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I have a list from which I would like to extract sublists:

lis = {{a,X,b},{c,d,e,X,f},{h,j,X}}

giving:

res = {{a},{c,d,e},{h,j}}

i.e., deleting each X and its following elements from each sublist. Likely asked and answered; any pointers will be greatly appreciated!

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    $\begingroup$ you could make Thanks Listable and save yourself the /@ :D $\endgroup$
    – Sascha
    Oct 5, 2016 at 19:59

4 Answers 4

Reset to default
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You can use ReplaceAll with an appropriate pattern in such cases

{{a, X, b}, {c, d, e, X, f}, {h, j, X}} /. {begin___, X, end___} :> {begin}

Note the use of named patterns of the BlankNullSequence variety that can stand for any sequence of zero or more expressions

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  • $\begingroup$ This is one elegant answer! beautiful! $\endgroup$
    – alex
    Apr 16 at 10:34
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lis = {{a, X, b}, {c, d, e, X, f}, {h, j, X}};
TakeWhile[#, # =!= X &] & /@ lis

{{a},{c,d,e},{h,j}}

From comment by Mr. J.M

First[Split[#, #2 =!= X &]] & /@ lis
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    $\begingroup$ In the same vein: First[Split[#, #2 =!= X &]] & /@ lis $\endgroup$
    – J. M.'s eventual burnout ♦
    Oct 5, 2016 at 19:06
  • $\begingroup$ @J.M. Maybe it deserve as a answer. :) $\endgroup$
    – yode
    Oct 6, 2016 at 7:05
  • $\begingroup$ I'd prefer you edit your answer instead, as it is too similar to yours. $\endgroup$
    – J. M.'s eventual burnout ♦
    Oct 6, 2016 at 7:20
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Using Position

lis = {{a, X, b}, {c, d, e, X, f}, {h, j, X}};

#[[1 ;; Position[#, X][[1, 1]] - 1]] & /@ lis

(*  {{a}, {c, d, e}, {h, j}}  *)
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    $\begingroup$ It's worth noting that this one assumes that each sublist does contain an X. $\endgroup$
    – Martin Ender
    Oct 6, 2016 at 12:05
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Using SequenceReplace:

lis = {{a, X, b}, {c, d, e, X, f}, {h, j, X}, {X, n, o}}

SequenceReplace[#, {a___, X, b___} :> a] & /@ lis

{{a}, {c, d, e}, {h, j}, {}}

If lists without an X need to be ignored, then it is better to execute:

DeleteCases[lis, _?(FreeQ[#, X] &)]

prior to running the modified list through the SequenceReplace command.

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