Re: Position of Last Element in List
- To: mathgroup at smc.vnet.net
- Subject: [mg94179] Re: Position of Last Element in List
- From: thomas <thomas.muench at gmail.com>
- Date: Sat, 6 Dec 2008 06:17:40 -0500 (EST)
- References: <ghb02f$oho$1@smc.vnet.net>
This is about 8-9 times faster, but only for numeric data. For non-
numeric data it is very slow:
lastPos1[v_List,elem_]:=SparseArray[Unitize[v-elem]-1]/.SparseArray
[_,_,_,p_]:>Flatten[p[[2,2]]][[-1]]
The trick here is that in its internal structure, SparseArray stores
the positions of non-zero elements within its forth argument (named p
above). Unitize[v-elem] is 0 where-ever elem occurs, 1 everywhere
else, and by subtracting 1 only your desired element becomes non-zero.
Unitize evaluates to 1 only for non-zero *numerical* data, otherwise
it remains unevaluated. This is what makes the whole thing slow for
non-numerical data.
Best,
Thomas
On Dec 5, 11:33 am, Raffy <ra... at mac.com> wrote:
> I'm looking for the fastest way to find the last position of an
> element at level 1 in a list.
>
> The fastest way, which is still extremely bloated and extremely slow,
> appears to be:
>
> lastPos[v_List,p_:True,default_:$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