1, 0]]
>
> Rather surprising, isn't it ? So I suppose all the subarrays ref=E9rence =
the
> same array :
>
>>>> id(t[0]), id(t[1]), id(t[2])
> (3077445996L, 3077445996L, 3077445996L)
>>>>
>
>
> So what is the right way to initialize to 0 a 2D array ? Is that way corr=
ect
> =A0:
>
>
>>>> t=3D[[0 for _ in range(2)] for _ in range(3)]
>
> It seems there is no more trouble now :
>
>>>> t
> [[0, 0], [0, 0], [0, 0]]
>>>> t[0][0]=3D1
>>>> t
> [[1, 0], [0, 0], [0, 0]]
>>>>
>
> Correct ?
> --
> http://mail.python.org/mailman/listinfo/python-list
>

--=20
Matteo Landi
http://www.matteolandi.net/
```
 0
landimatte (24)
6/17/2010 11:11:25 AM
```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 :
>
>  >>> t
> [[0, 0], [0, 0], [0, 0]]
>  >>> t[0][0]=1
>  >>> t
> [[1, 0], [0, 0], [0, 0]]
>  >>>
>
> Correct ?

```
 0
Boris
6/17/2010 1:56:32 PM
```candide <candide@free.invalid> 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 arrays are "objects" in Python.  Should the
multiplication operator know how to instantiate three arrays from a
single array instance?  What about an instance of a user-defined class?

> So I suppose all the subarrays reférence
> the same array :
>
>>>> id(t[0]), id(t[1]), id(t[2])
> (3077445996L, 3077445996L, 3077445996L)
>>>>
>

As they should.

>
> 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)]
>
> It seems there is no more trouble now :
>
>>>> t
> [[0, 0], [0, 0], [0, 0]]
>>>> t[0][0]=1
>>>> t
> [[1, 0], [0, 0], [0, 0]]
>>>>
>
> Correct ?

>>> 2d_zero_vector = lambda len: [[0, 0] for _ in range(len)]
>>> t = 2d_zero_vector(3)
>>> print t
[[0, 0], [0, 0], [0, 0]]
>>> t[0][0] = 1
>>> print t
[[1, 0], [0, 0], [0, 0], [0, 0]]

(Of course, if you're doing matrix math you'll probably want to work
with numpy which has a function for doing just this)
```
 0
J
6/17/2010 8:25:53 PM
```"J Kenneth King" <james@agentultra.com> wrote in message
news:87wrtxh0dq.fsf@agentultra.com...
> candide <candide@free.invalid> 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 t[0][0] to 1, not t[1][0] and t[2][0]
as well.

This behaviour is quite scary actually, especially when t[0]=42 *does* work
as expected, while t[0][0]=42 is apparently duplicated. It appears
inconsistent.

> I'd be surprised if the multiplication operator was aware of object
> constructors.  Even arrays are "objects" in Python.  Should the
> multiplication operator know how to instantiate three arrays from a
> single array instance?  What about an instance of a user-defined class?

Multiplication operators shouldn't need to be directly aware of any such
thing; it should just request that an object be duplicated without worrying
about how it's done.

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
list, then recursively duplicating each new element..)

--
Bartc

```
 0
bart
6/17/2010 11:20:30 PM
```On Thu, Jun 17, 2010 at 4:20 PM, bart.c <bartc@freeuk.com> wrote:
>
> "J Kenneth King" <james@agentultra.com> wrote in message
> news:87wrtxh0dq.fsf@agentultra.com...
>>
>> candide <candide@free.invalid> writes:
>>
>>> Let's the following code :
>>>
>>>>>> t=3D[[0]*2]*3
>>>>>> t
>>>
>>> [[0, 0], [0, 0], [0, 0]]
>>>>>>
>>>>>> t[0][0]=3D1
>>>>>> t
>>>
>>> [[1, 0], [1, 0], [1, 0]]
>>>
>>> Rather surprising, isn't it ?
>>
>> Not at all, actually.
>
> The code is clearly trying to set only t[0][0] to 1, not t[1][0] and t[2]=
[0]
> as well.
>
> This behaviour is quite scary actually, especially when t[0]=3D42 *does* =
work
> as expected, while t[0][0]=3D42 is apparently duplicated. It appears
> inconsistent.
>
>> I'd be surprised if the multiplication operator was aware of object
>> constructors. =A0Even arrays are "objects" in Python. =A0Should the
>> multiplication operator know how to instantiate three arrays from a
>> single array instance? =A0What about an instance of a user-defined class=
?
>
> Multiplication operators shouldn't need to be directly aware of any such
> thing; it should just request that an object be duplicated without worryi=
ng
> about how it's done.
>
> 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
> list, then recursively duplicating each new element..)
>
> --
> Bartc

It's the recursively duplicating each element that's the problem. How
do you know when to stop?
```
 0
Benjamin
6/17/2010 11:44:49 PM
```On Jun 17, 6:44=A0pm, Benjamin Kaplan <benjamin.kap...@case.edu> wrote:

> It's the recursively duplicating each element that's the problem. How
> do you know when to stop?

