Partition sublists of a list

Question

Let's say I have the following list:

l1={{2015, 5, 6, 13692}, {2015, 5, 7, 13715}, {2015, 5, 10,  13274}, 
  {2015, 5, 11, 13581}, {2015, 5, 12, 13609}};

How is it possible to rearrange so it becomes

l2={{{2015, 5, 6}, 13692}, {{2015, 5, 7}, 13715}, {{2015, 5, 10},  13274},
 {{2015, 5, 11}, 13581}, {{2015, 5, 12}, 13609}}

That is, I want to partition each sublist of l1 and make it look like l2.

Argument destructuring works well in this situation. After defining restructure[{a__, b_}] := {{a}, b}, restructure /@ l1 gives the desired result. — m_goldberg, Aug 08 '17 at 20:53

eldo · Answer 1 · 2017-08-08T20:11:36.653

8

l1 = {{2015, 5, 6, 13692}, {2015, 5, 7, 13715}, {2015, 5, 10, 13274}, {2015, 5, 11, 13581}, {2015, 5, 12, 13609}}

l1 /. {a__, b_?AtomQ} :> {{a}, b}

or

Replace[l1, {a__, b_} :> {{a}, b}, {1}]

{{{2015, 5, 6}, 13692}, {{2015, 5, 7}, 13715}, {{2015, 5, 10}, 13274}, {{2015, 5, 11}, 13581}, {{2015, 5, 12}, 13609}}

Another possibility with Part

{#[[1 ;; 3]], #[[4]]} & /@ l1

edited Aug 08 '17 at 20:11

answered Aug 08 '17 at 19:47

eldo

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Thanks a lot! Both work. The second answer with 'Replace' is fantastic. I understand it since it's more close to my level! – Dimitris Leivaditis Aug 08 '17 at 19:55

Kuba · Answer 2 · 2017-08-09T10:24:23.073

5

and a classic:

{{#, #2, #3}, #4} & @@@ l1

edited Aug 09 '17 at 10:24

answered Aug 09 '17 at 05:43

Kuba

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user1066 · Answer 3 · 2017-08-09T10:26:57.437

3

l1 // {#[[All, ;; 3]], #[[All, -1]]} & // Transpose

{{{2015, 5, 6}, 13692}, {{2015, 5, 7}, 13715}, {{2015, 5, 10}, 13274}, {{2015, 5, 11}, 13581}, {{2015, 5, 12}, 13609}}

edited Aug 09 '17 at 10:26

answered Aug 08 '17 at 23:41

user1066

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score 3 · Answer 4 · answered Feb 25 '24 at 00:20

list = 
 {{2015, 5, 6, 13692}, {2015, 5, 7, 13715}, {2015, 5, 10, 13274}, 
  {2015, 5, 11, 13581}, {2015, 5, 12, 13609}};

Using Comap (new in 14.0)

Comap[{Most, Last}] /@ list

{{{2015, 5, 6}, 13692}, {{2015, 5, 7}, 13715}, {{2015, 5, 10}, 13274}, {{2015, 5, 11}, 13581}, {{2015, 5, 12}, 13609}}

Using Query

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

(* same result *)

score 3 · Answer 5 · answered Feb 25 '24 at 02:43

list = {{2015, 5, 6, 13692}, {2015, 5, 7, 13715}, 
        {2015, 5, 10, 13274}, {2015, 5, 11, 13581},
        {2015, 5, 12, 13609}};

Using SlotSequence and MapApply:

Through[{Most, Last}@{##}] & @@@ list
({{{2015, 5, 6}, 13692}, {{2015, 5, 7}, 13715},
 {{2015, 5, 10}, 13274}, {{2015, 5, 11}, 13581}, 
 {{2015, 5, 12}, 13609}})

score 3 · Answer 6 · answered Feb 25 '24 at 05:31

l1 = {{2015, 5, 6, 13692}, {2015, 5, 7, 13715}, {2015, 5, 10, 
    13274}, {2015, 5, 11, 13581}, {2015, 5, 12, 13609}};

FlattenAt[#, -1] &@Partition[#, UpTo[3]] & /@ l1
FlattenAt[#, -1] &@TakeList[#, {3, 1}] & /@ l1  
FlattenAt[#, -1] &@TakeDrop[#, 3] & /@ l1
{Take[#, 3], Last@#} & /@ l1
SequenceCases[#, {a__, b_} :> Sequence @@ {{a}, b}] & /@ l1

Result:

{{{2015, 5, 6}, 13692}, {{2015, 5, 7}, 13715}, {{2015, 5, 10},
13274}, {{2015, 5, 11}, 13581}, {{2015, 5, 12}, 13609}}

score 2 · Answer 7 · answered Aug 09 '17 at 04:24

2

b = Map[List[Flatten@Partition[#, 3], Last[#]] &, l1]

answered Aug 09 '17 at 04:24

Alucard

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Partition sublists of a list

7 Answers7