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MathGroup Archive 2005

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Re: Modeling and Array Problem

  • To: mathgroup at smc.vnet.net
  • Subject: [mg58753] Re: Modeling and Array Problem
  • From: Curtis Osterhoudt <gardyloo at mail.wsu.edu>
  • Date: Sun, 17 Jul 2005 03:04:04 -0400 (EDT)
  • References: <200507160503.BAA14933@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

Hi Peter,

   You might want to check out the documentation for "ReplacePart"; I
think it'll do just what you want
(e.g., array = ReplacePart[array, 42, {1,1}] as per your example).

   On the topic of finding the data's curvature, I've written a package
called ListOps, which I use quite a bit. I've included it below. It
strikes me that if you apply "SuccessiveDifferential" a couple of times
to your data, it'll do what you want. Try ListPlot-ting the results!

               Good luck!

                         C. O.

(****** Save the following lines in a file called "ListOps.m" in a
folder called "Osterhoudt" in your Addons/ExtraPackages directory.******)
BeginPackage["Osterhoudt`ListOps`", {"Statistics`DataManipulation`"}]

Osterhoudt::usage = "'Osterhoudt' is just a directory in the Mathematica
directory, which contains the packages
which I write for Mathematica."

SuccessiveDifferences::usage = "SuccessiveDifferences[data] calculates
and returns the
successive differences of points in a discrete data list. This list must
be one-dimensional.
For example, SuccessiveDifferences[{1, 2, 3, 4, -8}] = {1, 1, 1, -12}."
\
SuccessiveSums::usage = "SuccessiveSums[data] calculates and returns the
successive sums of points in a discrete data list. This list must be
one-dimensional.
For example, SuccessiveSums[{1, 2, 3, 4, -8}] = {3, 5, 7, -4}."

SuccessiveAverages::usage = "SuccessiveAverages[data] calculates and
returns the
successive averages of points in a data set. The input list must be
one-dimensional.
For example, SuccessiveAverages[{1, 2, 3, 4, -8}] = {1.5, 2.5, 3.5, -2}."

SuccessiveDifferential::usage = "SuccessiveDifferential[data] calculates
and returns the
point-by-point derivative of points in a data set. The input list must
be two-dimensional."

Begin["`Private`"]

SuccessiveDifferences[data_] :=
         ListCorrelate[{-1, 1}, data];

SuccessiveSums[data_]:=
         ListCorrelate[{1, 1}, data];

SuccessiveAverages[data_]:=
         SuccessiveSums[data]/2;

SuccessiveDifferential[list_]:=
         Transpose[{SuccessiveAverages[Column[list, 1]],
         SuccessiveDifferences[Column[list,
2]]/SuccessiveDifferences[Column[list, 1]]
                         }
                        ];

End[]   (* Tells Mathematica to revert to the previous context (makes
ListOps` "private")  *)

EndPackage[]   (*Ends the package, prepending ListOps` to the context
search path.   *)

   (* Now, the symbols "data", and "list" are in the private context
ListOps`Private` and, as such, won't interfere wit
h other uses for these symbols, because they're no longer accessible by
their "short names"  *)

(*****************************************************************************************************************)


Sycamor at gmail.com wrote:

