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Re: Position


Length[t = RandomReal[{0, 5}, 10^6]]
1000000

AbsoluteTiming@Length[p = Position[t, _?(# > 2 &)]]
{4.348019, 599993}

AbsoluteTiming@Length[q = Pick[Range@Length@t, UnitStep[t - 2], 1]]
{0.716786, 599993}

Flatten@p === q
True

----- Costa Bravo <q13a27tt at aol.com> wrote:
> How can I find the indices  of all  elements greaters than e.g. 2 of a vector.
> 
> t=RandomReal[{0,5},100]
> Position[t,??]
> 
> --
>   Costa



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