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I have a list like this and want to split it into three sublists. One should contain only the first element the 2nd should contain 8 elements starting from 1.944 to 11 and the third should start from 50 to 1 and should have 7 elements in there. How can I split it? I have a data set of multiple lists of the same kind and wanna split them all the same way.

list ={{"Res ", Around[ 2.5, 0.03], Around[ 1.5, 0.66], Around[  0.07, 0.0], Around[ 2.76, 0.357], Around[  6.736, 2.14], Around[ 4.76, 0.8732], Around[ 12.7, 1.08], Around[ 33.7, 3.747], 50, 0.08, 0.01, 1.*10^-8, 4, 1, 1}, {"Res", Around[ 2.0, 0.07], Around[ 1.5, 0.01], Around[  0.21, 0.06], Around[  2.823, 0.68], Around[  6.5965, 2.49], Around[ 3.874, 1.0], Around[ 11.34, 1.1], Around[ 24.3, 3.02], 50, 0.08, 0.01, 1, 4.8, 1, 1}}
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    $\begingroup$ If alist = Alphabet[][[1 ;; 16]] then TakeList[alist, {1, 8, 7}] would get the desired result. $\endgroup$
    – Syed
    Commented Mar 30, 2023 at 13:33
  • $\begingroup$ It doesn't work, unfortunately. and If I have multiple lists of the same kind like an array then I should write the code for all rows. how should I do that ? $\endgroup$
    – Miss
    Commented Mar 30, 2023 at 13:40
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    $\begingroup$ Please load code (not images). Look at Map. $\endgroup$
    – Syed
    Commented Mar 30, 2023 at 13:44
  • $\begingroup$ Hi @Miss, your list has the correct format but I think there is a typo as it is not a valid list. There is a mixing of strings and lists. Pasting your list in mathematica gives a syntactically broken list. $\endgroup$
    – alex
    Commented Mar 30, 2023 at 14:28

1 Answer 1

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@Sayed's method using TakeList

I can confirm that @Syed's solution works perfectly well:

TakeList[#, {1, 8, 7}] & /@ list

enter image description here

Note

The reason it might not have worked for you is because you had to tweak a bit the values within TakeList.

Alternative method

Alternatively, a more verbose and less elegant solution: I think this should answer your question:

splitter[list_] := 
  list[[#[[1]] ;; #[[2]]]] & /@ {{1, 1}, {2, 9}, {10, -1}};

splitter[#] & /@ list

enter image description here

The secondary question

list1 = TakeList[#, {1, 8, 7}] & /@ list
list2 = Prepend[#[[2 ;;]], #[[1, 1]]] & /@ list1

enter image description here

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    $\begingroup$ Note that approach given by @Syed in comments is much simpler and gives identical results splitter[mylist] === TakeList[mylist, {1, 8, 6}] $\endgroup$
    – Bob Hanlon
    Commented Mar 30, 2023 at 14:15
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    $\begingroup$ I did see Syed's answer but my mind didn't register it as I had never used (or seen) TakeList until I saw your comment. Indeed TakeList is much more elegant of a solution. $\endgroup$
    – alex
    Commented Mar 30, 2023 at 14:19
  • $\begingroup$ {{"Res", Around[ 1.9, 0.1], Around[ 1.2, 0.03], Around[ 0.01, 0.09], Around[ 3.4, 0.6], Around[ 19.8, 2.3], Around[ 0.9, 0.4], Around[ 4.03, 0.38], Around[ 11.2, 3.6], 50, 0.8, 0.01, 5, 4.8, 1, 1}{"Res", Around[ 2.08, 0.8], Around[ 1.3, 0.3], Around[ 0.6, 0.9], Around[ 2.1, 0.4], Around[ 21.4, 1.7], Around[ 0.5, 0.4], Around[ 6.2, 0.4], Around[ 13.9, 2.6`], 50, 0.8, 0.01, 6, 4.8, 1, 1}} $\endgroup$
    – Miss
    Commented Mar 30, 2023 at 14:32
  • $\begingroup$ How can I apply this code to the entire array with multiple lists? $\endgroup$
    – Miss
    Commented Mar 30, 2023 at 15:01
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    $\begingroup$ Thanks a lot! I worked perfectly :). $\endgroup$
    – Miss
    Commented Mar 30, 2023 at 15:09

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