Sublist pattern matching

Question

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

ybeltukov · Accepted Answer

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}]

eldo · Answer

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}}

wxffles · Answer

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}}

kglr · Answer

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}} *)

4 Answers