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Re: [functional approach should give] an even faster way
*To*: mathgroup at smc.vnet.net
*Subject*: [mg85660] Re: [mg85648] [functional approach should give] an even faster way
*From*: Bob Hanlon <hanlonr at cox.net>
*Date*: Tue, 19 Feb 2008 01:45:42 -0500 (EST)
*Reply-to*: hanlonr at cox.net
Your second method is not 15% faster than your first method it is almost 6 times faster.
test = Table[RandomReal[{-0.1, 1.25}], {512}, {512}, {3}];
Note that the test data is skewed so that it is more likely to be out of bounds on the high side
data1 = test;
imax = Length[data1];
jmax = Length[data1[[1]]];
t1 = Timing[For[i = 1, i <= imax, i++, For[j = 1, j <= jmax, j++,
If[((data1[[i, j, 1]] < 0) || (data1[[i, j, 1]] >
1) ||
(data1[[i, j, 2]] < 0) || (data1[[i, j, 2]] >
1) ||
(data1[[i, j, 3]] < 0) || (data1[[i, j, 3]] > 1)),
data1[[i, j]] = {0., 0., 0.}]]]][[1]];
data2 = test;
t2 = Timing[
data2 = Map[
If[#[[1]] < 0 || #[[1]] > 1 ||
#[[2]] < 0 || #[[2]] >
1 || #[[3]] < 0 || #[[3]] > 1,
{0., 0., 0.}, #] &, data2, {2}];][[1]];
This approach can be written more compactly as
data3 = test;
t3 = Timing[data3 = Map[If[Min[#] < 0 || Max[#] > 1,
{0., 0., 0.}, #] &, data3, {2}];][[1]];
data4 = test;
The same method but checking first within the Or for the more likely out of bounds condition
t4 = Timing[data4 = Map[If[Max[#] > 1 || Min[#] < 0,
{0., 0., 0.}, #] &, data4, {2}];][[1]];
data1 == data2 == data3 == data4
True
{t1, t2, t3, t4}/t1
{1.,0.17293,0.18388,0.1759}
1/%
{1.,5.78268,5.43834,5.68505}
Note that method 4 (checking first for the more likely out of bounds condition) is slightly faster than method 3. Presumably the Or evaluation terminates as soon as a True condition is encountered. If you have prior knowledge of the data statistics this may help.
Bob Hanlon
---- 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.}
> ]
> ]
> ];
>
> 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.
>
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