Find position of sublists containing a given element at nth position

Question

I have a list of lists similar to this:

L = {{"a", "b", "c"}, {"x", "c", "y"}, {"i", "j", "h"}, {"x", "b", "z"}}

Each list within L happens to be of length 3. Suppose I need to find the position of the lists that have a particular element (say, "b") at the $n^{th}$ position. How can I do this efficiently?

Currently, I have an approach that works but I don't think it is very efficient inefficient (it creates a new list with the $n^{th}$ elements and then looks for the queried element, I'm sure there's a way to search L directly):

queriedElement = "b";

queriedPosition = 2;

occurencePositions = Flatten@Position[#[[queriedPosition]] & /@ L, queriedElement]//AbsoluteTiming

Which gives the correct answer:

{0.000035, {1, 4}}

Searching for an alternative efficient way because I need to do this in large lists.

Thanks!

Answer 1

Flatten @ Position[L, _List?(#[[2]] === "b" &)]

{1, 4}

Making it a function:

posF1 = Flatten @ Position[#, _List?(Function[x, x[[#2]] === #3])] &;

posF1[L, 2, "b"]

 {1, 4}

Alternatively,

posF2 = Flatten @ Position[#[[All, #2]], #3] &;

posF2[L, 2, "b"]

 {1, 4}

Also

posF3 = PositionIndex[#[[All, #2]]]@#3 &;

posF3[L, 2, "b"]

 {1, 4}

Answer 2

I tried a solution with Pick.

f[v_,p_,e_]:=Pick[Range[Length[v]], v[[All, p]], e]

where v is your L, p is your queriedPosition and e is your queriedElement.

Here is a timing test of Pick, along with the functions from @kglr.

Block[{element, position, list, posF1, posF2, posF3},
   element = "a";
   position = 1;
   posF1 = Flatten@Position[#, _List?(Function[x, x[[#2]] === #3])] &;
   posF2 = Flatten@Position[#[[All, #2]], #3] &;
   posF3 = PositionIndex[#[[All, #2]]]@#3 &;
   list = RandomChoice[CharacterRange["a", "j"], {2000, 3}];
   First /@ {
      RepeatedTiming[f[list, position, element]],
      RepeatedTiming[posF3[list, position, element]],
      RepeatedTiming[posF2[list, position, element]],
      RepeatedTiming[Flatten@Position[list, _List?(#[[position]] === element &)]],
      RepeatedTiming[posF1[list, position, element]]
   }
]

Sample output:

{0.000125, 0.000200, 0.0002303, 0.00248, 0.0030}