6
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A very short version of my list named "fire" is

fire = {{0, 0, 0, 20, 0, 0}, {0, 0, 0, 20, 0, 0}, {0, 0, 20, 20, 0, 0}}.

For the first sublist Length[TakeWhile[fire[[1]], # < 20 &]], calculates the number of preceding "0" before 20 shows up. In this case I get 3. The number of preceding 0 before 20 or the end of the sublist shows up is my ultimate goal.

Since my true list has 1000 of these sublists of six numbers I tried to map the function Length[TakeWhile[fire...]. However, MapApply or Map did not work. The problem seems to identy the subsequent elements of fire within the expression. Any help is highly appreciated.

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10
  • 2
    $\begingroup$ Length@First@SplitBy[#, 20] & /@ fire $\endgroup$
    – martin
    Aug 19, 2022 at 21:22
  • 1
    $\begingroup$ f[x_] := Length[TakeWhile[x, # < 20 &]]; f /@ fire $\endgroup$
    – martin
    Aug 19, 2022 at 21:23
  • $\begingroup$ Unclear: are 0s the only possible entries before the 20? $\endgroup$ Aug 19, 2022 at 21:27
  • 1
    $\begingroup$ Does the built-in work? LengthWhile[#, EqualTo[0]] & /@ fire? Assuming the only possible values are 0 or 20, at least, if the lists here are good examples. $\endgroup$
    – thorimur
    Aug 19, 2022 at 21:43
  • 1
    $\begingroup$ Let me give my 5 cents into this discussion: Position[fire[[1]], 20][[1, 1]] - 1 yields the expected 3. $\endgroup$ Aug 20, 2022 at 6:49

2 Answers 2

1
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list = {{0, 0, 0, 20, 0, 0}, {0, 0, 0, 20, 0, 0}, {0, 0, 20, 20, 0, 0}};

Using SequenceCount

SequenceCount[#, {0 .., Except[0]}, Overlaps -> True] & /@ list

{3, 3, 2}

Another test

SequenceCount[ {{1}, {0, 1}}, {0 .., Except[0]}, Overlaps -> True] & /@ list

{0, 1}

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1
$\begingroup$
list = {{0, 0, 0, 20, 0, 0}, {0, 0, 0, 20, 0, 0}, {0, 0, 20, 20, 0, 0}};

Using Replace at level 1 and SequencePosition:

Replace[list, x_ :> Length@SequencePosition[x, {0 .., 20}], {1}]

{3, 3, 2}

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