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I need to create a list that as an input receives the number $N$ of elements that it will contain and that returns:

{{_,_,#[[1]]},{_,_,#[[2]]},{_,_,#[[3]]},...{_,_,#[[N]]}}

I've tried using Hold and Release but haven't come up with a solution.

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  • $\begingroup$ How about ToExpression[Table [StringJoin["{_,_,#[[",ToString[k],"]]}"],{k,1,n}]] ? You might want to add an & at the end of your list. $\endgroup$
    – Peltio
    Oct 27, 2013 at 21:22
  • $\begingroup$ Can you explain your use case? Are you trying to generate some code? For what end? $\endgroup$
    – Szabolcs
    Oct 27, 2013 at 21:28
  • $\begingroup$ I am trying to generate this list to use it in a pattern match operation: Position[ToExpression /@ indParList, {{, _, #[[1]]}, {, _, #[[2]]}}] & /@ tuples in which the tuples list may vary in length and therefore the pattern match array should adapt to this variation $\endgroup$
    – chipdelmal
    Oct 27, 2013 at 21:30
  • 2
    $\begingroup$ This is a bad way of doing it. Why not just transpose your data and then use the last element? It will automatically work for all values of n. $\endgroup$
    – rm -rf ♦
    Oct 27, 2013 at 22:02
  • $\begingroup$ @rm-rf Agree, but perhaps n != Lenght@tuples[[_]]. It can be fixed easily though $\endgroup$
    – Dr. belisarius
    Oct 27, 2013 at 22:05

2 Answers 2

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Perhaps Thread[{_, _, #}] applied to each tuple thus:

SeedRandom[1];
data = Table[RandomInteger[1, {RandomInteger[{2, 4}], 3}], {100}];
tuples = Table[RandomInteger[1, RandomInteger[{2, 4}]], {5}];

Application:

pos = Position[data, Thread[{_, _, #}]] & /@ tuples
(* {{{38}, {82}}, {{33}, {43}, {47}, {94}}, {{58}}, {{37}, {85}, {88}}, {}} *)

Check:

Grid[
 MapThread[{Map[Last, Extract[data, #1], {2}], #2} &, {pos, tuples}],
 Alignment -> Left]
(* {{1, 1, 0, 0}, {1, 1, 0, 0}}                   {1, 1, 0, 0}
   {{1, 1, 1}, {1, 1, 1}, {1, 1, 1}, {1, 1, 1}}   {1, 1, 1}
   {{0, 0, 1, 0}}                                 {0, 0, 1, 0}
   {{1, 0, 0}, {1, 0, 0}, {1, 0, 0}}              {1, 0, 0}
   {}                                             {0, 0, 1, 1}}  *)
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Perhaps you could also consider:

(* Michael's data *)
SeedRandom[1];
data = Table[RandomInteger[1, {RandomInteger[{2, 4}], 3}], {100}];
tuples = Table[RandomInteger[1, RandomInteger[{2, 4}]], {5}];

Then:

Position[Last /@ Transpose /@ data, Alternatives @@ tuples]
(*
==> {{33}, {37}, {38}, {43}, {47}, {58}, {82}, {85}, {88}, {94}}
*)

or

Position[Last /@ Transpose /@ data, #] & /@ tuples
(*
==> {{{38}, {82}}, {{33}, {43}, {47}, {94}}, {{58}}, {{37}, {85}, {88}}, {}}
*)

or Simon's (slickest)

Position[data[[All, All, -1]], #] & /@ tuples
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  • $\begingroup$ I was waiting for rm-rf or you to post this. The Alternatives one will probably bog down on long lists of tuples. But the second one should be faster than mine. $\endgroup$
    – Michael E2
    Oct 28, 2013 at 12:58
  • $\begingroup$ @MichaelE2 I've seen his comment a few hours ago, but thought he wasn't going to post it. Anyway this is the first thing I would think of for this problem, and surely not pattern matching $\endgroup$
    – Dr. belisarius
    Oct 28, 2013 at 13:16
  • 1
    $\begingroup$ Could also use data[[All, All, -1]] $\endgroup$
    – Simon Woods
    Oct 28, 2013 at 14:12
  • $\begingroup$ @SimonWoods Nice one. Going to add it $\endgroup$
    – Dr. belisarius
    Oct 29, 2013 at 2:06

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