I'd like to mirror the manipulations (specifically deletion, insertion, and reassignment) done at each position of two equally-sized lists, based on the contents of only one of those lists. As an example say I have

l1 = {a, b, c, 4, e, f, 7, 8, i}
l2 = {1, 2, 3, 4, 5, 6, 7, 8, 9}

and then I want to remove all the numbers from the l1, then also remove elements at the same positions in l2 yielding


I can achieve this specific behavior by creating a selector list according to the conditions imposed on the first list and then applying Pick to both, like

sel[x_List] := Table[! MatchQ[ic, _?NumberQ], {ic, x}]


The above yields the correct l1` and l2`. The function sel can also be used in a more cumbersome fashion with Insert and Part for insertion and reassignment, but all of this seems janky and my heart insists there is a more elegant way. Is this the best that can be done?

I should also note that it cannot be assumed that elements being removed from both lists are equal, nor are the lists canonically ordered. The only safe assumptions are that the lists are equally sized, and that the operation performed at position n in l1 should be repeated at position n in l2 (but determined only by the value at position n in l1).

  • 2
    $\begingroup$ The obvious (and perhaps uncomfortable) solution is working with l = Transpose[{l1,l2}] .... $\endgroup$ Commented Nov 11, 2015 at 3:03
  • 1
    $\begingroup$ When using Pick this might be more straight forward: Pick[l2, NumericQ /@ l1, False] $\endgroup$ Commented Nov 11, 2015 at 3:06
  • $\begingroup$ ...and there's always Position[] to fall back on. $\endgroup$ Commented Nov 11, 2015 at 3:11

2 Answers 2


Maybe start with a basic test:

test=NumericQ /@ l1

This can be used with Pick

Pick[l2, test, False]

If you want positions you could always use Position directly

pos=Position[l1, _?NumericQ]
Delete[l1, pos]

or use Pick again to get positions and test for True or False depending on what you want to do:



List /@ Pick[Range[Length[l1]],test,True|False]

...depending on what format you require

  • $\begingroup$ With Position[], I imagined something like pos = Position[l1, x_ /; ! NumericQ[x], 1, Heads -> False]; Extract[#, pos] & /@ {l1, l2}. $\endgroup$ Commented Nov 11, 2015 at 3:19
  • $\begingroup$ @J.M. My answer was a bit terse but with Position I was thinking more in terms of using with Delete $\endgroup$ Commented Nov 11, 2015 at 3:35
  • $\begingroup$ Oh, of course; with Delete[] you don't need to negate the expression test. :) $\endgroup$ Commented Nov 11, 2015 at 3:59
  • $\begingroup$ I'm down voting because this answer seems to make an incorrect assumption about the generation of the dependent list. Also, it doesn't seem any simpler for the specific manipulation it address than the one presented in the question, but that's a matter of opinion. $\endgroup$
    – IPoiler
    Commented Nov 11, 2015 at 18:12
  • $\begingroup$ @IPoiler I based my answer on finding the positions because that was where the OP needed some work. What you do with the positions once found is straight forward using Delete, Insert and Part $\endgroup$ Commented Nov 11, 2015 at 21:51

In the vein of @belisariushassettled's comment, the lists can be joined as an array to have the rows operated on after Transpose.

l1 = {a, b, c, 4, e, f, 7, 8, i}
l2 = {1, 2, 3, 4, 5, 6, 7, 8, 9}


Example: Remove based on positions of l1 which contain integers:

ld=Transpose@DeleteCases[l, {_Integer, _}];


Example: Insert elements after positions in l1 which contain integers:

li=Transpose@Insert[l, {y, z}, Position[l, {_Integer, _}] + 1];


Example: Transform elements to letters at positions where l1 contains integers:

lr=Transpose@ReplaceAll[l, {x_Integer, y_} :> {FromLetterNumber[x],FromLetterNumber[x + 1]}];

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