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I have the following list

list = {
  {{"-", "-"}, {"+", "-"}, {"+", "-"}},
  {{"-", "-"}, {"+", "-"}, {"+", "+"}},
  {{"-", "-"}, {"+", "+"}, {"-", "-"}},
  {{"-", "-"}, {"+", "+"}, {"-", "+"}},
  {{"-", "+"}, {"-", "-"}, {"+", "-"}},
  {{"-", "+"}, {"-", "-"}, {"+", "+"}},
  {{"-", "+"}, {"-", "+"}, {"-", "-"}},
  {{"-", "+"}, {"-", "+"}, {"-", "+"}},
  {{"+", "-"}, {"+", "-"}, {"+", "-"}},
  {{"+", "-"}, {"+", "-"}, {"+", "+"}},
  {{"+", "-"}, {"+", "+"}, {"-", "-"}},
  {{"+", "-"}, {"+", "+"}, {"-", "+"}},
  {{"+", "+"}, {"-", "-"}, {"+", "-"}},
  {{"+", "+"}, {"-", "-"}, {"+", "+"}},
  {{"+", "+"}, {"-", "+"}, {"-", "-"}},
  {{"+", "+"}, {"-", "+"}, {"-", "+"}}
};

I wish to perform the following manipulations on this list:

  1. Replace first and last subelement in sublist. for example: the sublist {{"-","-"},{"+","-"},{"+","-"}} should be replace with the sublist {{0,"-"},{"+","-"},{"+",0}}.

  2. The new list produces duplicates if considering the symmetry which I would like to eliminate. For example: the sublists {{0,"-"},{"+","-"},{"+",0}} and {{0,"+"},{"-","+"},{"-",0}} are considered as duplicates.

Can anyone advise on how to perform these list manipulations?

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2
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New answer, with proper requirements:

list[[All, 1, 1]] = 0;
list[[All, -1, -1]] = 0;
DeleteDuplicates[list, Reverse[Reverse /@ #1] == #2 &]

(* Out:
{
 {{0, "-"}, {"+", "-"}, {"+", 0}},
 {{0, "-"}, {"+", "-"}, {"+", 0}},
 {{0, "-"}, {"+", "+"}, {"-", 0}}, 
 {{0, "+"}, {"-", "-"}, {"+", 0}},
 {{0, "-"}, {"+", "-"}, {"+", 0}},
 {{0, "-"}, {"+", "-"}, {"+", 0}}
}
*)

This is my original answer, in which I misunderstood the requirements:

replaced = MapAt[If[# == "+", 1, 0] &, list, {{All, 1, 1}, {All, -1, -1}}]
DeleteDuplicates[replaced, Reverse[Reverse /@ #1] == #2 &]

(* Out:
{{{0, "-"}, {"+", "-"}, {"+", 0}}, 
 {{0, "-"}, {"+", "-"}, {"+", 1}}, 
 {{0, "-"}, {"+", "+"}, {"-", 0}}, 
 {{0, "-"}, {"+", "+"}, {"-", 1}},
 {{0, "+"}, {"-", "-"}, {"+", 0}}, 
 {{0, "+"}, {"-", "-"}, {"+", 1}},
 {{0, "+"}, {"-", "+"}, {"-", 1}}, 
 {{1, "-"}, {"+", "-"}, {"+", 1}}, 
 {{1, "-"}, {"+", "+"}, {"-", 1}}, 
 {{1, "+"}, {"-", "-"}, {"+", 1}}}
*)
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5
  • $\begingroup$ why did you replace 0 and 1 and not just 0? $\endgroup$ – jarhead Jan 22 at 18:47
  • $\begingroup$ @jarhead Maybe I misunderstood. So you want the first and last element to become 0, independently of their original value? $\endgroup$ – MarcoB Jan 22 at 18:50
  • $\begingroup$ yes. for this case I wish both edges to be with 0 $\endgroup$ – jarhead Jan 22 at 18:53
  • $\begingroup$ @jarhead OK sorry, I think I fixed it. $\endgroup$ – MarcoB Jan 22 at 19:04
  • 1
    $\begingroup$ Is this answer correct? The first and last rows in the revised output are identical. $\endgroup$ – Shredderroy Jan 22 at 19:46
3
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A functional approach:

list // RightComposition[

    (* Replace the extreme elements with 0 *)
    MapAt[Replace[_ -> 0], {{All, 1, 1}, {All, -1, -1}}],

    (* Delete duplicates after the appropriate considerations *)
    DeleteDuplicatesBy[
        (* Disregard the order of the pairs *)
        Map[Apply[OrderlessPatternSequence]] /*
        (* Disregard the overall order *)
        Apply[OrderlessPatternSequence]
    ]

]

(*
    {
        {{0, "-"}, {"+", "-"}, {"+", 0}},
        {{0, "-"}, {"+", "+"}, {"-", 0}},
        {{0, "+"}, {"-", "-"}, {"+", 0}}
    }
*)

Notice that this gives fewer elements than the accepted answer.

Notice also that the composite function of DeleteDuplicatesBy can be replaced with a more elegant function (i.e. one that does not require the application of OrderlessPatternSequence twice):

DeleteDuplicatesBy[
    OperatorApplied[Apply, {1, 3, 2}][OrderlessPatternSequence, {0, 1}]
]

But that makes it harder to insert the explanatory comments.

EDIT

Corrected a misunderstanding on my part after @kglr pointed it out.

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2
  • $\begingroup$ The last one is a duplicate of the first, b/c OP says: "the sublists {{0,"-"},{"+","-"},{"+",0}} and {{0,"+"},{"-","+"},{"-",0}} are considered as duplicates." $\endgroup$ – kglr Jan 22 at 20:20
  • $\begingroup$ Thanks for pointing that out. I thought the order of the middle elements was important. Otherwise, the solution is a bit simpler. Please see the edit. $\endgroup$ – Shredderroy Jan 22 at 20:29
2
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This, like @Shredderoy's method, eliminates duplicates

rubeGoldberg = MapAt[0 &, {{;; , 1, 1}, {;; , -1, -1}}] @* 
 DeleteDuplicatesBy[Sort @* ({#, Reverse @ #} &) @* Most @* Rest @* Flatten]

rubeGoldberg @ list // Column

enter image description here

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