Thats easy, stack overflow! ;-)
```
 0
rantingrick
6/18/2010 12:48:06 AM
```On 06/18/10 09:20, bart.c wrote:
>
> "J Kenneth King" <james@agentultra.com> wrote in message
> news:87wrtxh0dq.fsf@agentultra.com...
>> candide <candide@free.invalid> 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 t[0][0] to 1, not t[1][0] and
> t[2][0]
> as well.
>
> This behaviour is quite scary actually, especially when t[0]=42 *does* work
> as expected, while t[0][0]=42 is apparently duplicated. It appears
> inconsistent.

I agree, the behavior is often quite inconvenient, but I disagree that
it is inconsistent. List multiplication behavior is consistent with the
tenet: "objects are never copied unless explicitly requested"

Peeking further:
t = [[0] * 2] * 3
print id(t[0]) == id(t[1])  # True
print id(t[0][0]) == id(t[1][0])  # True

so, it is consistent (though it is quite inconvenient).

>> I'd be surprised if the multiplication operator was aware of object
>> constructors.  Even arrays are "objects" in Python.  Should the
>> multiplication operator know how to instantiate three arrays from a
>> single array instance?  What about an instance of a user-defined class?
>
> Multiplication operators shouldn't need to be directly aware of any such
> thing; it should just request that an object be duplicated without
> worrying about how it's done.
>
> 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 list, then recursively duplicating each new element..)

That is inconsistent with the tenet. Moreover, the implicit copying
makes it quite difficult to reason about the program's behavior. How
would you propose this list should be copied:

class O(object):
def __init__(self):
self.attr = 0
self.attr += 1

b = [[O()] * 2] * 3
```
 0
Lie
6/18/2010 1:38:59 AM
```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 somebody would explicitly *want* that behaviour:

# Does it make sense to talk of anonymous aliases?
list_of_aliases = [[0]*2]*3
list_of_aliases[0][0] = 1
assert list_of_aliases[0][1] == 1

so it is a safe guess that anyone writing [[0]*2]*3 has probably made a
mistake.

However, I've certainly done something like this:

a = [0]*2
my_instance.items = a
a[0] = 1
assert my_instance.items[0] = 1

This is the same fundamental behaviour with the same cause: Python does
not copy objects unless you explicitly tell it to.

I cheerfully accept that the behaviour of [[0]*2]*3 is a Gotcha, but it
follows logically from Python's object model and assignment rules. If you
are surprised by it, it just goes to show that your understanding of
Python has at least one hole in it.

> This behaviour is quite scary actually, especially when t[0]=42 *does*
> work as expected, while t[0][0]=42 is apparently duplicated. It appears
> inconsistent.

Emphasis on the word "appears". It actually displays a deep consistency
with the language fundamentals.

If you're ever interviewing somebody for a position as Python developer,
this is a quick test to distinguish those who know the language from
those who know the language *well*.

>> I'd be surprised if the multiplication operator was aware of object
>> constructors.  Even arrays are "objects" in Python.  Should the
>> multiplication operator know how to instantiate three arrays from a
>> single array instance?  What about an instance of a user-defined class?
>
> Multiplication operators shouldn't need to be directly aware of any such
> thing; it should just request that an object be duplicated without
> worrying about how it's done.

The multiplication operator is not a duplicator (copier). It is a
*repetition* operator: repeat the object N times, not make N copies.

> I don't know how Python does things,

Well there you go :)

> but an object should either specify
> a special way of duplicating itself, or lend itself to some standard way
> of doing so.

import copy
copy.copy(obj)

Dicts have a copy() method as a shortcut, and for lists you can use
slicing:

L = [1,2,3]
Lcopy = L[:]

> (So for a list, it's just a question of copying the data in
> the list, then recursively duplicating each new element..)

There's nothing "just" about that. Consider:

L = [1, 2]
L.append(L)

How would you copy that?

The correct answer is:

x = copy.deepcopy(L)

--
Steven
```
 0
Steven
6/18/2010 5:36:33 AM
```Benjamin Kaplan wrote:
> On Thu, Jun 17, 2010 at 4:20 PM, bart.c <bartc@freeuk.com> 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 list, then recursively duplicating each new
>> element..)

> It's the recursively duplicating each element that's the problem. How
> do you know when to stop?

When you reach a primitive object (one not comprising other objects). (I
don't know if Python allows circular references, but that would give
problems anyway: how would you even print out such a list?)

--
Bartc

```
 0
bart
6/18/2010 10:00:39 AM
```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, [...]]
```
 0
Lie
6/18/2010 10:46:07 AM
```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 whatever logic print uses to know when to stop and show "...",
can also be used by copy handlers...

