Re: position of sequence of numbers in list

*To*: mathgroup at smc.vnet.net*Subject*: [mg106997] Re: position of sequence of numbers in list*From*: Ray Koopman <koopman at sfu.ca>*Date*: Sun, 31 Jan 2010 05:56:32 -0500 (EST)*References*: <hk17lo$ogf$1@smc.vnet.net>

On Jan 30, 4:11 am, 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 pne[vec,val] returns a list of the positions in vec that are NOT EQUAL to val. peq[vec,val] returns a list of the positions in vec that are EQUAL to val. It's slower than pne but faster than the built-in Position. Both pne and peq return simple lists, not the list-of-lists that Position returns. pne[vec_,val_] := SparseArray[vec,Automatic,val] /. SparseArray[_, _, _, p_] :> Flatten@p[[2,2]] peq[vec_,val_] := Block[{r = Range@Length@vec}, r[[pne[vec,val]]] = 0; SparseArray[r] /. SparseArray[_, _, _, p_] :> p[[3]]] list = Table[Random[Integer,{1,5}],{1*^6}]; Timing@Length[a = Intersection[Position[list,3],Position[list,4]-1]] {1.09 Second, 39674} Timing@Length[b = Position[Partition[list,2,1],{3,4}]] {1.4 Second, 39674} Timing@Length[c = Intersection[peq[list,3],peq[list,4]-1]] {0.44 Second, 39674} SameQ[a,b,Transpose@{c}] True