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I have data that, when imported to my notebook, is shown in the form:

list = {{1,2,3},{4,5,6},{7,8,9},{10,11,12}}

I want to transform this into:

list = {{{1,2},3},{{4,5},6},{{7,8},9},{{10,11},12}}

How can I do this?

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  • $\begingroup$ My intent is make the data in a form that permite me do a multi-dimensional interpolation. $\endgroup$ May 26, 2023 at 15:42

12 Answers 12

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Method 1 - No Map

With no Maps.

list = {{1, 2, 3}, {4, 5, 6}, {7, 8, 9}, {10, 11, 12}};
Transpose@{list[[;; , ;; 2]], list[[;; , 3]]}

{{{1, 2}, 3}, {{4, 5}, 6}, {{7, 8}, 9}, {{10, 11}, 12}}

Method 2 - Apply

Not technically Map, but still iterating.

{{#1, #2}, #3} & @@@ list

{{{1, 2}, 3}, {{4, 5}, 6}, {{7, 8}, 9}, {{10, 11}, 12}}

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11
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Another way

list /. {x_, y_, z_} -> {{x, y}, z}

{{{1, 2}, 3}, {{4, 5}, 6}, {{7, 8}, 9}, {{10, 11}, 12}}

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Through[{Most, Last}[#]] & /@ list
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7
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You can do Map[{Most[#], Last[#]} &, list] to produce the desired output.

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7
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also you can do

lst = {{1, 2, 3}, {4, 5, 6}, {7, 8, 9}, {10, 11, 12}}
{#[[;; -2]], #[[-1]]} & /@ lst

Mathematica graphics

Also possible (if you like to view things as matrices)

a = Map[List, lst[[All, {1, -2}]]]
b = lst[[All, -1]]
MapThread[Append, {a, b}]

Mathematica graphics

8 more different 6 more ways could be possible.

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7
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Using SequenceCases:

SequenceCases[list, {x_}:>{Most@x, Last@x}]
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A rule-based approach:

list = {{1, 2, 3}, {4, 5, 6}, {7, 8, 9}, {10, 11, 12}};
list /. {x__?NumericQ, y_?NumericQ} :> {{x}, y}

A simple Table-based approach:

Table[{list[[kk, 1 ;; 2]], list[[kk, 3]]}, {kk, 1, Length@list}]
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BlockMap[Join,Riffle[list[[All,;;2]],list[[All,3]]],{2}]

(* {{{1, 2}, 3}, {{4, 5}, 6}, {{7, 8}, 9}, {{10, 11}, 12}} *)

In addition

Fold[List,#]&/@list

(* {{{1, 2}, 3}, {{4, 5}, 6}, {{7, 8}, 9}, {{10, 11}, 12}} *)

Also

MapThread[{{#1,#2},#3}&,Transpose[list]]

(* {{{1,2},3},{{4,5},6},{{7,8},9},{{10,11},12}} *)
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5
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Drop and Append are sometimes useful. The following works:

Map[Append[{Drop[#, -1]}, Last[#]] &, list]
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  • $\begingroup$ Instead of using Drop, use Most. It is clearer and shorter $\endgroup$ May 28, 2023 at 8:41
4
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You can use Partition. All of the following work:

Map[{Flatten[Partition[#, 2]], #[[-1]]} &, list]

Map[{Flatten[Partition[#, 2]], #[[3]]} &, list]

Map[{First[Partition[#, 2]], #[[-1]]} &, list]

Map[{First[Partition[#, 2]], #[[3]]} &, list]

Map[{First[Partition[#, 2]], Last[#]} &, list]
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4
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The well known IA told me, with the exact question as prompt :

In[1]:= list = {{1, 2, 3}, {4, 5, 6}, {7, 8, 9}, {10, 11, 12}}
Out[1]= {{1, 2, 3}, {4, 5, 6}, {7, 8, 9}, {10, 11, 12}}

In[2]:= transformedList = Map[{Take[#, 2], Last[#]} &, list]
Out[2]= {{{1, 2}, 3}, {{4, 5}, 6}, {{7, 8}, 9}, {{10, 11}, 12}}

And it works. I never touched mathics before. It's not (so much) ironic but to remember that new fact.

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list = {{1, 2, 3}, {4, 5, 6}, {7, 8, 9}, {10, 11, 12}};

With Comap (new in 14.0)

Comap[{Most, Last}] /@ list

{{{1, 2}, {3}}, {{4, 5}, {6}}, {{7, 8}, {9}}, {{10, 11}, {12}}}

With TakeList (new in 11.2)

TakeList[#, {2, 1}] & /@ list

{{{1, 2}, {3}}, {{4, 5}, {6}}, {{7, 8}, {9}}, {{10, 11}, {12}}}

With Query (new in 10.0)

Query[All, {Most, Last}] @ list

{{{1, 2}, 3}, {{4, 5}, 6}, {{7, 8}, 9}, {{10, 11}, 12}}

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