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My code replaces each repeated element by "X".

ReplaceRepeated[{1, 2, 3, 2, 5, 2, 1, 5, 5, 2, 3, 3, 
    7}, {a1___, y_ /; y != #, a2___, y_, a3___} :> {a1, y, a2, #, 
     a3}] &@"X"

(*{1, 2, 3, "X", 5, "X", "X", "X", "X", "X", "X", "X", 7}*)

What is the most elegant code to do this?

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4 Answers 4

Reset to default
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ClearAll[replaceDuplicates]
replaceDuplicates[rep_: "X"] := Module[{f}, f[y_] := (f[y] = rep; y); f /@ #] &

Examples:

replaceDuplicates[] @ {1, 2, 3, 2, 5, 2, 1, 5, 5, 2, 3, 3, 7}

{1, 2, 3, "X", 5, "X", "X", "X", "X", "X", "X", "X", 7}

replaceDuplicates[Nothing] @ {1, 2, 3, 2, 5, 2, 1, 5, 5, 2, 3, 3, 7}

{1, 2, 3, 5, 7}

Alternatively:

ClearAll[replaceDuplicates2]
replaceDuplicates2[rep_: "X"] := Module[{y = #}, 
   y[[Join @@ (Rest /@ Values @ PositionIndex[y])]] = rep; y] &

Examples:

replaceDuplicates2[] @ {1, 2, 3, 2, 5, 2, 1, 5, 5, 2, 3, 3, 7}

{1, 2, 3, "X", 5, "X", "X", "X", "X", "X", "X", "X", 7}

replaceDuplicates2[Nothing] @ {1, 2, 3, 2, 5, 2, 1, 5, 5, 2, 3, 3, 7}

{1, 2, 3, 5, 7}
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    $\begingroup$ Very clever, and ten times faster than what I suggested. Cheers! $\endgroup$
    – Roman
    Apr 24, 2021 at 17:30
  • $\begingroup$ I am trying to understand what your function is really doing. $\endgroup$
    – azerbajdzan
    Apr 24, 2021 at 20:03
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    $\begingroup$ @azerbajdzan the heart of it is this definition f[y_] := (f[y] = rep; y) which will execute the first time it sees a value y and first perform the assignment f[y] = rep and then return y. That is why the values do not get replaced with "X" the first time they are seen. On the next time the value y is seen, because of the assignment f[y] = rep, the function simply returns y. All of this is wrapped in Module so it's cleaned up properly and then the function is mapped over the list of arguments. $\endgroup$
    – b3m2a1
    Apr 25, 2021 at 2:54
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    $\begingroup$ +1. One nitpick is that one would need to call Clear[f] (but not ClearAll!) at the end before returning the result, otherwise f will not be garbage-collected and will hang around indefinitely after the call to replaceDuplicates - and if you call it multiple times, then just as many different fs will be left hanging around. $\endgroup$
    – Leonid Shifrin
    Apr 26, 2021 at 11:56
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    $\begingroup$ Thank you @Leonid. Excellent point. $\endgroup$
    – kglr
    Apr 26, 2021 at 12:04
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FoldPairList[If[MemberQ[#1, #2], {"X", #1}, {#2, Append[##]}] &,
             {},
             {1, 2, 3, 2, 5, 2, 1, 5, 5, 2, 3, 3, 7}]

(*    {1, 2, 3, "X", 5, "X", "X", "X", "X", "X", "X", "X", 7}    *)

FoldPairList accumulates a list of "already seen" numbers, starting with the empty list. If the new number is in the "already seen" list, we emit "X" and keep the list unchanged; otherwise, we emit the new number and add it to the "already seen" list (with Append[##] being an abbreviation for Append[#1, #2]).

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You may use PositionIndex with ReplacePart.

With

x = {1, 2, 3, 2, 5, 2, 1, 5, 5, 2, 3, 3, 7};

and

replaceDups[expr_, rep_ : "X"] := 
 ReplacePart[expr, List /@ Flatten[Rest /@ Values@PositionIndex@expr] -> rep]

Then

replaceDups[x]

{1, 2, 3, "X", 5, "X", "X", "X", "X", "X", "X", "X", 7}

Is also faster than OP solution.

Hope this helps.

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Not that elegant but rather straight.

A = {1, 2, 3, 2, 5, 2, 1, 5, 5, 2, 3, 3, 7}

ReplacePart[A, {#} & /@ 
   Select[Range[Length[A]], MemberQ[Take[A, # - 1], A[[#]]] &] -> "X"]

Note that

Select[Range[Length[A]], MemberQ[Take[A, # - 1], A[[#]]] &]

is a set of indices those are duplicated.

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