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I am trying to get Position work with patterns. I have the following code:

dat = {"star", "u", "g", "r", "i", "z", "star2", "u", "g", "r", "i", 
   "z", "star3", "u", "g", "r", "i", "z", "star4", "u", "g", "r", "i",
    "z", "Astro", "u", "g", "r", "i", "z"};
Position[dat, "star" ~~ _]
Position[dat, RegularExpression["star."]]

Netheir returns anything. What I want is to return position of "star", "star2" etc. How do I properly use patterns with Position?

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  • $\begingroup$ @kglr No, this does not return the first one. Also it returns empty fields of dimensions of the original list. I would prefer make it work in the way Position works and also understand why this does not actually work $\endgroup$
    – atapaka
    Jun 13, 2016 at 22:01
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    $\begingroup$ As to the "why?"... Patterns and string patterns are two distinct syntactic forms. Position only supports the former. The reasons for the distinction are discussed in (8945). $\endgroup$
    – WReach
    Jun 14, 2016 at 14:20

2 Answers 2

Reset to default
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Position[dat, _String?(StringMatchQ[#, "star" ~~ ___] &)]

or

Position[dat, _String?(StringMatchQ[#, "star*"] &)]

or

Position[dat, _?(StringMatchQ[#, "star*"] &), Heads -> False]

or

Position[StringMatchQ[dat, "star*"], True] (*thanks: @TomD *)

{{1}, {7}, {13}, {19}}

Alternatively,

Pick[Range@Length@dat, StringMatchQ[#, "star*"] & /@ dat]

{1, 7, 13, 19}

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    $\begingroup$ Slight variant: Position[StringMatchQ[#, "star*"], True] &@dat $\endgroup$
    – user1066
    Jun 13, 2016 at 23:51
  • $\begingroup$ @TomD, much nicer, thank you. Updated with that alternative. $\endgroup$
    – kglr
    Jun 13, 2016 at 23:56
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For large lists, this should be snappy:

Pick[Range@Length@dat, StringTake[dat, UpTo@4], "star"]

and actually, taking advantage of listability,

Pick[Range@Length@dat, StringMatchQ[dat, "star*"]]

is a bit faster it seems...

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