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use a loop to create lists

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  use a loop to create lists martaamunar@gmail.com - 2013-04-10 01:40 -0700
    Re: use a loop to create lists Chris Angelico <rosuav@gmail.com> - 2013-04-10 18:52 +1000
    Re: use a loop to create lists Dave Angel <davea@davea.name> - 2013-04-10 08:36 -0400
    Re: use a loop to create lists rusi <rustompmody@gmail.com> - 2013-04-10 08:31 -0700
    Re: use a loop to create lists Thomas Goebel <Thomas.Goebel@ohm-hochschule.de> - 2013-04-10 15:55 +0200
    Fwd: use a loop to create lists Franz Kelnreiter <kelnreiter@gmail.com> - 2013-04-11 14:11 +0200
    Re: use a loop to create lists Thomas Goebel <Thomas.Goebel@ohm-hochschule.de> - 2013-04-11 14:57 +0200
    Re: use a loop to create lists Chris Angelico <rosuav@gmail.com> - 2013-04-11 23:22 +1000
    Re: Fwd: use a loop to create lists Thomas Goebel <Thomas.Goebel@ohm-hochschule.de> - 2013-04-11 15:43 +0200
    Re: use a loop to create lists Franz Kelnreiter <kelnreiter@gmail.com> - 2013-04-11 16:11 +0200
    Re: use a loop to create lists Thomas Goebel <Thomas.Goebel@ohm-hochschule.de> - 2013-04-11 16:33 +0200
    Re: Fwd: use a loop to create lists Franz Kelnreiter <kelnreiter@gmail.com> - 2013-04-11 16:33 +0200
    Re: use a loop to create lists Dennis Lee Bieber <wlfraed@ix.netcom.com> - 2013-04-11 18:17 -0400

#43256 — use a loop to create lists

Hi!

I would like to create a list containing lists. I need each list to have a differente name and i would like to use a loop to name the list. But as the name, is a string, i cannot asign it to a value... how can I do that??


global_list=[]
for i in range (20):
  ("list_"+i)=[]   #These would be the name of the list...
  global_list.append("list_"+i)

Thank you!!!!!

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#43258

On Wed, Apr 10, 2013 at 6:40 PM,  <martaamunar@gmail.com> wrote:
> Hi!
>
> I would like to create a list containing lists. I need each list to have a differente name and i would like to use a loop to name the list. But as the name, is a string, i cannot asign it to a value... how can I do that??
>
>
> global_list=[]
> for i in range (20):
>   ("list_"+i)=[]   #These would be the name of the list...
>   global_list.append("list_"+i)

They don't need unique names. Just use the same name inside the loop:

global_list=[]
for i in range(20):
  current_list=[]  # Presumably you do something with it here
  global_list.append(current_list)

That'll work fine, and you can reference the lists by their positions
in global_list.

ChrisA

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#43272

On 04/10/2013 04:40 AM, martaamunar@gmail.com wrote:
> Hi!
>
> I would like to create a list containing lists. I need each list to have a differente name and i would like to use a loop to name the list. But as the name, is a string, i cannot asign it to a value... how can I do that??
>
>
> global_list=[]
> for i in range (20):
>    ("list_"+i)=[]   #These would be the name of the list...
>    global_list.append("list_"+i)
>
> Thank you!!!!!
>

The fact that the content of the outer list also happens to be lists is 
irrelevant to your question.  I believe the real question here is how to 
create an arbitrary bunch of new names, and bind them to elements of he 
list.

In general, you can't.  But more importantly, in general you don't want 
to.  If you come up with a convoluted way to fake it, you'll have to use 
the same convoluted way to access those names, so there's no point.  As 
Chris says, just reference them by index.

On the other hand, if the outer list happens to have exactly 7 elements, 
and you know that ahead of time, then it may well make sense to assign 
names to them.  Not list_0 through list_6, but name, addr1, ec.

global_list = []
for i in range(7):
     global_list.append[]

first_name, mid_name, last_name, addr1, addr2, city, town = global_list

Incidentally, that's approximately what a collections.namedtuple is all 
about, giving names to items that are otherwise considered elements of a 
tuple.

-- 
DaveA

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#43280

On Apr 10, 1:40 pm, martaamu...@gmail.com wrote:
> Hi!
>
> I would like to create a list containing lists. I need each list to have a differente name and i would like to use a loop to name the list. But as the name, is a string, i cannot asign it to a value... how can I do that??
>
> global_list=[]
> for i in range (20):
>   ("list_"+i)=[]   #These would be the name of the list...
>   global_list.append("list_"+i)
>
> Thank you!!!!!


If as Chris points out you dont actually need names, a list should
suffice (with indexing serving as 'naming')
If however there are a fixed (at program-writing time) small set of
names, you can do as Dave suggests and use unpacking assignment. [Some
people like to use objects for this]
If however you need an arbitrary set of names, unknown at programming
time, a dictionary is typically what is appropriate:

For dictionary d
- get the value of name n with d[n]
- set (bind) name n to value v in dict d with d[n] = v

Which of these options you should follow is not clear from your
question; you need to tell us a bit more about what you are trying to
do.

