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Re: an even faster way to normalize a
*To*: mathgroup at smc.vnet.net
*Subject*: [mg85690] Re: an even faster way to normalize a
*From*: Ray Koopman <koopman at sfu.ca>
*Date*: Tue, 19 Feb 2008 02:01:10 -0500 (EST)
*References*: <fp9945$199$1@smc.vnet.net>
On Feb 17, 4:25 am, congruentialumina... at yahoo.com wrote:
> Hello UG:
>
> I have a 512x512 array of 3-tuples. I want to make any tuple with a
> value outside of 0 <--> 1, become {0.,0.,0.}.
>
> The first version has this loop:
>
> For[i = 1, i <= graphSize, i++,
> For[j = 1, j <= graphSize, j++,
> If[((sum[[i, j, 1]] < 0) || (sum[[i, j, 1]] > 1) ||
> (sum[[i, j, 2]] < 0) || (sum[[i, j, 2]] > 1) ||
> (sum[[i, j, 3]] < 0) || (sum[[i, j, 3]] > 1)),
> sum[[i, j]] = {0., 0., 0.}
> ]
> ]
> ];
>
> After scratching my head for a while I came up with this (equivalent)
> Map statement.
>
> sum = Map[
> If[#[[1]] < 0 || #[[1]] > 1 || #[[2]] < 0 || #[[2]] > 1 || #[[3]] <
> 0 || #[[3]] > 1, {0., 0., 0.}, #] &, sum, {2}];
>
> It is faster but only by about 15%.
>
> It is unreasonable to believe some other construction can accomplish
> this with a bigger payoff?
>
> Thanks in advance.
>
> Regards..Roger W.
sum = Table[Random[Real,{-.1,1.1}],{512},{512},{3}];
Timing[a = Map[If[#[[1]] < 0 || #[[1]] > 1 || #[[2]] < 0 || #[[2]] >
1 || #[[3]] < 0 || #[[3]] > 1, {0., 0., 0.}, #] &, sum, {2}];]
Timing[b = Map[If[#[[1]] < 0 || #[[3]] > 1 &[Sort@#],
{0., 0., 0.}, #] &, sum, {2}];]
a === b
{0.86 Second,Null}
{0.72 Second,Null}
True
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