With a list say {{x1, y1}, {x2, y2}, …} and a list {z1}, what is the best way to create a list {{x1, y1, z1}, {x2, y2, z1}, {x3, y3, z1}, …}?

  • $\begingroup$ Join[#, {z1}] & /@ {{x1, y1}, {x2, y2}} $\endgroup$ – wuyudi Jan 23 at 8:20

You can use PadRight:

    {{x1, y1}, {x2, y2}},
    {Automatic, 3},

{{x1, y1, z1}, {x2, y2, z1}}

Or ArrayPad:

    {{x1, y1}, {x2, y2}, {x3, y3}},

{{x1, y1, z1}, {x2, y2, z1}, {x3, y3, z1}}

n = 1000000;
a = RandomReal[{-1, 1}, {n, 2}];
b = ConstantArray[0., 1];
c = Join[a, ConstantArray[b, Length[a]], 2]; // RepeatedTiming // First


xy = {{x1, y1}, {x2, y2}, ...}
z = {z1, z2, ...}
xyz = Partition[Flatten[Riffle[xy, z]], 3]

and you're done. Riffle also works with just one z-value: it'll do exactly what you asked, after re-reading your question more carefully.


First, Riffle[list1, list2 **or** element, so just z1 works, too] makes

{ {x1, y1}, z1, {x2, y2}, z2, ...}

then, Flatten[list] makes

{ x1, y1, z1, x2, y2, z2, ... }

finally, Partition[list, 3] turns it into

{ {x1, y1, z1}, {x2, y2, z2}, ...}
  • $\begingroup$ This is NOT what the question asked though. The question wanted to add a single constant z1 element to the right of each sublist. $\endgroup$ – MarcoB Jan 22 at 23:03
  • 1
    $\begingroup$ @MarcoB I think the question is open to question. The OP uses the term "a list {z1}` rather than just z1 and the above answer works whether there's a single term in the list ({z1}) or multiple terms ({z1,z2,z3}) or just a single constant (z1). I also say that because of the odd use of compilation rather than concatenation. $\endgroup$ – JimB Jan 22 at 23:57
  • $\begingroup$ @JimB I see. On the other hand, in their post OP included a "desired output" with a constant z1, and they accepted an answer using a constant z1... I ready the "list {z1}" to mean "a list containing a single element, z1". Plenty of functions in MMA return a list even when the output is a single element, for consistency, so I could see the requirement being described that way in a simplified MWE. Either way, I'll edit and retract the downvote, since it's clearly not as cut-and-dried to everybody as it was to me on first read. $\endgroup$ – MarcoB Jan 23 at 2:22

If z1 is a list (as indicated in the question), then does the following do what you want:

z1 = {1, 2, 3};
xy = {{x1, y1}, {x2, y2}, {x3, y3}};
Flatten[{#, z1}] & /@ xy
(* {{x1, y1, 1, 2, 3}, {x2, y2, 1, 2, 3}, {x3, y3, 1, 2, 3}} *)

Or should the output be as follows:

z = {z1, z2, z3};
xy = {{x1, y1}, {x2, y2}, {x3, y3}};
Flatten[{#, zz}] & /@ xy /. zz -> z
(* {{x1, y1, {z1, z2, z3}}, {x2, y2, {z1, z2, z3}}, {x3, y3, {z1, z2, z3}}} *)

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