       Re: Value and position in list

• To: mathgroup at smc.vnet.net
• Subject: [mg33423] Re: Value and position in list
• From: Jens-Peer Kuska <kuska at informatik.uni-leipzig.de>
• Date: Thu, 21 Mar 2002 09:27:00 -0500 (EST)
• Organization: Universitaet Leipzig
• References: <a79ckp\$7uk\$1@smc.vnet.net>
• Sender: owner-wri-mathgroup at wolfram.com

```Hi,

minpos = Compile[{{lst, _Real, 2}},
Module[{i = 1},
MapIndexed[If[First[#] < lst[[i, 1]], i = First[#2], 0] &, lst];
{{i, 1}}
]
]

minpos[data] // Timing

should be faster.

Regards
Jens

Hugh Goyder wrote:
>
> I have a list where each element is a list of three values
> {{v11,v12,v13},{v21,v22,v23}...}.
> What is the fastest method for finding the position and value of the
> element which has the smallest first element.
>
> I have the following but is it possible to do this more quickly, perhaps
> without two sorts through the data?
>
> In:=
> \$Version
>
> Out=
> 4.1 for Microsoft Windows (June 13, 2001)
>
> In:=
> data=Table[{Random[],Random[]-0.5,10(Random[]-0.5)},{10000}];
>
> In:=
> Timing[a=Min[Transpose[data][]];p=Position[data,a];{a,p}]
>
> Out=
> {0.33 Second,{0.0000316723,{{1706,1}}}}
>
> I must also find the element which has the smallest second element greater
> than zero. Here are my attempts so far. Is there a faster method?
>
> In:=
> Timing[a=Infinity;i=0;
>   While[i++;i<Length[data],If[0<data[[i,2]]<a,a=data[[i,2]];p=i]];{a,p}]
>
> Out=
> {1.31 Second,{0.0000126855,1134}}
>
> In:=
> Timing[a=Min[Transpose[data][]/.v_/;v\[LessEqual] 0 \[Rule] Infinity];
>   p=Position[data,a];{a,p}]
>
> Out=
> {0.71 Second,{0.0000126855,{{1134,2}}}}
>
> In:=
> Timing[a=Min[Select[Transpose[data][],#>0&]];p=Position[data,a];{a,p}]
>
> Out=
> {0.61 Second,{0.0000126855,{{1134,2}}}}
>
> Thank you for your ideas
>
> Hugh Goyder

```

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