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Is there any way to simplify the following pattern:

{p___, {a___, x_, x_, b___}, q___} :> {p, {a, x, b}, q}

Ie. where I remove duplicated elements within sublists?

Edit: Just to be clear, I have the feeling p and q could be removed here, but well I'm not certain :)

Thanks

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  • $\begingroup$ Think. If you remove the p and q and you happened to have two adjacent sublists that were identical, what would happen then? Something like {{3,4},{5,7,7,2},{5,7,7,2},{2,8}} Compare that example with and without the p and q and see what happens. Then think about it until you can explain what happened. Then try all this using /. and //. It all depends on what you want to accomplish. $\endgroup$
    – Bill
    Oct 9, 2014 at 19:56
  • $\begingroup$ Well exactly, that's why I kept them. But since p and q remain at the same place I thought I could have removed them. Besides, it assumes here that I'm working at the second level of my main list, but what if I didn't know which level they were at? Anyway, it might just be the most simple expression but I had the intuition it wasn't so. And btw it's a repeated pattern //. here. $\endgroup$
    – Literal
    Oct 9, 2014 at 20:07
  • $\begingroup$ You could look at DeleteDuplicates or Split + First $\endgroup$
    – mfvonh
    Oct 9, 2014 at 20:51

5 Answers 5

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Several demonstrative examples:

list = {{1, 2, 3}, {1, 2, 3}, {1, 2, 2, 3}, {1, 3, 3, 5}};

list /. {p___, {a___, x_, x_, b___}, q___} :> {p, {a, x, b}, q}
(* {{1, 2, 3}, {1, 2, 3}, {1, 2, 3}, {1, 3, 3, 5}} *)

list /. {a___, x_, x_, b___} :> {a, x, b}
(* {{1, 2, 3}, {1, 2, 2, 3}, {1, 3, 3, 5}} *)

Replace[list, {a___, x_, x_, b___} :> {a, x, b}, {1}]
(* {{1, 2, 3}, {1, 2, 3}, {1, 2, 3}, {1, 3, 5}} *)

I think the last one is what you are looking for.

If you want to delete many sequential duplicates with patterns you can use

Replace[list, {a___, Repeated[x_, {2, ∞}], b___} :> {a, x, b}, {1}]
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  • $\begingroup$ Sweet. Yeah the last answer with the levelspec argument is what I was looking for. It's too bad it can't be expressed using the /. syntax though. One last question, I'm not certain why the first pattern (the which I submitted doesn't catch the {1, 3, 3, 5} list. $\endgroup$
    – Literal
    Oct 9, 2014 at 21:10
  • $\begingroup$ @Neuschwanstein It is because the whole list matches the pattern and /. applies the substitution, but this substitution eliminate only one duplicate. You should use //. (ReplaceRepeated) to apply replacement repeatedly. Note, that ReplaceRepeated is very slow for big lists due to matching the whole list a lot of times. $\endgroup$
    – ybeltukov
    Oct 9, 2014 at 21:28
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Map ReplaceAll at Level 1:

# /. {a___, x_, x_, b___} :> {a, x, b} & /@ list
(* {{1, 2, 3}, {1, 2, 3}, {1, 2, 3}, {1, 3, 5}} *)

Map DeleteDuplicates at Level 1:

DeleteDuplicates /@ list
(* {{1, 2, 3}, {1, 2, 3}, {1, 2, 3}, {1, 3, 5}} *)

Alternative methods to ReplaceRepeated to get {{1, 2, 3}, {1, 3, 5}}:

FixedPoint[# /. {a___, x_, x_, b___} :> {a, x, b} &, list]
(* {{1, 2, 3}, {1, 3, 5}} *)

Map[DeleteDuplicates, list, {0, 1}]
(* {{1, 2, 3}, {1, 3, 5}} *)

And to get {{1, 2, 3}, {1, 2, 2, 3}, {1, 3, 3, 5}}:

Map[# /. {a___, x_, x_, b___} :> {a, x, b} &, list, {0}]
(* {{1, 2, 3}, {1, 2, 2, 3}, {1, 3, 3, 5}} *)

or

Map[DeleteDuplicates, list, {0}]
(* {{1, 2, 3}, {1, 2, 2, 3}, {1, 3, 3, 5}} *)
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If I don't missunderstand your question this could be a good candidat for ReplaceRepeated

list = {1, 1, 2, 3, 3, 3, 1, 1, 7};

list //. {head___, x_, x_, tail___} :> {head, x, tail}

{1, 2, 3, 1, 7}

To scrutinise @ ybeltukov's excellent answer and your nice but ambiguous question

matrix = {{1, 2, 3}, {1, 2, 3}, {1, 2, 2, 3}, {1, 3, 3, 5}};

matrix //. {head___, x_, x_, tail___} :> {head, x, tail}

{{1, 2, 3}, {1, 3, 5}}

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Another option to consider is to only (repeatedly) replace if the repeated item is an atom:

rule = {a___, x_?AtomQ, x_, b___} :> {a, x, b}

In action:

{{1, 2, 3}, {1, 2, 3}, {1, 2, 2, 3}, {1, 3, 3, 5}} //. rule

{{1, 2, 3}, {1, 2, 3}, {1, 2, 3}, {1, 3, 5}}

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  • Duplicated elements in a list can be removed by DeleteDuplicates without changing list order. Duplicated elements can be non-consecutive.

  • Using Union will sort the list which may or may not be desired.

  • Sequence* functions were introduced in 2018 and are a natural fit for the task or removing consecutively appearing elements.

Entity["WolframLanguageSymbol"
, "SequenceReplace"]["DateIntroduced"]

enter image description here

Clear["Global`*"]
list = {
   {1, 2, 3}
   , {1, 2, 3}
   , {1, 2, 2, 3}
   , {1, 3, 3, 5}
   , {1, 3, 4, 5, 5, 5, {5}, {5}, 6, 7, 8}
   , {1, 1, 2, 2, 3, 4, 6, 6, 4, 5, 5, 9}
   };

where a few additional test cases have been introduced to demonstrate removal of repeated/sublisted items.

The following deletes any number of similar and consecutive elements in list but not inside the sublists.

SequenceReplace[list, {x_ ..} ->  x]

{{1, 2, 3}, {1, 2, 2, 3}, {1, 3, 3, 5}, {1, 3, 4, 5, 5, 5, {5}, {5},
6, 7, 8}, {1, 1, 2, 2, 3, 4, 6, 6, 4, 5, 5, 9}}

Defining this as a function and applying it to each sublist:

f = SequenceReplace[#, {x_ ..} ->  x] &
f /@ list

{{1, 2, 3}, {1, 2, 3}, {1, 2, 3}, {1, 3, 5}, {1, 3, 4, 5, {5}, 6, 7,
8}, {1, 2, 3, 4, 6, 4, 5, 9}}

More selectivity with repetition count can be achieved using options for Repeated.

Try: f@(f /@ list)

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