Re: speed
- To: mathgroup at smc.vnet.net
- Subject: [mg15652] Re: speed
- From: Paul Abbott <paul at physics.uwa.edu.au>
- Date: Sat, 30 Jan 1999 04:28:47 -0500 (EST)
- Organization: University of Western Australia
- References: <77utg2$jnk@smc.vnet.net> <78pbv2$d0t@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
Margit wrote:
> I have a question concerning the numerical speed of Mathematica:
> I performed the following calculation
>
> x=0.3;Print[x]; Do[x=x*1.003456,{i,1,1000000}];Print[x]
>
> with Mathematica, it took about 150 seconds.
Reducing the number of iterations by 100, one sees that Do is slow:
In[1]:= x = 0.3; Do[x = x*1.003456, {i, 1, 100000}] // Timing // First
Out[1]= 8.66 Second
In[2]:= x
Out[2]=
149
2.04424 10
Using Nest -- which is the natural way of coding this type of operation
-- one see that Nest runs over 30 times faster.
In[3]:= Nest[1.003456# &, 0.3, 100000] // Timing Out[3]=
149
{0.27 Second, 2.04424 10 }
> The same calculation is performed by TurboPascal in 0.4 seconds.
But is the answer returned by TurboPascal sensible? Mathematica gets
1497
6.474677709 10
Do you get an overflow with TurboPascal?
> I guess that the difference is caused by the working precision. Does
> anyone know what is the reason for this long duration and how it can
> be changed?
Part of the difference is that Nest operations in Mathematica are much
faster than Do loops. You can pseudo-compile your function (which then
only uses machine numbers) and get a slight speed increase (whilst the
result is still a machine number)
In[4]:= g = Compile[{a, x, n}, Nest[Function[y, a y], x, n]]; In[5]:=
g[1.003456, 0.3, 100000] // Timing Out[5]=
149
{0.22 Second, 2.04424 10 }
Cheers,
Paul
____________________________________________________________________
Paul Abbott Phone: +61-8-9380-2734
Department of Physics Fax: +61-8-9380-1014
The University of Western Australia Nedlands WA 6907
mailto:paul at physics.uwa.edu.au AUSTRALIA
http://www.physics.uwa.edu.au/~paul
God IS a weakly left-handed dice player
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