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

[TP]

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

[Ian Kelly]

On Wed, Sep 26, 2012 at 3:20 PM, TP <T...@frenoespam.fr.invalid> 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.

[Paul Rubin]

TP <T...@frenoespam.fr.invalid> 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]]

[88888 Dihedral]

TP於 2012年9月27日星期四UTC+8上午5時25分04秒寫道：

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

Paul Rubin於 2012年9月27日星期四UTC+8上午5時43分58秒寫道：

I used numpy before.

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

[88888 Dihedral]

88888 Dihedral於 2012年9月27日星期四UTC+8上午6時07分35秒寫道：

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.

[Tim Chase]

On 09/26/12 17:28, 88888 Dihedral wrote:
> 88888 Dihedral於 2012年9月27日星期四UTC+8上午6時07分35秒寫道：

>>>> 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. :-P

-tkc

[88888 Dihedral]

Tim Chase於 2012年9月27日星期四UTC+8上午6時44分42秒寫道：

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]

Tim Chase於 2012年9月27日星期四UTC+8上午6時44分42秒寫道：

[88888 Dihedral]

Tim Chase於 2012年9月27日星期四UTC+8上午6時44分42秒寫道：

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]

Tim Chase於 2012年9月27日星期四UTC+8上午6時44分42秒寫道：

[Ramchandra Apte]

88888 Dihedral is a bot.

[Ramchandra Apte]

On Thursday, 27 September 2012 04:14:42 UTC+5:30, Tim Chase wrote:

[88888 Dihedral]

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]

On Saturday, September 29, 2012 9:46:22 PM UTC+8, Ramchandra Apte wrote:

[Ian Kelly]

On Sat, Sep 29, 2012 at 11:01 AM, 88888 Dihedral
<dihedr...@googlemail.com> 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?

[Chris Angelico]

On Sun, Sep 30, 2012 at 3:18 AM, Ian Kelly <ian.g...@gmail.com> wrote:
> On Sat, Sep 29, 2012 at 11:01 AM, 88888 Dihedral
> <dihedr...@googlemail.com> 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)

[88888 Dihedral]

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]

On Sunday, September 30, 2012 1:19:22 AM UTC+8, Ian wrote: