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I have a list like this:

l={{2,3},1,{4,7,2,1},3,{2,2,3,3},0,{4,4,2,1},4}

and I want to replace the natural numbers outside the sublists (at level 1) with an equal number of 0s:

replace[l]=={{2,3},0,{4,7,2,1},0,0,0,{2,2,3,3},{4,4,2,1},0,0,0,0}
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  • 2
    $\begingroup$ Depend on @rhermans If[Head[#] === List, #, Splice@ ConstantArray[0, #]] & /@ l $\endgroup$
    – cvgmt
    Commented Feb 19, 2023 at 12:24
  • $\begingroup$ Variant on other answers: MapAt[Splice@ConstantArray[0, #]&, #, Position[#, _Integer, 1]]&@l $\endgroup$
    – user1066
    Commented Apr 6 at 14:10

7 Answers 7

Reset to default
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Using Replace:

l = {{2, 3}, 1, {4, 7, 2, 1}, 3, {2, 2, 3, 3}, 0, {4, 4, 2, 1}, 4}

Replace[l, i_Integer :> Sequence @@ ConstantArray[0, i], 1]

Using SequenceReplace:

SequenceReplace[l, i : {_Integer} :> Sequence @@ ConstantArray[0, i]]

Using Sow/Reap:

First@Last@
  Reap@Scan[
    If[IntegerQ[#]
      , Sow[Splice@ConstantArray[0, #]]
      (*,Sow@Splice[Table[0,#]]*)
      , Sow@#] &, l
    , {1}
    ]

Using MapAt:

MapAt[Splice[Table[0, #]] &, l
 , Position[l, i_Integer, 1]
 ]

Result:

{{2, 3}, 0, {4, 7, 2, 1}, 0, 0, 0, {2, 2, 3, 3}, {4, 4, 2, 1}, 0, 0, 0, 0}

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10
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Similar with @rhermans

If[ListQ[#], #, Splice@ConstantArray[0, #]] & /@ l

{{2, 3}, 0, {4, 7, 2, 1}, 0, 0, 0, {2, 2, 3, 3}, {4, 4, 2, 1}, 0, 0, 0, 0}

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10
$\begingroup$ 
l /. {x : {__Integer} :> x, i_Integer :> Splice[Table[0, i]]}

{{2, 3}, 0, {4, 7, 2, 1}, 0, 0, 0, {2, 2, 3, 3}, {4, 4, 2, 1}, 0, 0, 0, 0}

ReplaceAll >> Details:

ReplaceAll looks at each part of expr, tries all the rules on it, and then goes on to the next part of expr. The first rule that applies to a particular part is used; no further rules are tried on that part or on any of its subparts.

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8
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Another way to do this is to use SplitBy:

Join @@ (If[Length[Level[#, {-1}]] == 1, ConstantArray[0, #], #] & /@ SplitBy[l, Length])

(*{{2, 3}, 0, {4, 7, 2, 1}, 0, 0, 0, {2, 2, 3, 3}, {4, 4, 2, 1}, 0, 0, 0, 0}*)

Or using SplitBy and ReplaceAll:

Join @@ (SplitBy[l, Length] /. {{x_Integer} :> ConstantArray[0, x]})

(*{{2, 3}, 0, {4, 7, 2, 1}, 0, 0, 0, {2, 2, 3, 3}, {4, 4, 2, 1}, 0, 0, 0, 0}*)
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3
$\begingroup$ 
list = {{2, 3}, 1, {4, 7, 2, 1}, 3, {2, 2, 3, 3}, 0, {4, 4, 2, 1}, 4};

Using SequenceCases and Splice (new in 12.1)

Catenate @ SequenceCases[list, {a_, b_} :> {a, Splice @ Table[0, b]}]

{{2, 3}, 0, {4, 7, 2, 1}, 0, 0, 0, {2, 2, 3, 3}, {4, 4, 2, 1}, 0, 0, 0, 0}

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3
$\begingroup$ 
list = {{2, 3}, 1, {4, 7, 2, 1}, 3, {2, 2, 3, 3}, 0, {4, 4, 2, 1}, 4};

Using SubsetReplace (new in 12.1)

SubsetReplace[list, {a_Integer} :> Splice @ Table[0, a]]

{{2, 3}, 0, {4, 7, 2, 1}, 0, 0, 0, {2, 2, 3, 3}, {4, 4, 2, 1}, 0, 0, 0, 0}

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2
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Another way using ReplaceList:

ReplaceList[l, {___, a_List, b__Integer, ___} :> Splice@{a, Splice@Table[0, b]}]

{{2, 3}, 0, {4, 7, 2, 1}, 0, 0, 0, {2, 2, 3, 3}, {4, 4, 2, 1}, 0, 0, 0, 0}

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