(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, and not (1,2,3,(1,2,3,(1,2,3,...))) )

--
bartc

```
 0
bart
6/18/2010 11:07:38 AM
```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,
> and not (1,2,3,(1,2,3,(1,2,3,...))) )

I find that behaviour a big surprise. You asked to append the list L, not
a copy of the list L. So why is this "simpler" language making a copy

If you asked for:

L:=(1,2,3)
M:=(0,1)
M append:= L

does it also append a copy of L instead of L? If so, how do you append
the original rather than wastefully making a copy? If L is huge, making a
copy before appending will be slow, and potentially fail.

--
Steven
```
 0
Steven
6/18/2010 3:03:34 PM
```"Steven D'Aprano" <steve@REMOVE-THIS-cybersource.com.au> wrote in message
news:4c1b8ac6\$0\$14148\$c3e8da3@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:= L
>> print L
>>
>> output:  (1,2,3,(1,2,3))
>>
>> which is exactly what I'd expect,
>> and not (1,2,3,(1,2,3,(1,2,3,...))) )
>
> I find that behaviour a big surprise. You asked to append the list L, not
> a copy of the list L. So why is this "simpler" language making a copy
> without being asked?
>
> If you asked for:
>
> L:=(1,2,3)
> M:=(0,1)
> M append:= L
>
> does it also append a copy of L instead of L?

It make a copy.

> If so, how do you append
> the original rather than wastefully making a copy?

I don't think it can, without perhaps doing something with explicit
pointers.

> If L is huge, making a
> copy before appending will be slow, and potentially fail.

I don't know whether L append:=L requires 3 times the space of L, or 2
times, during the operation. But it should be doable using just twice the
space.

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:

a1=[1,2,3]
a1.append(a1)

a2=[1,2,3]
b=[1,2,3]
a2.append(b)

a3=[1,2,3]
a3.append([1,2,3])

print ("a1 = ",a1)
print ("a2 = ",a2)
print ("a3 = ",a3)

Here, a1 ends up with a different result from a2, a3, even though the same
value is being appended to the same list of numbers in each case. And it
might sometimes bite you in the arse as the OP demonstrated:

L=[1,2,3]
M=[0,1]
M.append(L)

print (M)     # output: [0, 1, [1, 2, 3]]

L[1]=31416

print (M)     # output: [0, 1, [1, 31416, 3]], yikes!

--
Bartc

```
 0
bartc (786)
6/18/2010 3:40:35 PM
```This is an OpenPGP/MIME signed message (RFC 2440 and 3156)
--------------enig3BB2A9F215CDFED19118EACA
Content-Type: text/plain; charset=ISO-8859-1
Content-Transfer-Encoding: quoted-printable

On 6/18/10 8:40 AM, bart.c wrote:
> I suppose there are pros and cons to both approaches; copying all the t=
ime
> at least avoids some of the odd effects and inconsistencies you get usi=
ng
> Python:

What inconsistencies? All your examples are perfectly consistent. Its
just consistent to different ideals. Python never copies implicitly; and
every object you create has a concrete identity in and of itself,
utterly separate from its potential equality.

> a1=3D[1,2,3]
> a1.append(a1)

The "a1" object is a distinct object unto itself; you are appending said
distinct object onto the end of itself. Entirely doable, even if you
don't usually want to. Recursive, but doable if that's what you want. If
you wished to append a copy, you must-- as always, consistently--
explicitly copy it.

I.e.,

a1.append(a1[:])

> a2=3D[1,2,3]
> b=3D[1,2,3]
> a2.append(b)

a2 is a distinct object from b; that the two objects are equal means
nothing. So you're appending an object to the end of a2 which happens to
be equal to it.

> a3=3D[1,2,3]
> a3.append([1,2,3])

This is just another way of writing the previous example; that in one
you are naming the object [1,2,3] and in the other you are not, doesn't
mean anything. A named vs unnamed object in Python behaves exactly the sa=
me.

> print ("a1 =3D ",a1)
> print ("a2 =3D ",a2)
> print ("a3 =3D ",a3)
>=20
> Here, a1 ends up with a different result from a2, a3, even though the s=
ame
> value is being appended to the same list of numbers in each case.=20

There's the rub: the VALUE is not being appended. The *actual object* is.=

Consider:

>>> print a1 is a2
False

> And it
> might sometimes bite you in the arse as the OP demonstrated:
>=20
> L=3D[1,2,3]
> M=3D[0,1]
> M.append(L)
>=20
> print (M)     # output: [0, 1, [1, 2, 3]]
>=20
> L[1]=3D31416
>=20
> print (M)     # output: [0, 1, [1, 31416, 3]], yikes!

That might bite you on your arse if you think Python implicitly copies;
but since the rule is very simple -- it /never/ implicitly copies -- and
that it objects are discrete entities and not just names of certain
values, it more likely then not will be beneficial to you quite often
down the road. You'll -want- that to be how things work. Eventually.

--=20

Stephen Hansen
... Also: Ixokai
... Mail: me+list/python (AT) ixokai (DOT) io
... Blog: http://meh.ixokai.io/

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