# using "*" to make a list of lists with repeated (and independent) elements

Discussion in 'Python' started by TP, Sep 26, 2012.

1. ### TPGuest

Hi everybody,

I have tried, naively, to do the following, so as to make lists quickly:

>>> a=[0]*2
>>> a

[0, 0]
>>> a[0]=3
>>> a

[3, 0]

All is working fine, so I extended the technique to do:

>>> a=[[0]*3]*2
>>> a

[[0, 0, 0], [0, 0, 0]]
>>> a[0][0]=2
>>> a

[[2, 0, 0], [2, 0, 0]]

The behavior is no more expected!
The reason is probably that in the first case, 0 is an integer, not a list,
so Python copies two elements that are independent.
In the second case, the elements are [0,0,0], which is a list; when Python
copies a list, he copies in fact the *pointer* to the list, such that we
obtain this apparently strange behavior.

Is it the correct explanation?
In these conditions, how to make this list [[0,0,0],[0,0,0]] with "*"
without this behavior?

Thanks,

TP

TP, Sep 26, 2012

2. ### Ian KellyGuest

Re: using "*" to make a list of lists with repeated (and independent)elements

On Wed, Sep 26, 2012 at 3:20 PM, TP <> wrote:
> Hi everybody,
>
> I have tried, naively, to do the following, so as to make lists quickly:
>
>>>> a=[0]*2
>>>> a

> [0, 0]
>>>> a[0]=3
>>>> a

> [3, 0]
>
> All is working fine, so I extended the technique to do:
>
>>>> a=[[0]*3]*2
>>>> a

> [[0, 0, 0], [0, 0, 0]]
>>>> a[0][0]=2
>>>> a

> [[2, 0, 0], [2, 0, 0]]
>
> The behavior is no more expected!
> The reason is probably that in the first case, 0 is an integer, not a list,
> so Python copies two elements that are independent.
> In the second case, the elements are [0,0,0], which is a list; when Python
> copies a list, he copies in fact the *pointer* to the list, such that we
> obtain this apparently strange behavior.

Mostly correct. When you do [foo] * 3 it extends the list with the
*same objects* no matter what type they are. In the case of integers,
it doesn't matter that it's the same objects, because integers are
immutable. Lists are mutable, however, and so it becomes apparent
that the same objects are repeated when you try to modify one of the
lists.

> In these conditions, how to make this list [[0,0,0],[0,0,0]] with "*"
> without this behavior?

Use a list comprehension:

a = [[0] * 3 for _ in range(2)]

This way the expression `[0] * 3` is re-evaluated at each position in
the outer list, rather than evaluated just once and then copied.

Ian Kelly, Sep 26, 2012

3. ### Paul RubinGuest

TP <> writes:
> copies a list, he copies in fact the *pointer* to the list ....
> Is it the correct explanation?

Yes, that is correct.

> In these conditions, how to make this list [[0,0,0],[0,0,0]] with "*"
> without this behavior?

>>> a = [[0]*3 for i in xrange(2)]
>>> a[0][0]=2
>>> a

[[2, 0, 0], [0, 0, 0]]

Paul Rubin, Sep 26, 2012
4. ### 88888 DihedralGuest

TPæ–¼ 2012å¹´9æœˆ27æ—¥æ˜ŸæœŸå››UTC+8ä¸Šåˆ5æ™‚25åˆ†04ç§’å¯«é“ï¼š
> Hi everybody,
>
>
>
> I have tried, naively, to do the following, so as to make lists quickly:
>
>
>
> >>> a=[0]*2

>
> >>> a

>
> [0, 0]
>
> >>> a[0]=3

>
> >>> a

>
> [3, 0]
>
>
>
> All is working fine, so I extended the technique to do:
>
>
>
> >>> a=[[0]*3]*2

>
> >>> a

>
> [[0, 0, 0], [0, 0, 0]]
>
> >>> a[0][0]=2

>
> >>> a

>
> [[2, 0, 0], [2, 0, 0]]
>
>
>
> The behavior is no more expected!
>
> The reason is probably that in the first case, 0 is an integer, not a list,
>
> so Python copies two elements that are independent.
>
> In the second case, the elements are [0,0,0], which is a list; when Python
>
> copies a list, he copies in fact the *pointer* to the list, such that we
>
> obtain this apparently strange behavior.
>
>
>
> Is it the correct explanation?
>
> In these conditions, how to make this list [[0,0,0],[0,0,0]] with "*"
>
> without this behavior?
>
>
>
> Thanks,
>
>
>
> TP

def zeros(m,n):
for i in xrange(m):
for j in xrange(n):
a[j]=0
return a

>>> a=zeros(3,2)
>>> a

[[0, 0, 0], [0, 0, 0], [0, 0, 0], [0, 0, 0]]
>>>

I think this is what you want.

