using "*" to make a list of lists with repeated (and independent) elements
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From:
TP <T... @frenoespam.fr.invalid> Date: Wed, 26 Sep 2012 23:20:10 +0200
Local: Wed, Sep 26 2012 5:20 pm
Subject: using "*" to make a list of lists with repeated (and independent) elements
Hi everybody,
I have tried, naively, to do the following, so as to make lists quickly:
>>> a=[0]*2
[3, 0]
All is working fine, so I extended the technique to do:
>>> a=[[0]*3]*2
[[0, 0, 0], [0, 0, 0]]
>>> a[0][0]=2
[[2, 0, 0], [2, 0, 0]]
The behavior is no more expected!
Is it the correct explanation?
Thanks,
TP
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From:
Ian Kelly <ian.g.ke... @gmail.com> Date: Wed, 26 Sep 2012 15:39:32 -0600
Local: Wed, Sep 26 2012 5:39 pm
Subject: Re: using "*" to make a list of lists with repeated (and independent) elements
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
> All is working fine, so I extended the technique to do:
>>>> a=[[0]*3]*2
> The behavior is no more expected!
> In these conditions, how to make this list [[0,0,0],[0,0,0]] with "*"
Use a list comprehension:
a = [[0] * 3 for _ in range(2)]
This way the expression `[0] * 3` is re-evaluated at each position in
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From:
Paul Rubin <no.em... @nospam.invalid> Date: Wed, 26 Sep 2012 14:43:58 -0700
Local: Wed, Sep 26 2012 5:43 pm
Subject: Re: using "*" to make a list of lists with repeated (and independent) elements
TP <T
... @frenoespam.fr.invalid> writes:
> copies a list, he copies in fact the *pointer* to the list ....
> Is it the correct explanation?
> In these conditions, how to make this list [[0,0,0],[0,0,0]] with "*"
>>> a = [[0]*3 for i in xrange(2)]
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From:
88888 Dihedral <dihedral88... @googlemail.com> Date: Wed, 26 Sep 2012 14:45:58 -0700 (PDT)
Local: Wed, Sep 26 2012 5:45 pm
Subject: Re: using "*" to make a list of lists with repeated (and independent) elements
TP於 2012年9月27日星期四UTC+8上午5時25分04秒寫道：
> 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
>>> a=zeros(3,2)
[[0, 0, 0], [0, 0, 0], [0, 0, 0], [0, 0, 0]]
I think this is what you want.
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From:
88888 Dihedral <dihedral88... @googlemail.com> Date: Wed, 26 Sep 2012 15:07:35 -0700 (PDT)
Local: Wed, Sep 26 2012 6:07 pm
Subject: Re: using "*" to make a list of lists with repeated (and independent) elements
Paul Rubin於 2012年9月27日星期四UTC+8上午5時43分58秒寫道：
> 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]]
Python is not lisp but python can emulate the lisp behaviors.
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From:
88888 Dihedral <dihedral88... @googlemail.com> Date: Wed, 26 Sep 2012 15:28:15 -0700 (PDT)
Local: Wed, Sep 26 2012 6:28 pm
Subject: Re: using "*" to make a list of lists with repeated (and independent) elements
88888 Dihedral於 2012年9月27日星期四UTC+8上午6時07分35秒寫道：
> Paul Rubin於 2012年9月27日星期四UTC+8上午5時43分58秒寫道：
> > 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]]
> I used numpy before.
> Python is not lisp but python can emulate the lisp behaviors.
If one wants to tranlate to C, this is the style.
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From:
Tim Chase <python.l... @tim.thechases.com> Date: Wed, 26 Sep 2012 17:45:19 -0500
Local: Wed, Sep 26 2012 6:45 pm
Subject: Re: using "*" to make a list of lists with repeated (and independent) elements
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):
> If one wants to tranlate to C, this is the style.
