# Finding a subsequence in a list

I have a list and I want to find (in this particular case the first) appearance of a any of some subsequences, of possible different lengths. None of the subsequences is a subsequence of each other. In my particular case I could do this translating the list to a string and using StringPosition. But I could do it because all elements on my list were one-character-long. Before realizing this I had implemented a not-nearly-one-liner that did the trick without recurring to Strings. It didn't do any useless comparison but it did lots of useless coping of the list as a whole, and it turned out to be 50 times slower than the StringPosition version. It can be improved, avoiding that issue, making it even less one-liner. The task just seems too easy to describe so as to be so not-easy to program well... Is there an efficient way to do it for the general case? "Find the first appearance of one of many subsequences (possible different lengths, perhaps could be patterns, or not) in a list"

(Wow, I think I just thought of a good way, I'll give it a shot... If it works I'll auto-answer. But I'd still like your input, I'm afraid I'm missing some options)

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may be you can look at SequenceAlignment and the other related sequence alignments. There is also multiple alignment. I think SmithWaterman is supposed to be a good algorithm for this. – Nasser Jan 29 '12 at 14:38
I asked something very similar here: stackoverflow.com/questions/8740033/… – Szabolcs Jan 29 '12 at 14:40
@Szabolcs, thanks. I'll read it now. What should I do? Close this question? Or leave it because it hasn't been asked heeere? – Rojo Jan 29 '12 at 14:42
@Rojo Leave it, people shouldn't be expected to check SO before posting. I posted my favourite solution as an answer, and credited the original answerer. – Szabolcs Jan 29 '12 at 14:43
For packed arrays, the fastest method I am aware of is the seqposC function from this answer: stackoverflow.com/questions/8364804/… – Leonid Shifrin Jan 29 '12 at 15:12
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I asked the same question on StackOverflow recently, and the answer that is now my favourite came from Jan Pöschko (modified):

findSubsequence[list_, {ss__}] :=
ReplaceList[list, {pre___, ss, ___} :> Length[{pre}] + 1]


This will find all positions of ss in list. Example:

findSubsequence[Range[50] ~Mod~ 17, {4, 5, 6}]


{4, 21, 38}

Despite using patterns, this solution runs very quickly, even for packed arrays. Please see the question I linked to for more possibilities.

findSubsequence[list : h_[__], _[ss__]] :=
ReplaceList[list, h[pre___, ss, ___] :> Length[{pre}] + 1]


Allowing such forms as:

x = Hold[1 + 1, 2 + 1, 3 + 1, 4 + 1, 2 + 1, 3 + 1, 1 + 1, 2 + 1, 3 + 1];

findSubsequence[x, Hold[2 + 1, 3 + 1]]


{2, 5, 8}

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Very nice solution and interesting thread you linked to – Rojo Jan 29 '12 at 15:07
Since you mentioned packed arrays: my function seqposC from this answer stackoverflow.com/questions/8364804/…, is 6-8 times faster than the method you described. The speed comparison is here: stackoverflow.com/questions/8740033/…. This is not to detract from the elegance of the latter. – Leonid Shifrin Jan 29 '12 at 15:10

An idea. Could be extended for more than just one result I suppose, but not as it is right now

Module[{tag, pList}, SetAttributes[pList, Flat];
findFirstSubSequence[l_List, subs_] := (Clear[pList];
pList[Sequence @@ subs] := tag;
Position[pList @@ l, tag, {1}, 1, Heads -> False][[1, 1]])
]


Doesn't seem to be faster than your version Szabolcs, but since I've been going around Flat lately I felt the need

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 Interesting, but complicated. – Sjoerd C. de Vries Jan 29 '12 at 21:25