88888 Dihedral, Sep 26, 2012
5. ### 88888 DihedralGuest

Paul Rubinæ–¼ 2012å¹´9æœˆ27æ—¥æ˜ŸæœŸå››UTC+8ä¸Šåˆ5æ™‚43åˆ†58ç§’å¯«é“ï¼š
> TP <> writes:
>
> > copies a list, he copies in fact the *pointer* to the list ....

>
> > Is it the correct explanation?

>
>
>
> Yes, that is correct.
>
>
>
> > In these conditions, how to make this list [[0,0,0],[0,0,0]] with "*"

>
> > without this behavior?

>
>
>
> >>> a = [[0]*3 for i in xrange(2)]

>
> >>> a[0][0]=2

>
> >>> a

>
> [[2, 0, 0], [0, 0, 0]]

I used numpy before.

Python is not lisp but python can emulate the lisp behaviors.

88888 Dihedral, Sep 26, 2012
6. ### 88888 DihedralGuest

88888 Dihedralæ–¼ 2012å¹´9æœˆ27æ—¥æ˜ŸæœŸå››UTC+8ä¸Šåˆ6æ™‚07åˆ†35ç§’å¯«é“ï¼š
> Paul Rubinæ–¼ 2012å¹´9æœˆ27æ—¥æ˜ŸæœŸå››UTC+8ä¸Šåˆ5æ™‚43åˆ†58ç§’å¯«é“ï¼š
>
> > TP <> writes:

>
> >

>
> > > copies a list, he copies in fact the *pointer* to the list ....

>
> >

>
> > > Is it the correct explanation?

>
> >

>
> >

>
> >

>
> > Yes, that is correct.

>
> >

>
> >

>
> >

>
> > > In these conditions, how to make this list [[0,0,0],[0,0,0]] with "*"

>
> >

>
> > > without this behavior?

>
> >

>
> >

>
> >

>
> > >>> a = [[0]*3 for i in xrange(2)]

>
> >

>
> > >>> a[0][0]=2

>
> >

>
> > >>> a

>
> >

>
> > [[2, 0, 0], [0, 0, 0]]

>
>
>
> I used numpy before.
>
>
>
> Python is not lisp but python can emulate the lisp behaviors.

def zeros(m,n):
a=[]
for i in xrange(m):
a.append([0]*n)
return a

If one wants to tranlate to C, this is the style.

88888 Dihedral, Sep 26, 2012
7. ### Tim ChaseGuest

Re: using "*" to make a list of lists with repeated (and independent)elements

On 09/26/12 17:28, 88888 Dihedral wrote:
> 88888 Dihedralæ–¼ 2012å¹´9æœˆ27æ—¥æ˜ŸæœŸå››UTC+8ä¸Šåˆ6æ™‚07åˆ†35ç§’å¯«é“ï¼š
>>>> In these conditions, how to make this list [[0,0,0],[0,0,0]] with "*"
>>>> without this behavior?
>>> >>> a = [[0]*3 for i in xrange(2)]
>>> >>> a[0][0]=2
>>> >>> a
>>> [[2, 0, 0], [0, 0, 0]]

>
> def zeros(m,n):
> a=[]
> for i in xrange(m):
> a.append([0]*n)
> return a
>
> If one wants to tranlate to C, this is the style.

But this is Python, so why the heck would anybody want to emulate
*C* style? It could also be written in an assembly-language style,
COBOL style, or a Fortran style...none of which are particularly
valuable.

Besides, a C-style would allocate a single array of M*N slots and
then calculate 2d offsets into that single array.

