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RE: Re: Newbie Question
*To*: mathgroup at smc.vnet.net
*Subject*: [mg32776] RE: [mg32728] Re: [mg32686] Newbie Question
*From*: "Wolf, Hartmut" <Hartmut.Wolf at t-systems.com>
*Date*: Sat, 9 Feb 2002 05:11:50 -0500 (EST)
*Sender*: owner-wri-mathgroup at wolfram.com
> -----Original Message-----
> From: Sseziwa Mukasa [mailto:mukasa at jeol.com]
To: mathgroup at smc.vnet.net
> Sent: Thursday, February 07, 2002 11:10 AM
> To: mathgroup at smc.vnet.net
> Subject: [mg32776] [mg32728] Re: [mg32686] Newbie Question
>
>
> "Brunsman, Kenneth J" wrote:
>
> >
> > I can parse out the column I need using Take [data [[All,
> Column #]]] --- no
> > big deal. So far, so good.
> >
>
> I'm not sure what you're using Take for, data[[All,Column #]]
> should be
> sufficient.
>
> >
> > Now here's the problem --- How do I get % Differences
> between any of 10,000
> > items in that list? I need to take % Differences between
> adjacent pairs,
> > i.e. n and n-1, as well as items n and n-m where m can
> range from 1 to
> > 10,000.
> >
>
> Do you want all the differences n-m (m = 1..n-1) at once or just for a
> particular m? For an individual m I got good results with
>
> f[x_,m_Integer]:=Block[{a=Drop[x,m]},(a-Drop[x,-m])/a]
>
> On a PowerMac with an 800MHz G4 processor it takes 0.02
> seconds to evaluate
> f[x,1] where x is a list of length 10000.
> Table[f[x,i],{i,Length[x]-1}] will
> of course give you all the differences for m = 1..n-1.
>
> >
> > Further, I need to make sure that this code runs fast
> because I'm about to
> > run this on a data set of 10^8 data points (financial times
> series). I
> > could do this standing on my head in Fortran, but I'm bound
> and determined
> > to learn Mathematica if it takes the rest of my unnatural life.
> >
>
> The key difference between Fortran programming and
> Mathematica in my opinion is
> that writing explicit loops in Mathematica is generally
> inefficient, so is
> indexing elements of lists and arrays. Take advantage of the
> fact that many
> operators will automatically apply themselves to every
> element in a list. Also
> avoid making copies as much as possible by using lambda
> functions and mapping.
> Finally, variables are best used to eliminate common
> subexpressions as in the
> Block statement in the function above.
>
> Incidentally though, keeping a list of 10^8 elements in
> memory is probably not
> a good idea, the list should probably be broken up and
> processed in smaller
> pieces to avoid spending all your time swapping memory to disk.
>
> > This is my first time attempting list processing and all
> I'm doing is
>
> > screwing up royally.
> >
>
> Functional programming and pattern matching are not generally
> practiced outside
> of academic programming exercises and they are a very
> different paradigm from
> traditional imperative programming (object oriented techniques being a
> different kettle of fish entirely). Practice makes perfect
> of course. There
> are many excellent books that not only teach efficient
> programming style in
> Mathematica but generally do so while applying to techniques
> to domain specific
> problems. I don't do any financial series analysis myself
> but a quick search
> on Amazon.com for "mathematica finance" turns up 6 titles on
> using Mathematica
> for economic and financial modeling.
>
> Regards,
>
> Sseziwa Mukasa
>
>
Sseziwa,
this is the right way to deal with the problem, I think. Here
just a little modification of this idea, and an other one following
a suggestion from Andrzej Kozlowski:
In[1]:= ll = Table[Random[Real, {0, 100}], {300}];
your suggestion:
In[2]:=
f[x_, m_Integer] := Block[{a = Drop[x, m]}, (a - Drop[x, -m])/a]
In[3]:=
r1 = Table[f[ll, i], {i, Length[ll] - 1}];
my variant:
In[4]:=
r2 = (#1 - #2)/#1 & @@@
Drop[NestList[{Drop[First[#], 1], Drop[Last[#], -1]} &, {ll, ll},
Length[ll] - 1], 1];
In[5]:= r1 == r2
Out[5]= True
The idea was: dropping the first or last element of a list
is faster than dropping half of it.
Another idea (trying to use efficient list operations):
In[6]:=
g[x_, y_] := (y - x)/y; g[_] := Sequence[];
In[7]:=
r3 = Drop[ListCorrelate[ll, ll, {1, 1}, {}, g, List], 1];
In[8]:= r3 == r2
Out[8]= True
Timing however excludes this variant; compare the other two:
In[9]:= ll = Table[Random[Real, {0, 100}], {4000}];
In[16]:=
(r = Table[f[ll, i], {i, Length[ll] - 1}]); // Timing
Out[16]=
{20.57 Second, Null}
In[17]:= Remove[r];
In[18]:=
(r = (#1 - #2)/#1 & @@@
NestList[{Drop[First[#], 1], Drop[Last[#], -1]} &,
{Drop[ll, 1], Drop[ll, -1]},
Length[ll] - 2]); // Timing
Out[18]=
{13.359 Second, Null}
In[19]:= Remove[r];
Your variant however is more economic with memory, and I was
not able to run the test with a list of 10000. (400 MHz P II
Notebook, 192 MB real memory). I estimate that 10000 will take
about 3 minutes. This would mean 10^8 data will at least take
several days, perhaps too much to make any predictions.
Anyways I agree with you that this is not the right task for
Mathematica, producing a vaste of numbers of little information
content. From which you finally will have to read off something,
what? how?
--
Hartmut
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