Re: Find Position of many elements in a large list.

*To*: mathgroup at smc.vnet.net*Subject*: [mg127695] Re: Find Position of many elements in a large list.*From*: Peter Pein <petsie at dordos.net>*Date*: Wed, 15 Aug 2012 03:35:46 -0400 (EDT)*Delivered-to*: l-mathgroup@mail-archive0.wolfram.com*Delivered-to*: l-mathgroup@wolfram.com*Delivered-to*: mathgroup-newout@smc.vnet.net*Delivered-to*: mathgroup-newsend@smc.vnet.net*References*: <k0d4bu$m8f$1@smc.vnet.net>

Am 14.08.2012 11:05, schrieb benp84 at gmail.com: > I have a sorted, 1-dimensional list X of 1,000,000 integers, and a sorted, 1-dimensional list Y of 10,000 integers. Most, but not all, of the elements of Y are also elements of X. I'd like to know the positions of the elements in X that are also in Y. What's the fastest way to compute this? > > I have an algorithm in mind but it requires lots of custom code and I'm wondering if there's a clever way to do it with built-in functions. Thanks. > Well, using the fact that the huge list (x) is sorted, I got a faster one. binpos[xl_, y0_] := Block[{mid = BitShiftRight[Length[xl]], sel, pos}, If[mid === 0, Boole[xl === {y0}], If[xl[[mid]] <= y0, sel = Drop; pos = mid, sel = Take; pos = 0]; pos + binpos[sel[xl, mid], y0] ]]; Reap[Fold[Drop[#1, Sow[binpos[##]]] &, x, Intersection[x, y]]][[2, 1]] // Accumulate needs only 20% of the time needed by Position[x,Alternatives@@Intersection[x,y]]//Flatten I'm sure, this can be slightly optimized. Peter