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Transforming a list to list of lists using a specific rule


Splitting up delimited data in listsSelecting a list from a list of listsPartitioning time series data in sublists by their durationFinding pairs where the intersection of them is empty set from a nested listWay to generate all multisetsGrouping by total length of elements groupedHow to split list into segments based on first element of sublist, and remove ineligible segmentsGathering a List efficientlyRandomly merge elements from paired lists to a new paired listSelecting elements satisfying a condition













4












$begingroup$


Say, we have a list: l = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14 and a corresponding list: v = 2,3,2,3,4. We want to split l into sublists of certain length of consecutive elements ofl and v basically tells us what that length should be. So, in this case, our output would be: 



1,2,3,4,5,6,7,8,9,10,11,12,13,14


I have written a function that works:



g[l_List] := Module[split,
split =Table[, i,1,Length @ l];
split[[1]] = Table[i,i,1,First @ l];
Table[split[[i+1]] = Table[j,j,Last @ split[[i]] + 1 ,Total @ l[[1;;i+1]]], 
i,1,Length@l - 1];
split
]

g@v



1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14




But I'm sure this could be done in a nicer, and more importantly, more efficient way. Any hints?






list-manipulation







share|improve this question









$endgroup$







  • 1




    $begingroup$
    It might be worth thinking about what happens when Total@v does not equal Length@l.
    $endgroup$
    – geordie
    May 8 at 12:07






  • 1




    $begingroup$
    In my case that is always the case, but yeah, it's a valid point.
    $endgroup$
    – amator2357
    May 8 at 12:19















4












$begingroup$


Say, we have a list: l = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14 and a corresponding list: v = 2,3,2,3,4. We want to split l into sublists of certain length of consecutive elements ofl and v basically tells us what that length should be. So, in this case, our output would be: 



1,2,3,4,5,6,7,8,9,10,11,12,13,14


I have written a function that works:



g[l_List] := Module[split,
split =Table[, i,1,Length @ l];
split[[1]] = Table[i,i,1,First @ l];
Table[split[[i+1]] = Table[j,j,Last @ split[[i]] + 1 ,Total @ l[[1;;i+1]]], 
i,1,Length@l - 1];
split
]

g@v



1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14




But I'm sure this could be done in a nicer, and more importantly, more efficient way. Any hints?






list-manipulation







share|improve this question









$endgroup$







  • 1




    $begingroup$
    It might be worth thinking about what happens when Total@v does not equal Length@l.
    $endgroup$
    – geordie
    May 8 at 12:07






  • 1




    $begingroup$
    In my case that is always the case, but yeah, it's a valid point.
    $endgroup$
    – amator2357
    May 8 at 12:19













4












4








4


1



$begingroup$


Say, we have a list: l = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14 and a corresponding list: v = 2,3,2,3,4. We want to split l into sublists of certain length of consecutive elements ofl and v basically tells us what that length should be. So, in this case, our output would be: 



1,2,3,4,5,6,7,8,9,10,11,12,13,14


I have written a function that works:



g[l_List] := Module[split,
split =Table[, i,1,Length @ l];
split[[1]] = Table[i,i,1,First @ l];
Table[split[[i+1]] = Table[j,j,Last @ split[[i]] + 1 ,Total @ l[[1;;i+1]]], 
i,1,Length@l - 1];
split
]

g@v



1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14




But I'm sure this could be done in a nicer, and more importantly, more efficient way. Any hints?






list-manipulation







share|improve this question









$endgroup$




Say, we have a list: l = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14 and a corresponding list: v = 2,3,2,3,4. We want to split l into sublists of certain length of consecutive elements ofl and v basically tells us what that length should be. So, in this case, our output would be: 



1,2,3,4,5,6,7,8,9,10,11,12,13,14


I have written a function that works:



g[l_List] := Module[split,
split =Table[, i,1,Length @ l];
split[[1]] = Table[i,i,1,First @ l];
Table[split[[i+1]] = Table[j,j,Last @ split[[i]] + 1 ,Total @ l[[1;;i+1]]], 
i,1,Length@l - 1];
split
]

g@v



1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14




But I'm sure this could be done in a nicer, and more importantly, more efficient way. Any hints?






list-manipulation




list-manipulation






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked May 8 at 10:30









amator2357amator2357

66810




66810







  • 1




    $begingroup$
    It might be worth thinking about what happens when Total@v does not equal Length@l.
    $endgroup$
    – geordie
    May 8 at 12:07






  • 1




    $begingroup$
    In my case that is always the case, but yeah, it's a valid point.
    $endgroup$
    – amator2357
    May 8 at 12:19












  • 1




    $begingroup$
    It might be worth thinking about what happens when Total@v does not equal Length@l.
    $endgroup$
    – geordie
    May 8 at 12:07






  • 1




    $begingroup$
    In my case that is always the case, but yeah, it's a valid point.
    $endgroup$
    – amator2357
    May 8 at 12:19







1




1




$begingroup$
It might be worth thinking about what happens when Total@v does not equal Length@l.
$endgroup$
– geordie
May 8 at 12:07




$begingroup$
It might be worth thinking about what happens when Total@v does not equal Length@l.
$endgroup$
– geordie
May 8 at 12:07




1




1




$begingroup$
In my case that is always the case, but yeah, it's a valid point.
$endgroup$
– amator2357
May 8 at 12:19




$begingroup$
In my case that is always the case, but yeah, it's a valid point.
$endgroup$
– amator2357
May 8 at 12:19










1 Answer
1






active

oldest

votes


















8












$begingroup$

you can use this built-in 



 TakeList[l,v]



1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14




the above solution works with version 11.3



If you have an older version try 



FoldPairList[TakeDrop,l,v]



1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14







share|improve this answer











$endgroup$












  • $begingroup$
    Brilliant, thank you!
    $endgroup$
    – amator2357
    May 8 at 10:34











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oldest

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1 Answer
1






active

oldest

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active

oldest

votes






active

oldest

votes









8












$begingroup$

you can use this built-in 



 TakeList[l,v]



1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14




the above solution works with version 11.3



If you have an older version try 



FoldPairList[TakeDrop,l,v]



1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14







share|improve this answer











$endgroup$












  • $begingroup$
    Brilliant, thank you!
    $endgroup$
    – amator2357
    May 8 at 10:34















8












$begingroup$

you can use this built-in 



 TakeList[l,v]



1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14




the above solution works with version 11.3



If you have an older version try 



FoldPairList[TakeDrop,l,v]



1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14







share|improve this answer











$endgroup$












  • $begingroup$
    Brilliant, thank you!
    $endgroup$
    – amator2357
    May 8 at 10:34













8












8








8





$begingroup$

you can use this built-in 



 TakeList[l,v]



1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14




the above solution works with version 11.3



If you have an older version try 



FoldPairList[TakeDrop,l,v]



1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14







share|improve this answer











$endgroup$



you can use this built-in 



 TakeList[l,v]



1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14




the above solution works with version 11.3



If you have an older version try 



FoldPairList[TakeDrop,l,v]



1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14








share|improve this answer














share|improve this answer



share|improve this answer








edited May 8 at 10:34

























answered May 8 at 10:32









J42161217J42161217

5,280425




5,280425











  • $begingroup$
    Brilliant, thank you!
    $endgroup$
    – amator2357
    May 8 at 10:34
















  • $begingroup$
    Brilliant, thank you!
    $endgroup$
    – amator2357
    May 8 at 10:34















$begingroup$
Brilliant, thank you!
$endgroup$
– amator2357
May 8 at 10:34




$begingroup$
Brilliant, thank you!
$endgroup$
– amator2357
May 8 at 10:34

















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