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#43332

* On 10/04/2013 10:40, martaamunar@gmail.com wrote:
> Hi!
> 
> I would like to create a list containing lists. I need each list to
> have a differente name and i would like to use a loop to name the
> list.
> [...]
> global_list=[]
> for i in range (20):
>     ("list_"+i)=[]   #These would be the name of the list...
>     global_list.append("list_"+i)

You can use a dictionary instead of a list of lists:

global_list = {'_'.join(['list', str(i)]):[] for i in range(20)}
global_list_1 = global_list['list_0']

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#43356

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---------- Forwarded message ----------
From: Franz Kelnreiter <kelnreiter@gmail.com>
Date: Thu, Apr 11, 2013 at 2:09 PM
Subject: Re: use a loop to create lists
To: Thomas.Goebel@ohm-hochschule.de


On Thu, Apr 11, 2013 at 1:46 PM, Thomas Goebel <
Thomas.Goebel@ohm-hochschule.de> wrote:

> * On 11/04/2013 13:25, Franz Kelnreiter wrote:
> > On Wed, Apr 10, 2013 at 3:55 PM, Thomas Goebel wrote:
> >>
> >> global_list = {'_'.join(['list', str(i)]):[] for i in range(20)}
> >> global_list_1 = global_list['list_0'] --
> >
> > Sorry Thomas, but you had a typo or in any case a wrong syntax
> > concept in your last posting. If you'd like to use list
> > comprehension it should be written as:
> >
> > mydict = {'_'.join(['list', str(i)]):[x for x in range(20)]}
>
> Hi Franz,
>
> the difference between your and my code is that
>
> global_list = {'_'.join(['list', str(i)]):[] for i in range(20)}
>
> creates a dict 'global_list' which has 20 keys named from 'list_0' to
> 'list_19'. The value for all keys is an empty list. If you want to
> create i.e. 20 keys which value is a list with 20 ints you have to use
>
> global_list = ({'_'.join(['list', str(i)]):[a for a in range(20)] for
>     i in range(20)})
>
> Your code creates a dict with one key 'list_19' which value is a list
> with 20 ints if you replace str(i) with str(a).
>
> Regards, Tom
>

Tom,
Thanks for your explanation, I think I know what you want to do and I would
very much like to understand your code in detail - maybe I am too stupid -
but when I execute the value part of your code construct:

[a for a in range(20)] for i in range(20)

I get a syntax error, as I exepected (Python 2.6.4 (r264:75708, Oct 26
2009, 08:23:19)).

So how can you get me on the right direction to make your code running on
my machine?

Thank you, Franz

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#43362

* On 11/04/2013 14:11, Franz Kelnreiter wrote:
> On Thu, Apr 11, 2013 at 1:46 PM, Thomas Goebel wrote:
>>
>> the difference between your and my code is that
>>
>> global_list = {'_'.join(['list', str(i)]):[] for i in range(20)}
>>
>> creates a dict 'global_list' which has 20 keys named from 'list_0' to
>> 'list_19'. The value for all keys is an empty list. If you want to
>> create i.e. 20 keys which value is a list with 20 ints you have to use
>>
>> global_list = ({'_'.join(['list', str(i)]):[a for a in range(20)] for
>>     i in range(20)})
>
> Thanks for your explanation, I think I know what you want to do and I would
> very much like to understand your code in detail - maybe I am too stupid -
> but when I execute the value part of your code construct:

This code
> [a for a in range(20)] for i in range(20)

won't run because you you didn't define 'i'!

[a for a in range(3)]

will return a list
[0, 1, 2]


To get a dict with n keys we can use a for-loop, too:

d = {}
for n in range(3):
    # Create keys 'list_0' to 'list_n'
    d['list_' + str(n)] = []

So we finally get:
d.keys()
['list_2', 'list_1', 'list_0']

d.values()
[[], [], []]


Now we create a dict with n keys and for every key n we set the value
to a list of m ints:

e = {}
for n in range(3):
    # Create keys 'list_0' to 'list_n'
    e['list_' + str(n)] = []
    for m in range(3):
        # Create a list [0, 1, 2] for every key of e
        e['list_' + str(n)].append(m)

The results is as follows:
e.keys()
['list_2', 'list_1', 'list_0']

e.values()
[[0, 1, 2], [0, 1, 2], [0, 1, 2]]

Which is the same as:
f = {'list_' + str(n):[m for m in range(3)] for n in range(3)}

I double checked the code and everything is working fine for me. Maybe
you forgot some braces?

