# [Tutor] problem solving with lists

Sat Mar 19 10:34:00 EDT 2022

```So what's wrong  with Dennis Bieber's answer?

On Sat, Mar 19, 2022 at 9:58 AM <marcus.luetolf at bluewin.ch> wrote:

>
> I am not an professional programmer. Programming with pyhon is a hobby in
> the sense of life long learning.
>
> And other than a course in C I have no knowledge of other programming
> languages. I am just trying to solve
>
> my problem with what little I know about python. That’s why I turned to
> tutor for help. I have found an
>
> algorithm and all required code steps but the very last one,  admittedly
> helped by most wellcome
>
>
>
>
> dn’s « definitions » are correct but I reduced the problem solving to
>
> >all_letters = ‘abcdefghi’
>
> >number_of_lists = 4
>
> >number_of_sublists per list = 3
>
> >number_per_sublist = 3
>
>
>
> But dn’s «results obtained»  is not what I need, I replaced them by
> itertools function combinations.
>
>
>
> dn’s last section I don’t understand.
>
> Second to Alan G.’s replay :
>
> My problem ist to create a set of unique sublists of 3 items/letters out
> of a list of 9 items/letters, later to be extended to
>
> sublists of 4 items/letters out of a list of 16 items/letters.
>
> I am using letters as items for the sake of readability. Once the code
> works letters will be substituted by names of people
>
> for the final goal is to set up  «unique» teams.
>
>
>
> «Unique» means that other than single items/letters,  pairs of
> items/letters (pair1, pair2, pair3) can appear only once in
>
> all sublists. A sublist could be understood as a combination of
> pair1+pair2+pair3. At the end there should be only 12 sublists
>
> in 4 lists left.
>
>
>
> Sorry for the length of this replay, Marcus.
>
>
>
>
>
> Von: alan.gauld at yahoo.co.uk <alan.gauld at yahoo.co.uk>
> Gesendet: Samstag, 19. März 2022 13:25
> An: marcus.luetolf at bluewin.ch
> Cc: tutor at python.org
> Betreff: Re: AW: AW: AW: [Tutor] problem solving with lists
>
>
>
>
>
>
>
> On 19 Mar 2022 10:57, marcus.luetolf at bluewin.ch <mailto:
> marcus.luetolf at bluewin.ch>  wrote:
>
> Many thanks again.
>
>
>
>
> >lst = [['a', 'b', 'c'], ['d', 'e', 'f'], ['a', 'b', 'g'], ['b', 'c', 'h']]
>
> >sub_lst = []
>
>
>
> >for p in lst:
>
> >    pair1 = p[0:2]
>
> >    pair2 = list(p[0]+ p[2])
>
> >    pair3 = p[1:]
>
> >    if pair1 not in sub_lst or pair2 not in sub_lst or pair3 not in
> sub_lst:
>
> >        sub_lst.append(p)
>
> >print(sub_lst)
>
>
>
>
> >if pair1 not in sub_lst or pair2 not in sub_lst or pair3 not in sub_lst:
>
>
>
> Still yields all four sublists instead of eliminating the last 2 sublists
> , ['a', 'b', 'g'], ['b', 'c', 'h'],
>
> since ‘a’, ‘b’ and ‘b’, ’c’ were already in the first 2 sublists.
>
> I think python does not allow to check if an item is in a list with only
> slices of the item to be checked ??
>
> the wrong question.
>
>
>
> But I still don't really understand what you are trying to do. Have you
> looked at Dennis replies, his approach is more likely to get you what you
> want....
>
>
>
> Alan g
>
> Von: alan.gauld at yahoo.co.uk <mailto:alan.gauld at yahoo.co.uk>  <
> alan.gauld at yahoo.co.uk <mailto:alan.gauld at yahoo.co.uk> >
> Gesendet: Freitag, 18. März 2022 19:16
> An: marcus.luetolf at bluewin.ch <mailto:marcus.luetolf at bluewin.ch>
> Cc: tutor at python.org <mailto:tutor at python.org>
> Betreff: Re: AW: AW: [Tutor] problem solving with lists
>
>
>
>
>
>
>
> On 18 Mar 2022 15:12, marcus.luetolf at bluewin.ch <mailto:
> marcus.luetolf at bluewin.ch>  wrote:
>
> …many thanks for your quick replay.
>
> When I used a for loop :
>
>
>
> >lst = [['a', 'b', 'c'], ['d', 'e', 'f'], ['a', 'b', 'g'], ['b', 'c', 'h']]
>
> >sub_lst = []
>
> >for p in lst:
>
> >    pair1 = lst[p][0:2]
>
>
>
>
>
> P is not an index its the actual sublist. So you just need
>
>
>
> Pair1 = p[0:2]
>
>
>
>
>
> >
>
> > ected. But I don’t understand
>
> what you mean with
>
>
> If pair 1 not empty
>
> Or pair2 not empty
>
> Or pair3 not in newlist
>
>
>
> What is the difference between pair not beeing empty or not beeing in
> new_list ?
>
>
>
> Pair not empty means pair contains anything other than the empty list.
>
> Pair not in newlist means pair is not empty and it's contents are not I.
> Newlist. Very different!
>
> How should I define the if condition properly ?
>
>
>
>
>
> You need, I assume,
>
>
>
> If Pair1 not in newlist
>
> Or pair2 not in newlist
>
> Or pair3 not in newlist
>
>
>
> Alan g.
>
>
>
>
>
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