-tkc

Tim Chase, Sep 26, 2012
8. ### 88888 DihedralGuest

Tim Chaseæ–¼ 2012å¹´9æœˆ27æ—¥æ˜ŸæœŸå››UTC+8ä¸Šåˆ6æ™‚44åˆ†42ç§’å¯«é“ï¼š
> On 09/26/12 17:28, 88888 Dihedral wrote:
>
> > 88888 Dihedralæ–¼ 2012å¹´9æœˆ27æ—¥æ˜ŸæœŸå››UTC+8ä¸Šåˆ6æ™‚07åˆ†35ç§’å¯«é“ï¼š

>
> >>>> In these conditions, how to make this list [[0,0,0],[0,0,0]] with "*"

>
> >>>> without this behavior?

>
> >>> >>> a = [[0]*3 for i in xrange(2)]

>
> >>> >>> a[0][0]=2

>
> >>> >>> a

>
> >>> [[2, 0, 0], [0, 0, 0]]

>
> >

>
> > def zeros(m,n):

>
> > a=[]

>
> > for i in xrange(m):

>
> > a.append([0]*n)

>
> > return a

>
> >

>
> > If one wants to tranlate to C, this is the style.

>
>
>
> But this is Python, so why the heck would anybody want to emulate
>
> *C* style? It could also be written in an assembly-language style,
>
> COBOL style, or a Fortran style...none of which are particularly
>
> valuable.
>
>
>
> Besides, a C-style would allocate a single array of M*N slots and
>
> then calculate 2d offsets into that single array.
>
>
>
> -tkc

I don't think a lot programmers can write assembly programs well
for different instruction sets of cpus.

Of course if GCC was not supportd in manny platforms free
of charge, then I won't recommend this style of
programming in python.

88888 Dihedral, Sep 26, 2012
9. ### 88888 DihedralGuest

Re: using "*" to make a list of lists with repeated (and independent)elements

Tim Chaseæ–¼ 2012å¹´9æœˆ27æ—¥æ˜ŸæœŸå››UTC+8ä¸Šåˆ6æ™‚44åˆ†42ç§’å¯«é“ï¼š
> On 09/26/12 17:28, 88888 Dihedral wrote:
>
> > 88888 Dihedralæ–¼ 2012å¹´9æœˆ27æ—¥æ˜ŸæœŸå››UTC+8ä¸Šåˆ6æ™‚07åˆ†35ç§’å¯«é“ï¼š

>
> >>>> In these conditions, how to make this list [[0,0,0],[0,0,0]] with "*"

>
> >>>> without this behavior?

>
> >>> >>> a = [[0]*3 for i in xrange(2)]

>
> >>> >>> a[0][0]=2

>
> >>> >>> a

>
> >>> [[2, 0, 0], [0, 0, 0]]

>
> >

>
> > def zeros(m,n):

>
> > a=[]

>
> > for i in xrange(m):

>
> > a.append([0]*n)

>
> > return a

>
> >

>
> > If one wants to tranlate to C, this is the style.

>
>
>
> But this is Python, so why the heck would anybody want to emulate
>
> *C* style? It could also be written in an assembly-language style,
>
> COBOL style, or a Fortran style...none of which are particularly
>
> valuable.
>
>
>
> Besides, a C-style would allocate a single array of M*N slots and
>
> then calculate 2d offsets into that single array.
>
>
>
> -tkc

I don't think a lot programmers can write assembly programs well
for different instruction sets of cpus.

Of course if GCC was not supportd in manny platforms free
of charge, then I won't recommend this style of
programming in python.

88888 Dihedral, Sep 26, 2012
10. ### 88888 DihedralGuest

Tim Chaseæ–¼ 2012å¹´9æœˆ27æ—¥æ˜ŸæœŸå››UTC+8ä¸Šåˆ6æ™‚44åˆ†42ç§’å¯«é“ï¼š
> On 09/26/12 17:28, 88888 Dihedral wrote:
>
> > 88888 Dihedralæ–¼ 2012å¹´9æœˆ27æ—¥æ˜ŸæœŸå››UTC+8ä¸Šåˆ6æ™‚07åˆ†35ç§’å¯«é“ï¼š

>
> >>>> In these conditions, how to make this list [[0,0,0],[0,0,0]] with "*"

>
> >>>> without this behavior?

>
> >>> >>> a = [[0]*3 for i in xrange(2)]

>
> >>> >>> a[0][0]=2

>
> >>> >>> a

>
> >>> [[2, 0, 0], [0, 0, 0]]

>
> >

>
> > def zeros(m,n):

>
> > a=[]

>
> > for i in xrange(m):

>
> > a.append([0]*n)

>
> > return a

>
> >

>
> > If one wants to tranlate to C, this is the style.

