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Re: [functional approach should give] an even faster way to normalize
*To*: mathgroup at smc.vnet.net
*Subject*: [mg85681] Re: [functional approach should give] an even faster way to normalize
*From*: Szabolcs Horvát <szhorvat at gmail.com>
*Date*: Tue, 19 Feb 2008 01:56:32 -0500 (EST)
*Organization*: University of Bergen
*References*: <fp9945$199$1@smc.vnet.net>
congruentialuminaire 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.}
> ]
> ]
> ];
Please don't use For[] for iteration. It is so ugly and unreadable. :-(
Use Do[] instead, which does exactly the same thing, and doesn't even
need to be nested.
>
> 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?
Try this:
Clip[data, {0, 1}] /. 1 -> 0; // Timing
(Note the space between /. and 1 !!)
On my computer (which has a Pentium-M processor) the For[] version takes
3.8 s, the Map[] version ~0.9 s, and the Clip[] version ~0.5 s. But
please understand that *any* timing less than a few seconds is
completely unreliable (especially with Mathematica's Timing[] on a
Windows platform). So all that these results tell us is that the Map[]
version is surely a few times faster than For[]. But please don't make
quantitative conclusions about the performance of the Clip[] version vs
the Map[] version.
Instead, measure the real application which uses this code fragment and
is too slow for you needs. Or just repeat it with Do[command,
{bigNumber}] if you're really curious.
Szabolcs
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