Besides, a C-style would allocate a single array of M*N slots and
-tkc
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From:
88888 Dihedral <dihedral88... @googlemail.com> Date: Wed, 26 Sep 2012 15:53:40 -0700 (PDT)
Local: Wed, Sep 26 2012 6:53 pm
Subject: Re: using "*" to make a list of lists with repeated (and independent) elements
Tim Chase於 2012年9月27日星期四UTC+8上午6時44分42秒寫道：
> 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
Of course if GCC was not supportd in manny platforms free
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From:
88888 Dihedral <dihedral88... @googlemail.com> Date: Wed, 26 Sep 2012 15:53:40 -0700 (PDT)
Local: Wed, Sep 26 2012 6:53 pm
Subject: Re: using "*" to make a list of lists with repeated (and independent) elements
Tim Chase於 2012年9月27日星期四UTC+8上午6時44分42秒寫道：
> 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
Of course if GCC was not supportd in manny platforms free
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From:
88888 Dihedral <dihedral88... @googlemail.com> Date: Thu, 27 Sep 2012 14:24:00 -0700 (PDT)
Subject: Re: using "*" to make a list of lists with repeated (and independent) elements
Tim Chase於 2012年9月27日星期四UTC+8上午6時44分42秒寫道：
> 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
I thnik this is very clear about the situation in entangled references.
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From:
88888 Dihedral <dihedral88... @googlemail.com> Date: Thu, 27 Sep 2012 14:24:00 -0700 (PDT)
Subject: Re: using "*" to make a list of lists with repeated (and independent) elements
Tim Chase於 2012年9月27日星期四UTC+8上午6時44分42秒寫道：
> 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
I thnik this is very clear about the situation in entangled references.
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From:
Ramchandra Apte <maniandra... @gmail.com> Date: Sat, 29 Sep 2012 06:46:22 -0700 (PDT)
Local: Sat, Sep 29 2012 9:46 am
Subject: 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 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
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From:
Ramchandra Apte <maniandra... @gmail.com> Date: Sat, 29 Sep 2012 06:46:22 -0700 (PDT)
Local: Sat, Sep 29 2012 9:46 am
Subject: 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 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
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From:
88888 Dihedral <dihedral88... @googlemail.com> Date: Sat, 29 Sep 2012 10:01:30 -0700 (PDT)
Local: Sat, Sep 29 2012 1:01 pm
Subject: 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 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 is a bot.
I prefer the def way with a name chosen.
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From:
88888 Dihedral <dihedral88... @googlemail.com> Date: Sat, 29 Sep 2012 10:01:30 -0700 (PDT)
Local: Sat, Sep 29 2012 1:01 pm
Subject: 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 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 is a bot.
I prefer the def way with a name chosen.
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Newsgroups: comp.lang.python
From:
Ian Kelly <ian.g.ke... @gmail.com> Date: Sat, 29 Sep 2012 11:18:48 -0600
Local: Sat, Sep 29 2012 1:18 pm
Subject: Re: using "*" to make a list of lists with repeated (and independent) elements
On Sat, Sep 29, 2012 at 11:01 AM, 88888 Dihedral
<dihedral88
... @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.
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From:
Chris Angelico <ros... @gmail.com> Date: Sun, 30 Sep 2012 03:41:04 +1000
Local: Sat, Sep 29 2012 1:41 pm
Subject: Re: using "*" to make a list of lists with repeated (and independent) elements
On Sun, Sep 30, 2012 at 3:18 AM, Ian Kelly <ian.g.ke
... @gmail.com> wrote:
> On Sat, Sep 29, 2012 at 11:01 AM, 88888 Dihedral
> <dihedral88
... @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
class Agrippa(object):
enemy = Agrippa()
ChrisA
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From:
88888 Dihedral <dihedral88... @googlemail.com> Date: Sat, 29 Sep 2012 11:50:52 -0700 (PDT)
Local: Sat, Sep 29 2012 2:50 pm
Subject: 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
> <dihedral88... @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?
a=[1,2,3]
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From:
88888 Dihedral <dihedral88... @googlemail.com> Date: Sat, 29 Sep 2012 11:50:52 -0700 (PDT)
Local: Sat, Sep 29 2012 2:50 pm
Subject: 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
> <dihedral88... @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?
a=[1,2,3]
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