>
>
>
> But this is Python, so why the heck would anybody want to emulate
>
> *C* style? It could also be written in an assembly-language style,
>
> COBOL style, or a Fortran style...none of which are particularly
>
> valuable.
>
>
>
> Besides, a C-style would allocate a single array of M*N slots and
>
> then calculate 2d offsets into that single array.
>
>
>
> -tkc

a=[1, 2,3]
b=[a]*4
print b
a[1]=4
print b

I thnik this is very clear about the situation in entangled references.

88888 Dihedral, Sep 27, 2012
11. ### 88888 DihedralGuest

Re: using "*" to make a list of lists with repeated (and independent)elements

Tim Chaseæ–¼ 2012å¹´9æœˆ27æ—¥æ˜ŸæœŸå››UTC+8ä¸Šåˆ6æ™‚44åˆ†42ç§’å¯«é“ï¼š
> On 09/26/12 17:28, 88888 Dihedral wrote:
>
> > 88888 Dihedralæ–¼ 2012å¹´9æœˆ27æ—¥æ˜ŸæœŸå››UTC+8ä¸Šåˆ6æ™‚07åˆ†35ç§’å¯«é“ï¼š

>
> >>>> In these conditions, how to make this list [[0,0,0],[0,0,0]] with "*"

>
> >>>> without this behavior?

>
> >>> >>> a = [[0]*3 for i in xrange(2)]

>
> >>> >>> a[0][0]=2

>
> >>> >>> a

>
> >>> [[2, 0, 0], [0, 0, 0]]

>
> >

>
> > def zeros(m,n):

>
> > a=[]

>
> > for i in xrange(m):

>
> > a.append([0]*n)

>
> > return a

>
> >

>
> > If one wants to tranlate to C, this is the style.

>
>
>
> But this is Python, so why the heck would anybody want to emulate
>
> *C* style? It could also be written in an assembly-language style,
>
> COBOL style, or a Fortran style...none of which are particularly
>
> valuable.
>
>
>
> Besides, a C-style would allocate a single array of M*N slots and
>
> then calculate 2d offsets into that single array.
>
>
>
> -tkc

a=[1, 2,3]
b=[a]*4
print b
a[1]=4
print b

I thnik this is very clear about the situation in entangled references.

88888 Dihedral, Sep 27, 2012
12. ### Ramchandra ApteGuest

On Thursday, 27 September 2012 04:14:42 UTC+5:30, Tim Chase wrote:
> On 09/26/12 17:28, 88888 Dihedral wrote:
>
> > 88888 Dihedralæ–¼ 2012å¹´9æœˆ27æ—¥æ˜ŸæœŸå››UTC+8ä¸Šåˆ6æ™‚07åˆ†35ç§’å¯«é“ï¼š

>
> >>>> In these conditions, how to make this list [[0,0,0],[0,0,0]] with "*"

>
> >>>> without this behavior?

>
> >>> >>> a = [[0]*3 for i in xrange(2)]

>
> >>> >>> a[0][0]=2

>
> >>> >>> a

>
> >>> [[2, 0, 0], [0, 0, 0]]

>
> >

>
> > def zeros(m,n):

>
> > a=[]

>
> > for i in xrange(m):

>
> > a.append([0]*n)

>
> > return a

>
> >

>
> > If one wants to tranlate to C, this is the style.

>
>
>
> But this is Python, so why the heck would anybody want to emulate
>
> *C* style? It could also be written in an assembly-language style,
>
> COBOL style, or a Fortran style...none of which are particularly
>
> valuable.
>
>
>
> Besides, a C-style would allocate a single array of M*N slots and
>
> then calculate 2d offsets into that single array.
>
>
>
> -tkc

88888 Dihedral is a bot.

Ramchandra Apte, Sep 29, 2012
13. ### Ramchandra ApteGuest

Re: using "*" to make a list of lists with repeated (and independent)elements

On Thursday, 27 September 2012 04:14:42 UTC+5:30, Tim Chase wrote:
> On 09/26/12 17:28, 88888 Dihedral wrote:
>
> > 88888 Dihedralæ–¼ 2012å¹´9æœˆ27æ—¥æ˜ŸæœŸå››UTC+8ä¸Šåˆ6æ™‚07åˆ†35ç§’å¯«é“ï¼š

>
> >>>> In these conditions, how to make this list [[0,0,0],[0,0,0]] with "*"

>
> >>>> without this behavior?

