I have a data set that looks like this:

lis = {a,{{b,c},{d,e}},m,{p,q}}

Each single element of the list ("a" and "m" in this example) is followed by one or more sublists consisting of two elements ("{b,c}", "{d,e}" and "{p,q}" in this example). I would like to tag each sublist pair with its preceding single element, turning the list into:

lis2 = {{b,c,a},{d,e,a},{p,q,m}}

I would be interested in an efficient way of doing this, and thank you in advance for any ideas.


3 Answers 3


Split with a custom test is the first thing that comes to mind. And then Map to assemble the results.

lis = {a, {{b, c}, {d, e}}, m, {p, q}};
splitlis = Split[lis, Not[ListQ[#1]] && ListQ[#2] &];
mapfun[{x_, y_}] := Map[Append[#, x] &, Partition[Flatten[y], 2]];
Flatten[Map[mapfun, splitlis], 1]

(* {{b, c, a}, {d, e, a}, {p, q, m}} *)

  • $\begingroup$ There's nothing in the question to suggest that the initial split couldn't just be Partition[lis, 2]. $\endgroup$
    – wxffles
    Commented Nov 2, 2016 at 3:30
  • $\begingroup$ Works like a charm. Thanks! $\endgroup$
    – Suite401
    Commented Nov 2, 2016 at 5:30

Assuming that the lis has a recurring pattern of a Symbol followed by either a list of lists or a vector, two definitions must be employed for processing.

f[{a_Symbol, b_?VectorQ}] := {Sequence @@ b, a}
f[{a_Symbol, b : {__List}}] := (Append[#, a] & /@ b ) // 

 , {a_Symbol, b_List} :> f[{a, b}]

{{b, c, a}, {d, e, a}, {p, q, m}}



  {{a_, b : {__List}} :> Splice[Append[a] /@ b], 
   {a_, b : {__}} :> Append[a] @ b}]
{{b, c, a}, {d, e, a}, {p, q, m}}


If[ListQ @ First @ #2, Splice[Append[#] /@ #2], Append[#] @ #2] & @@@ 
  Split[lis, ListQ @ #2 &]
{{b, c, a}, {d, e, a}, {p, q, m}}


{a___, b_, takeOver[c_], d___} ^:= 
   {a, If[ListQ @ First @ c, Splice[Append[b] /@ c], Append[b] @ c], d}

Replace[lis, x_List :> takeOver[x], {1}]
{{b, c, a}, {d, e, a}, {p, q, m}}
  • $\begingroup$ The first two sublists are different compared to what's stated in the OP. $\endgroup$
    – Syed
    Commented Apr 27, 2023 at 9:40
  • $\begingroup$ Thank you @Syed. I can't see a simple fix; so deleting for now. $\endgroup$
    – kglr
    Commented Apr 27, 2023 at 10:38
  • $\begingroup$ The third variation is much above my current skill level. Thanks for this comprehensive answer. $\endgroup$
    – Syed
    Commented May 2, 2023 at 3:44

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