Re: Efficiency and ReplacePart?
- To: mathgroup at smc.vnet.net
- Subject: [mg85949] Re: [mg85896] Efficiency and ReplacePart?
- From: DrMajorBob <drmajorbob at bigfoot.com>
- Date: Thu, 28 Feb 2008 02:55:16 -0500 (EST)
- References: <6285672.1204145042313.JavaMail.root@m08>
- Reply-to: drmajorbob at longhorns.com
Two more options, one of which seems faster:
n = 500000;
ftemp = Table[{RandomReal[1, {2}], RandomReal[1, {2}]}, {n}];
First@Timing[one = ftemp /. {{a_, b_}, {c_, d_}} :> {{a, b}, {0, 0}}]
First@Timing[two = Map[{#[[1]], {0, 0}} &, ftemp]]
First@Timing[three = Map[ReplacePart[#, 2 -> {0, 0}] &, ftemp]]
First@Timing[four = Map[MapAt[{0, 0} &, #, 2] &, ftemp]]
First@Timing[five = (ftemp[[All, 2]] *= 0; ftemp)]
one == two == three == four == five
0.943361
1.31027
3.20808
1.40886
0.536861
True
Bobby
On Wed, 27 Feb 2008 03:23:36 -0600, W. Craig Carter <ccarter at mit.edu> =
wrote:
>
> If someone can answer the following question, I will have
> learned something about efficiency in mathematica...
>
> I'd like each second part of a list to a zero-list:
>
> ftemp = Table[{RandomReal[1, {2}], RandomReal[1, {2}]}, {50000}];
>
> Compare:
> 1)
> Timing[ftemp /. {{a_, b_}, {c_, d_}} :> {{a, b}, {0, 0}}][[1]]
>
> 2)
> Timing[Map[(# = {#[[1]], {0, 0}} &), ftemp]][[1]]
>
> 3) Timing[Map[ReplacePart[#, 2 -> {0, 0}] &, ftemp]][[1]]
>
> (*
> or if you like:
> times[n_] :=
> Module[{ftemp =
> Table[{RandomReal[1, {2}], RandomReal[1, {2}]}, {n}]},
> {Timing[ftemp /. {{a_, b_}, {c_, d_}} :> {{a, b}, {0,
> 0}}][[1]],
> Timing[Map[(# = {#[[1]], {0, 0}} &), ftemp]][[1]]}]
>
> ListPlot[Table[times[i], {i, 100, 10000, 100}]]
> *)
>
>
> 1 is faster than 2 is faster than 3. Why?
>
> Craig
>
>
>
-- =
DrMajorBob at bigfoot.com