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Is there a way to write a function where the input is a vector and the output is the elements in the vector that are repeated?

I was trying to use something along the lines:

f[x_] := DeleteCases[Gather[x], Length[#1] != 1]

Of course this doesn't work, giving me this output:

input: f[{1, 2, 2, 1, 4, 3, 5, 5, 2, 7, 8, 9, 0, 11}] 
output: {{1, 1}, {2, 2, 2}, {4}, {3}, {5, 5}, {7}, {8}, {9}, {0}, {11}}

If I was able to delete the elements with Length 1, I could delete easily construct the output

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    $\begingroup$ Maybe f[x_] := Select[Gather[x], Length[#] != 1 &][[All, 1]]. $\endgroup$
    – JimB
    Apr 23 at 22:27
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DeleteCases requires a pattern, not a function, as second argument:

f[x_] := DeleteCases[Gather[x], {_}]

f[{1, 2, 2, 1, 4, 3, 5, 5, 2, 7, 8, 9, 0, 11}]
(*    {{1, 1}, {2, 2, 2}, {5, 5}}    *)
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You may use Tally and ConstantArray.

With

lst = {1, 2, 2, 1, 4, 3, 5, 5, 2, 7, 8, 9, 0, 11}

and

repeatedElements[lst_List] := ConstantArray @@@ DeleteCases[Tally[lst], {_, 1}]

then

repeatedElements[lst]
{{1, 1}, {2, 2, 2}, {5, 5}}

Hope this helps.

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list = {1, 2, 2, 1, 4, 3, 5, 5, 2, 7, 8, 9, 0, 11};

Gather + Cases

ClearAll[duplicateElements1]
duplicateElements1 = Cases[{_, __}] @* Gather;

duplicateElements1 @ list
{{1, 1}, {2, 2, 2}, {5, 5}}

Counts + Select + KeyValueMap:

ClearAll[duplicateElements2]
duplicateElements2 = KeyValueMap[ConstantArray] @* Select[GreaterThan @ 1] @* Counts;

duplicateElements2 @ list
{{1, 1}, {2, 2, 2}, {5, 5}}

GroupBy + Values

ClearAll[duplicateElements3]
duplicateElements3 = Values @ GroupBy[#, Identity, If[Length@# > 1, #, Nothing] &] &;

duplicateElements3 @ list
{{1, 1}, {2, 2, 2}, {5, 5}}
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You could use ResourceFunction["Duplicates"]

list = {1, 2, 2, 1, 4, 3, 5, 5, 2, 7, 8, 9, 0, 11};
Gather@ResourceFunction["Duplicates"]@list

(* result: {{1, 1}, {2, 2, 2}, {5, 5}} *)

And here's another way using undocumented function GeneralUtilities`FindDuplicates

With[{c = Counts@list},
 ConstantArray[#, c[#]] & /@ GeneralUtilities`FindDuplicates[list]
]
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