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I want to get a list from a parent list (containing sub-lists of unequal length) using Cases and a matching pattern. The parent list is

data = {{x,y,z,a},{a,x^m,y^n},{a,b^m}, {a,x,b,c}}

I want to pick the matching patterns (x^_,y^_) from the sub-lists and keep them in a similar list format. Using Cases (at level spec 2), I could almost achieve that but unfortunately, it writes in a single list (instead of keeping the patterns in the same sub-list from where they belong). Eg,

Cases[data,Alternatives@@{x|x^_,y|y^_},2]

gives me

{x, y, x^m, y^n, x}

whereas, the output I want

{{x, y}, {x^m, y^n}, {}, {x}}

such that the final findings corresponding to the matched pattern result similar format (even if it is an empty sub-list).

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    $\begingroup$ Suppose you Map[Cases[#,...yourpattern...]&,data] That should use Cases on each of your sublists and return the results from each of those as a list and then all those lists will be returned in an enclosing list. $\endgroup$
    – Bill
    Mar 13, 2022 at 0:27

3 Answers 3

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Just map the Cases expression over the list:

Cases[Alternatives[x | x^_ | y | y^_]] /@ data
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    $\begingroup$ Alternatives is redundant, so: Cases[x | x^_ | y | y^_] /@ data $\endgroup$
    – Rabbit
    Mar 14, 2022 at 20:04
  • $\begingroup$ lol! I simplified the disjunction but then didn't discard the wrapper! $\endgroup$
    – lericr
    Mar 14, 2022 at 20:09
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Pick performs the sort of operation the OP seeks, but it can be rather tricky (to me).

Pick[
 data,
 data /. {(x | y)^_. -> True, Except[_List | List] -> False}]

(*  {{x, y}, {x^m, y^n}, {}, {x}}  *)

Note: It will pick all form at all levels.

To get everything down only to level 2:

level = 2;
data2 = {{x, y, z, a}, {a, x^m, y^n}, {a, b^m}, {a, x, b, c, {x^2}}};
Pick[data2,
 Replace[data2, {(x | y)^_. -> True}, level] /. 
  Except[True | List | _List] -> False]

(*  {{x, y}, {x^m, y^n}, {}, {x, {}}}  *)
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    $\begingroup$ Here's an example of its trickiness: Pick[data, data, (x | y)^_.] has a {1} instead of the {} in the output. $\endgroup$
    – Michael E2
    Mar 13, 2022 at 1:36
  • $\begingroup$ I like the improvement dot . :) $\endgroup$
    – BabaYaga
    Mar 22, 2022 at 12:54
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data = {{x,y,z,a},{a,x^m,y^n},{a,b^m}, {a,x,b,c}}

DeleteCases[data, Except[x | x^_ | y | y^_], {2}]

{{x, y}, {x^m, y^n}, {}, {x}}

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