       Re: How to find the index of a maximal element in a list?

• To: mathgroup at smc.vnet.net
• Subject: [mg72958] Re: How to find the index of a maximal element in a list?
• From: Peter Pein <petsie at dordos.net>
• Date: Fri, 26 Jan 2007 07:18:37 -0500 (EST)
• References: <epa2rr\$iue\$1@smc.vnet.net>

```Valter Sorana schrieb:
> I may have a mental block, but I cannot find anything better than
>
> Position[listName,Max[listName]]
>
> that traverses the list twice - once to find the maximum and once to find where it is.
>
> Isn't there a way to get both the index and the max value in one go?
>
> (of course one could write a loop that does this, but I want to avoid loops)
>
> Thanks,
>
> Valter.
>

Hi Valter,

if your data is in the range which can be handled by compiled functions, you
might want to try a compiled Scan[].

testit[f_, r_: {3,7}]:=
(SeedRandom;
((data=Table[Random[],{10^#1}];Timing[f[data]])&)/@(Range@@r));

testit[Position[#,Max[#],1,1][[1,1]]&]
Out=
{{0.    Second, 243},
{0.    Second, 5935},
{0.062 Second, 72435},
{0.312 Second, 238526},
{6.797 Second, 5922868}}

This came to my mind, but it sorts the indices of the list and runs longer:

testit[Last[Ordering[#]]&]

{{0.    Second, 243},
{0.    Second, 5935},
{0.046 Second, 72435},
{0.625 Second, 238526},
{9.172 Second, 5922868}}

My suggestion:

testit[Compile[{{lst,_Real,1}},
Module[{mx=First[lst],ix=0,n=1},
Scan[(If[#>mx,mx=#;ix=n];n++)&,Rest[lst]];
ix],
{{mx,_Real},{ix|n,_Integer}}]]

{{0.    Second, 242},
{0.    Second, 5934},
{0.031 Second, 72434},
{0.281 Second, 238525},
{2.828 Second, 5922867}}

But do not try it uncompiled:

testit[Module[{mx=First[#],ix=1,n=1},
Scan[(If[#>mx,mx=#;ix=n];n++)&,Rest[#]]; ix]&]

{{ 0.015 Second, 242},
{ 0.047 Second, 5934},
{ 0.453 Second, 72434},
{ 4.469 Second, 238525},
{45.484 Second, 5922867}}

hth,
Peter

```

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