[Tutor] seed & random !?

[Andrei Kulakov]

On Sun, Oct 14, 2001 at 03:45:52AM -0700, necati kaya wrote:
> Thank you for your suggestion, but it is half of what
> I need: 
> I want to create random lists that have the same
> elements as a given list, but not in the same order
> with original and also with previous generated lists.

Oh.. okay then.. simply append them to a list of random lists and test if
it's there or not.. see below.

> 
> What you suggested is really useful for me to create
> the total number of permutations of original list, but
> it gives also previous generated list order:
> 
> list_1 : ['1', '2', '3', '4']
> 
> 0 -> ['1', '4', '3', '2']
> 1 -> ['4', '3', '2', '1']
> 2 -> ['1', '3', '2', '4']  ___
> 4 -> ['1', '3', '4', '2']     \
> 6 -> ['2', '1', '3', '4']      \  Both are same !!
> 7 -> ['3', '1', '2', '4']      /
> 8 -> ['1', '3', '2', '4']  ___/
> 9 -> ['4', '3', '2', '1']
> 
> ################################
> from random import shuffle
> from copy import copy
> 
> list_1 = ['1', '2', '3', '4']
lists = [list_1]
> list_2 = copy(list_1)
> print 'list_1 :', list_1
> print
> for i in range(10):
>     shuffle(list_2)
      if list_2 not in lists:
          l = copy(list_2)
          lists.append(l)

But.. do you need *all* permutations? If so, there must be a more
efficient and straightforward way to do this..

> 
> #################################
> 
> Regards,
> umit
> 
> 
> --- Andrei Kulakov <sill@optonline.net> wrote:
> > On Sun, Oct 14, 2001 at 02:27:21AM -0700, necati
> > kaya wrote:
> > > Hi,
> > > I am trying to generate different new_list
> > randomly in
> > > each step of the loop from original_list, but I
> > can
> > > not. As the example;
> > > original_list=['1','2','3','4'] and my program
> > produce
> > > those new_lists randomly, but
> > > 
> > > ['4', '2', '3', '1']   ==>  
> > > ['1', '4', '2', '3']         Both are same
> > > ['4', '2', '3', '1']   ==>
> > > ['3', '2', '1', '4']
> > > ['2', '4', '3', '1']
> > > 
> > > I tried to use seed function, but it gave me all
> > the
> > > same results. Do you have any suggestion to have
> > > different new_list which is generated from
> > > original_list in every step. 
> > 
> > If I understood right, you want to create random
> > lists that have the same
> > elements as a given list, but not in the same
> > order.. if that's true, you
> > can do this with my_list.shuffle:
> > 
> > >>> from random import shuffle
> > >>> l
> > [3, 4, 2, 1]
> > >>> from copy import copy
> > >>> l2 = copy(l)
> > >>> shuffle(l2)
> > >>> l2
> > [2, 4, 3, 1]
> > >>> l
> > [3, 4, 2, 1]
> > >>> for i in range(10):
> > ...  shuffle(l2)
> > ...  if l != l2:
> > ...   print l2
> > ... 
> > [3, 1, 4, 2]
> > [1, 2, 4, 3]
> > [4, 3, 2, 1]
> > [2, 1, 4, 3]
> > [1, 3, 4, 2]
> > [4, 3, 2, 1]
> > [2, 1, 4, 3]
> > [3, 4, 1, 2]
> > [4, 1, 2, 3]
> > [2, 1, 4, 3]
> > >>> l
> > [3, 4, 2, 1]
> > 
> > Instead of printing, you can append it to a list of
> > random lists..
> > 
> > Hope that's what you wanted :P
> > 
> > > Regards,
> > > umit
> > > 
> > > Here is my code:
> > > 
> > > 
> > > from random import Random
> > > import random
> > > 
> > > original_list=['1','2','3','4']
> > > for j in range(0, 5):
> > >     temp_list = original_list[:]
> > >     new_list=[]
> > >     for i in range(0, len(original_list)):
> > > #            random.seed()             
> > > #??????????????
> > >             choosen_item =
> > random.choice(temp_list)
> > >             new_list.append(choosen_item)
> > >             temp_list.remove(choosen_item)
> > >     print new_list
> > >     print
> > > 
> > > 
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