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