Re: position of sequence of numbers in
- To: mathgroup at smc.vnet.net
- Subject: [mg107031] Re: position of sequence of numbers in
- From: JH <jlucio at ubu.es>
- Date: Mon, 1 Feb 2010 06:10:56 -0500 (EST)
- References: <hk3vv8$etb$1@smc.vnet.net>
On 31 ene, 14:18, Bob Hanlon <hanl... at cox.net> wrote:
> This is wrong. I obviously didn't test this. Instead, use
>
> myList = {3, 1, 2, 5, 3, 4, 6, 2, 8, 9, 3, 4}
>
> {3,1,2,5,3,4,6,2,8,9,3,4}
>
> Select[Position[myList, 3],
> MemberQ[Position[myList, 4] - 1, #] &]
>
> {{5}, {11}}
>
> Bob Hanlon
>
> ---- Bob Hanlon <hanl... at cox.net> wrote:
>
> =============
>
> Position[list, Sequence[3, 4]]
>
> Bob Hanlon
>
> ---- JB <jke... at gmail.com> wrote:
>
> =============
> Hi,
>
> What is the most efficient way to find the position of the beginning
> of a sequence of numbers from a list?
>
> I found a couple of ways:
>
> find 3,4 in list={1,2,3,4,5};
>
> 1. pos=Intersection[Position[list,3],(Position[list,4])+1]
>
> 2. pos=Position[Partition[list,2,1],{3,4}]
>
> Are there other ways to do this?
> What is the best way when dealing with large lists?
>
> Thanks,
> JB
Hello
I suggest this:
myList = RandomInteger[{0, 8}, 1000000]; (* A list with one million
integer numbers between 0 and 6 *)
mySeq = myList[[1000000 - 6 ;; 1000000 - 2]]; (* Sequence to find
in the list: we take a portion of the list, for example, of length 5
almost at the end *)
myList2 =
Table[myList[[i ;; i + Length[mySeq] - 1]], {i, Length[myList] -
Length[mySeq] + 1}]; (* we generate a list with all portions of
myList with the size of mySeq *)
Position[myList2, mySeq] (* Here you have the positions you were
searching *)
JH