P.S: I'm running on python 2.7.4:
'2.7.4 (default, Apr  6 2013, 19:54:46) [MSC v.1500 32 bit (Intel)]'

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#43366

On Thu, Apr 11, 2013 at 10:57 PM, Thomas Goebel
<Thomas.Goebel@ohm-hochschule.de> wrote:
> [a for a in range(3)]
>
> will return a list
> [0, 1, 2]

Simplification possible: That's the same as:

list(range(3))

> f = {'list_' + str(n):[m for m in range(3)] for n in range(3)}

Meaning that this can be simplified too:

f = {'list_' + str(n):list(range(3)) for n in range(3)}

ChrisA

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#43369

* On 11/04/2013 14:11, Franz Kelnreiter wrote:
> On Thu, Apr 11, 2013 at 1:46 PM, Thomas Goebel wrote:
> 
>> global_list = {'_'.join(['list', str(i)]):[] for i in range(20)}
>
> Thanks for your explanation, I think I know what you want to do and I would
> very much like to understand your code in detail - maybe I am too stupid -
> but when I execute the value part of your code construct:
> 
> [a for a in range(20)] for i in range(20)
> 
> I get a syntax error, as I exepected (Python 2.6.4 (r264:75708, Oct 26
> 2009, 08:23:19)).

Sorry Franz,

as you are using python 2.6 you have to use

d1 = dict(('list_' + str(i), []) for i in range(3))
d2 = dict(('list_' + str(i), [m for m in range(3)]) for i in range(3))

like stated here:
http://stackoverflow.com/questions/1747817/python-create-a-dictionary-with-list-comprehension

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#43372

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On Thu, Apr 11, 2013 at 2:57 PM, Thomas Goebel <
Thomas.Goebel@ohm-hochschule.de> wrote:

> ...
> Which is the same as:
> f = {'list_' + str(n):[m for m in range(3)] for n in range(3)}
>

Thomas, thank you for your patience and your long explanation! Now I
understand better this shorthand expression of one single 'pythonic' ;)
line.
But didnt you miss square brackets:

f = {'list_' + str(n):[[m for m in range(3)] for n in range(3)]}

??
Otherwise the code wont work for me...

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#43374

* On 11/04/2013 16:11, Franz Kelnreiter wrote:
> On Thu, Apr 11, 2013 at 2:57 PM, Thomas Goebel <
> Thomas.Goebel@ohm-hochschule.de> wrote:
> 
>> ...
>> Which is the same as:
>> f = {'list_' + str(n):[m for m in range(3)] for n in range(3)}
>>
> [...]
> But didnt you miss square brackets:
> 
> f = {'list_' + str(n):[[m for m in range(3)] for n in range(3)]}

If i try your code i get a dict with key 'list_2' which value is a
list of three lists.

Maybe this occurs because we're using different python versions?! I
never used python<2.7!

Have you tried this one

d2 = dict(('list_' + str(i), list(range(3))) for i in range(3))

with Chris' simplification:

* On 11/04/2013 15:22, Chris Angelico wrote:
> On Thu, Apr 11, 2013 at 10:57 PM, Thomas Goebel wrote:
>> [a for a in range(3)]
>>
>> will return a list
>> [0, 1, 2]
>
> Simplification possible: That's the same as:
>
> list(range(3))
>
>> f = {'list_' + str(n):[m for m in range(3)] for n in range(3)}
>
> Meaning that this can be simplified too:
>
> f = {'list_' + str(n):list(range(3)) for n in range(3)}

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#43375

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On Thu, Apr 11, 2013 at 3:43 PM, Thomas Goebel <
Thomas.Goebel@ohm-hochschule.de> wrote:

> ...
> > I get a syntax error, as I exepected (Python 2.6.4 (r264:75708, Oct 26
> > 2009, 08:23:19)).
>
> Sorry Franz,
>
> as you are using python 2.6 you have to use
>
> d1 = dict(('list_' + str(i), []) for i in range(3))
> d2 = dict(('list_' + str(i), [m for m in range(3)]) for i in range(3))
>
> like stated here:
>
> http://stackoverflow.com/questions/1747817/python-create-a-dictionary-with-list-comprehension
> --
>

Thank you, Thomas, for your last posting and your help, that makes
everything clear for me now!

All best, Franz

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#43404

On Thu, 11 Apr 2013 16:11:13 +0200, Franz Kelnreiter
<kelnreiter@gmail.com> declaimed the following in
gmane.comp.python.general:

> On Thu, Apr 11, 2013 at 2:57 PM, Thomas Goebel <
> Thomas.Goebel@ohm-hochschule.de> wrote:
> 
> > ...
> > Which is the same as:
> > f = {'list_' + str(n):[m for m in range(3)] for n in range(3)}
> >
> 
> Thomas, thank you for your patience and your long explanation! Now I
> understand better this shorthand expression of one single 'pythonic' ;)
> line.
> But didnt you miss square brackets:
> 
> f = {'list_' + str(n):[[m for m in range(3)] for n in range(3)]}
> 
	Without the second set of square brackets, the result is a generator
expression/comprehension -- which came in a few versions after list
comprehensions did.

	However, your position in locking the "n" inside the right hand, but
you need "n" on the left too...

	Try moving the first [ to the beginning -- {["list....

	(NO PROMISES -- I DIDN'T TEST}
-- 
	Wulfraed                 Dennis Lee Bieber         AF6VN
        wlfraed@ix.netcom.com    HTTP://wlfraed.home.netcom.com/

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