>
> >>> >>> a = [[0]*3 for i in xrange(2)]

>
> >>> >>> a[0][0]=2

>
> >>> >>> a

>
> >>> [[2, 0, 0], [0, 0, 0]]

>
> >

>
> > def zeros(m,n):

>
> > a=[]

>
> > for i in xrange(m):

>
> > a.append([0]*n)

>
> > return a

>
> >

>
> > If one wants to tranlate to C, this is the style.

>
>
>
> But this is Python, so why the heck would anybody want to emulate
>
> *C* style? It could also be written in an assembly-language style,
>
> COBOL style, or a Fortran style...none of which are particularly
>
> valuable.
>
>
>
> Besides, a C-style would allocate a single array of M*N slots and
>
> then calculate 2d offsets into that single array.
>
>
>
> -tkc

88888 Dihedral is a bot.

Ramchandra Apte, Sep 29, 2012
14. ### 88888 DihedralGuest

On Saturday, September 29, 2012 9:46:22 PM UTC+8, Ramchandra Apte wrote:
> On Thursday, 27 September 2012 04:14:42 UTC+5:30, Tim Chase wrote:
>
> > On 09/26/12 17:28, 88888 Dihedral wrote:

>
> >

>
> > > 88888 Dihedralæ–¼ 2012å¹´9æœˆ27æ—¥æ˜ŸæœŸå››UTC+8ä¸Šåˆ6æ™‚07åˆ†35ç§’å¯«é“ï¼š

>
> >

>
> > >>>> In these conditions, how to make this list [[0,0,0],[0,0,0]] with "*"

>
> >

>
> > >>>> without this behavior?

>
> >

>
> > >>> >>> a = [[0]*3 for i in xrange(2)]

>
> >

>
> > >>> >>> a[0][0]=2

>
> >

>
> > >>> >>> a

>
> >

>
> > >>> [[2, 0, 0], [0, 0, 0]]

>
> >

>
> > >

>
> >

>
> > > def zeros(m,n):

>
> >

>
> > > a=[]

>
> >

>
> > > for i in xrange(m):

>
> >

>
> > > a.append([0]*n)

>
> >

>
> > > return a

>
> >

>
> > >

>
> >

>
> > > If one wants to tranlate to C, this is the style.

>
> >

>
> >

>
> >

>
> > But this is Python, so why the heck would anybody want to emulate

>
> >

>
> > *C* style? It could also be written in an assembly-language style,

>
> >

>
> > COBOL style, or a Fortran style...none of which are particularly

>
> >

>
> > valuable.

>
> >

>
> >

>
> >

>
> > Besides, a C-style would allocate a single array of M*N slots and

>
> >

>
> > then calculate 2d offsets into that single array.

>
> >

>
> >

>
> >

>
> > -tkc

>
>
>
> 88888 Dihedral is a bot.

Don't you get it why I avoided the lambda one liner as a functon.

I prefer the def way with a name chosen.

88888 Dihedral, Sep 29, 2012
15. ### 88888 DihedralGuest

Re: using "*" to make a list of lists with repeated (and independent)elements

On Saturday, September 29, 2012 9:46:22 PM UTC+8, Ramchandra Apte wrote:
> On Thursday, 27 September 2012 04:14:42 UTC+5:30, Tim Chase wrote:
>
> > On 09/26/12 17:28, 88888 Dihedral wrote:

>
> >

>
> > > 88888 Dihedralæ–¼ 2012å¹´9æœˆ27æ—¥æ˜ŸæœŸå››UTC+8ä¸Šåˆ6æ™‚07åˆ†35ç§’å¯«é“ï¼š

>
> >

>
> > >>>> In these conditions, how to make this list [[0,0,0],[0,0,0]] with "*"

>
> >

>
> > >>>> without this behavior?

>
> >

>
> > >>> >>> a = [[0]*3 for i in xrange(2)]

>
> >

>
> > >>> >>> a[0][0]=2

>
> >

>
> > >>> >>> a

>
> >

>
> > >>> [[2, 0, 0], [0, 0, 0]]

>
> >

>
> > >

>
> >

>
> > > def zeros(m,n):

>
> >

>
> > > a=[]

>
> >

>
> > > for i in xrange(m):

>
> >

>
> > > a.append([0]*n)

>
> >

>
> > > return a

>
> >

>
> > >

>
> >

>
> > > If one wants to tranlate to C, this is the style.

