Re: Position of Last Element in List
- To: mathgroup at smc.vnet.net
- Subject: [mg94183] Re: [mg94130] Position of Last Element in List
- From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
- Date: Sat, 6 Dec 2008 06:18:25 -0500 (EST)
- References: <200812051033.FAA25138@smc.vnet.net>
On 5 Dec 2008, at 19:33, Raffy wrote:
> I'm looking for the fastest way to find the last position of an
> element at level 1 in a list.
>
> The fastest way, which is still extremely bloated and extremely slow,
> appears to be:
>
> lastPos[v_List,p_:True,default_:$Failed]:=(
> $Temp=Position[Reverse[v],p,{1},1,Heads->False];
> If[Length[$Temp]>0,$Temp[[1,1]],default]
> );
>
> Any suggestions?
>
> Use the following code to compare against lastPos:
>
> v = Range[1000000];
> Do[lastPos[v,900000],{100}]//Timing
>
This is slightly faster on my 2.2 gigahertz MacBook:
lastPos1[v_List, p_] :=
With[{m = Last[Split[Unitize[v - p]]]}, If[m[[-1]] === 0, 1,
Length[m] + 1]]
In[3]:= v = Range[1000000];
Do[lastPos[v, 900000], {100}] // Timing
Out[4]= {11.4684, Null}
In[5]:=
Do[lastPos1[v, 900000], {100}] // Timing
Out[5]= {9.56562, Null}
Note that if p is not present instead of returning $Failed this
returns the length of the list + 1.
Andrzej Kozlowski
Andrzej Kozlowski
- References:
- Position of Last Element in List
- From: Raffy <raffy@mac.com>
- Position of Last Element in List