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Given the two lists below, is there an in-built command or otherwise neat way of accomplishing the desired output.

list = {{1, 2}, {5, 2}, {9, 3}, {6, 5}};
x = {x1, x2, x3, x4, x5, x6, x7};
desiredOutput = {{1, 2, x1}, {5, 2, x2}, {9, 3, x3}, {6, 5, x4}};

My attempt seems ugly:

{list[[#]][[1]], list[[#]][[2]], x[[#]]} & /@ (list // Length // Range)
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  • $\begingroup$ closely related: 51407 $\endgroup$
    – Kuba ♦
    Jun 21, 2016 at 11:00
  • $\begingroup$ Also related: (189), (13748), (14494) $\endgroup$
    – Mr.Wizard
    Jun 21, 2016 at 11:09

11 Answers 11

Reset to default
14
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Example

Code

list = {{1, 2}, {5, 2}, {9, 3}, {6, 5}};
x = {x1, x2, x3, x4, x5, x6, x7};

MapThread[Append, {list, x[[;; Length  @ list]]}]

Output

{{1, 2, x1}, {5, 2, x2}, {9, 3, x3}, {6, 5, x4}}

Reference

Append
MapThread

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3
  • $\begingroup$ Why does MapThread[Append, {list, x[[;; Length @ list]]}] work but not MapThread[Append, {list, x[[;; list // Length]]}] work? Even though list // Length and Length@list give the same result? $\endgroup$
    – Tom
    Jun 21, 2016 at 11:10
  • 1
    $\begingroup$ @Tom consider ;; list // Length and ;; (list // Length). $\endgroup$
    – Kuba ♦
    Jun 21, 2016 at 11:13
  • $\begingroup$ Oh ok, I suppose it's my lack of understanding of the order in which Mathematica computes things which tripped me up. $\endgroup$
    – Tom
    Jun 21, 2016 at 11:15
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There are so many ways to handle a problem like this and which one is preferred with depend on style, performance, the type and shape of your data, ease of recollection, etc., but here are several more:

Join[list, x ~Take~ Length[list] ~Partition~ 1, 2]

Riffle[Flatten @ list, x, {3, -1, 3}] ~Partition~ 3

PadRight[list, {Automatic, 3}, List /@ x]

And one inspired by J.M.'s use of Flatten

{list, x} ~Flatten~ {2} // Cases[{{x__}, y_} :> {x, y}]
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    $\begingroup$ Yes, that is making nice use of Riffle right there. Good when one can read the documentation. :) $\endgroup$
    – gwr
    Jun 21, 2016 at 11:25
  • $\begingroup$ @gwr Thank you! $\endgroup$
    – Mr.Wizard
    Jun 21, 2016 at 11:30
11
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Shortest so far:

i = 1; list /. {a_, b_} :> {a, b, x[[i++]]}
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    $\begingroup$ +1, but after you add a Module to correctly localize i it will be longer than PadRight. :^) $\endgroup$
    – Mr.Wizard
    Jun 21, 2016 at 11:29
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Transpose[Join[Transpose[list], {Take[x, Length[list]]}]]

should be quite fast for long lists. On my work desktop,

rand = RandomInteger[{1, 10}, {10^5, 2}];
xar = Array[x, 10^6];
AT = AbsoluteTiming;

AT[l1 = Transpose[Join[Transpose[rand], {Take[xar, Length[rand]]}]];]
(* {0.032721, Null} *)

AT[l2 = MapThread[Append, {rand, xar[[;; Length@rand]]}];]
(* {0.112556, Null} *)

AT[l3 = Table[{rand[[k, 1]], rand[[k, 2]], xar[[k]]}, {k, 1, 
     Length[rand]}];]
(* {1.975830, Null} *)

l1 == l2 == l3
(* True *)

EDIT: Some more timings just for fun, in no particular order :)

AT[l4 = (i = 1; rand /. {a_, b_} :> {a, b, xar[[i++]]});]
(* {0.161642, Null} *)

AT[l5 = ArrayFlatten[{{rand, {#} & /@ xar[[1 ;; Length[rand]]]}}];]
(* {0.186327, Null} *)

AT[l6 = Append @@@ 
    DeleteCases[Flatten[{rand, xar}, {{2}, {1}}], {_}];]
(* {1.054091, Null} *)

AT[l7 = MapIndexed[Join[#1, xar[[#2]]] &, rand];]
(* {0.277814, Null} *)

AT[l8 = Join[rand, xar~Take~Length[rand]~Partition~1, 2];]
(* {0.083558, Null} *)

AT[l9 = Riffle[Flatten@rand, xar, {3, -1, 3}]~Partition~3;]
(* {0.028951, Null} *)

AT[l10 = PadRight[rand, {Automatic, 3}, List /@ xar];]
(* {0.315211, Null} *)
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6
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How about this using Table?

Table[{list[[k, 1]], list[[k, 2]], x[[k]]}, {k, 1, Length[list]}]
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5
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Using MapIndexed...

MapIndexed[Join[#1, x[[#2]]] &, list]
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4
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Yet another possibility, using Flatten[] as a "generalized Transpose[]":

Append @@@ DeleteCases[Flatten[{list, x}, {{2}, {1}}], {_}]
   {{1, 2, x1}, {5, 2, x2}, {9, 3, x3}, {6, 5, x4}}
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    $\begingroup$ Shorter: Append @@@ Cases[Flatten[{list, x}, {2}], {_, _}] (and +1) $\endgroup$
    – Mr.Wizard
    Jun 21, 2016 at 14:38
  • $\begingroup$ Yes, that's more compact, @Mr. Wizard. Nevertheless, I still feel the need to remind myself of what Flatten[] is doing by specifying all levels explicitly in the second argument. But that's just my crutch... $\endgroup$
    – J. M.'s lack of A.I. ♦
    Jun 21, 2016 at 14:43
  • $\begingroup$ A perfectly valid crutch, IMHO. $\endgroup$
    – Mr.Wizard
    Jun 21, 2016 at 14:44
1
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list = {{1, 2}, {5, 2}, {9, 3}, {6, 5}}; 
x = {x1, x2, x3, x4, x5, x6, x7};
ArrayFlatten[{{list, {#} & /@ x[[1 ;; 4]]}}]

{{1, 2, x1}, {5, 2, x2}, {9, 3, x3}, {6, 5, x4}}
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1
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I haven't seen this yet, but it may be degenerate with someone else's answer.

Append @@@ Partition[Riffle[list, x], 2]
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1
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ArrayFlatten[{{list, Transpose[{x[[;; Length@list]]}]}}]

See here for some interesting comparisons by Timo

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0
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I think the preferred way to append columns X to matrix A is probably:

Join[X, A, 2]

Adding a single column is slightly less tidy

Join[list, Transpose[{x}], 2]
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