>
> >

>
> >

>
> >

>
> > But this is Python, so why the heck would anybody want to emulate

>
> >

>
> > *C* style? It could also be written in an assembly-language style,

>
> >

>
> > COBOL style, or a Fortran style...none of which are particularly

>
> >

>
> > valuable.

>
> >

>
> >

>
> >

>
> > Besides, a C-style would allocate a single array of M*N slots and

>
> >

>
> > then calculate 2d offsets into that single array.

>
> >

>
> >

>
> >

>
> > -tkc

>
>
>
> 88888 Dihedral is a bot.

Don't you get it why I avoided the lambda one liner as a functon.

I prefer the def way with a name chosen.

88888 Dihedral, Sep 29, 2012
16. ### Ian KellyGuest

Re: using "*" to make a list of lists with repeated (and independent)elements

On Sat, Sep 29, 2012 at 11:01 AM, 88888 Dihedral
<> wrote:
>
> Don't you get it why I avoided the lambda one liner as a functon.
>
> I prefer the def way with a name chosen.

Certainly, but the Bresenham line algorithm is O(n), which is why it
is so superior to quicksort that is O(n log n). Of course you might
try the Capo Ferro optimization, but I find that Thibault cancels out
Capo Ferro, don't you?

Ian Kelly, Sep 29, 2012
17. ### Chris AngelicoGuest

Re: using "*" to make a list of lists with repeated (and independent)elements

On Sun, Sep 30, 2012 at 3:18 AM, Ian Kelly <> wrote:
> On Sat, Sep 29, 2012 at 11:01 AM, 88888 Dihedral
> <> wrote:
>>
>> Don't you get it why I avoided the lambda one liner as a functon.
>>
>> I prefer the def way with a name chosen.

>
> Certainly, but the Bresenham line algorithm is O(n), which is why it
> is so superior to quicksort that is O(n log n). Of course you might
> try the Capo Ferro optimization, but I find that Thibault cancels out
> Capo Ferro, don't you?

Unless ...

class Agrippa(object):
students = []
def __init__(self):
students.append(self)

enemy = Agrippa()

ChrisA
(maybe I'm pushing this a bit too far)

Chris Angelico, Sep 29, 2012
18. ### 88888 DihedralGuest

On Sunday, September 30, 2012 1:19:22 AM UTC+8, Ian wrote:
> On Sat, Sep 29, 2012 at 11:01 AM, 88888 Dihedral
>
> <> wrote:
>
> >

>
> > Don't you get it why I avoided the lambda one liner as a functon.

>
> >

>
> > I prefer the def way with a name chosen.

>
>
>
> Certainly, but the Bresenham line algorithm is O(n), which is why it
>
> is so superior to quicksort that is O(n log n). Of course you might
>
> try the Capo Ferro optimization, but I find that Thibault cancels out
>
> Capo Ferro, don't you?

OK! I'll illustrate the lazy aspect of the python interpreter furthermore.

a=[1,2,3]
b=[a]*4 # different from a sliced copy as [list(a)]*4
print b
a[1]=4
print b
#-----
a=666 # type morphed
print b

88888 Dihedral, Sep 29, 2012
19. ### 88888 DihedralGuest

Re: using "*" to make a list of lists with repeated (and independent)elements

On Sunday, September 30, 2012 1:19:22 AM UTC+8, Ian wrote:
> On Sat, Sep 29, 2012 at 11:01 AM, 88888 Dihedral
>
> <> wrote:
>
> >

>
> > Don't you get it why I avoided the lambda one liner as a functon.

>
> >

>
> > I prefer the def way with a name chosen.

>
>
>
> Certainly, but the Bresenham line algorithm is O(n), which is why it
>
> is so superior to quicksort that is O(n log n). Of course you might
>
> try the Capo Ferro optimization, but I find that Thibault cancels out
>
> Capo Ferro, don't you?

OK! I'll illustrate the lazy aspect of the python interpreter furthermore.

a=[1,2,3]
b=[a]*4 # different from a sliced copy as [list(a)]*4
print b
a[1]=4
print b
#-----
a=666 # type morphed
print b

88888 Dihedral, Sep 29, 2012