>Hello,
>
>I am a high school student and am interning/volunteering at a
>university over the summer.  My ultimate goal is to model the movement
>of charge in a Nickel Metal Hydride Sphere using Mathematica.  This
>goal is beyond my ability as it requires calculus and differential
>equations and so forth.  But I am to progress as best I can, using
>iterative processes to replace calculus where possible.  The professor
>I am working with has started me with the simpler task of finding the
>curvature of a set of data points (in this first easy case, the 'data'
>is the values of 101 points of the x^2 function).
>
>While programming, I have found it necessary to change the value of
>certain elements of an array of ordered pairs, but have been unable to
>do so.
>
>In[1]:= array = {{1,1},{2,2},{3,3},{4,4},{5,5}};
>In[2]:= array[[1]][[1]]
>Out[2]= 1
>In[2]:= array[[1]][[1]] = 42
>Out[2]= Set::setps: test in assignment of part is not a symbol.
>
>I suppose the error arises when I attempt to set the numerical contents
>of the array equal to a certain value.  Is there anyway to simply
>overwrite the element at a certain address?  Why does the syntax used
>above seem to work for lists?
>
>In[3]:= list = {1,2,3,4,5};
>In[4}:= list[[1]]
>Out[4]= 1
>In[5]:= list[[1]] = 42;
>In[6]:= list
>Out[6]= {42,2,3,4,5}
>
>I have included what I have written so far at the end of this message
>to put my problem in context.  As I am new to Mathematica, I do not
>know hoe to write elegant, concise programs using the array of powerful
>built in functions, so I appologize for my clunky code.  There is
>probably a much easier way to accomplish my task.
>
>I would appreciate any help on this matter.
>
>
>Thank you,
>
>Peter Hedman
>
>
>
>
>
>
>Bellow is what I have written so far.  The comments are mostly meant
>for myself, and probably do not make much sense.
>
>g = 1;
>	(*sets value of liberated proton constant*)
>
>d=0.1;
>	(*sets value of diffusion constant*)
>
>v= 0.1;
>	(*sets value of boundry velocity*)
>
>a = 0.1;
>	(*sets the value of constant \[Alpha]*)
>
>j = 0;
>
>	(*sets the initial value of j.  This step shouldn't be necessary*)
>
>rprevb=rprev = Table[0z, {z,0,100}];
>
>r = 0; s = 0;
>
>(Do[rprev[([q])] = {r, s^2}; (r++); (s++), {q, 1, 101}];)
>
>	(*These three lines create the initial list of ordered pairs*)
>
>Do[rprevb[[q]]={0,0}, {q,1,101}]
>
>jmax = 100;
>	(*initially sets the maximum horizontal range*)
>
>iter[n_]:= If[
>      n==jmax , (rprev[[n+1]][[2]] + v*(g+rprev[[n+1]][[2]]) -
>          d*(rprev[[n+2]][[2]]-rprev[[n+1]][[2]])), (rprev[[n+1]][[2]]+
>          a*(rprev[[n]][[2]]-2rprev[[n+1]][[2]]+rprev[[n+2]][[2]]))] ;
>
>(*defines transformation function*)
>
>dim =Dimensions[rprev];
>
>dimb = Dimensions[rprevb];
>
>Print["Dimensions of initial list = ",dim]
>
>Print["Dimensions of initial list = ",dimb]
>
>initialplot = ListPlot[rprev]
>
>Do[Do[rprevb[[(j+1)]][[2]]=iter[j]; Print[rprevb]; {j,0,jmax,1}];
>  rprev=rprevb; Print["time = ",t]; Print["jmax = ",jmax ];
>ListPlot[rprev];
>  jmax--,{t,1,10}]
>
>(*With every iteration of the outside loop,
>  jmax is decremented by one.  As iterations progress to flatten the
>curve,
>  the boundry moves.  Should these two processes take place in this
>way?
>      It appears this is not the ideal model, but it will work for
>now.*)
>
>
>  
>

-- 
PGP Key ID: 0x235FDED1
Please avoid sending me Word or PowerPoint attachments.
http://www.gnu.org/philosophy/no-word-attachments.html



Sycamor at gmail.com wrote:

>Hello,
>
>I am a high school student and am interning/volunteering at a
>university over the summer.  My ultimate goal is to model the movement
>of charge in a Nickel Metal Hydride Sphere using Mathematica.  This
>goal is beyond my ability as it requires calculus and differential
>equations and so forth.  But I am to progress as best I can, using
>iterative processes to replace calculus where possible.  The professor
>I am working with has started me with the simpler task of finding the
>curvature of a set of data points (in this first easy case, the 'data'
>is the values of 101 points of the x^2 function).
>
>While programming, I have found it necessary to change the value of
>certain elements of an array of ordered pairs, but have been unable to
>do so.
>
>In[1]:= array = {{1,1},{2,2},{3,3},{4,4},{5,5}};
>In[2]:= array[[1]][[1]]
>Out[2]= 1
>In[2]:= array[[1]][[1]] = 42
>Out[2]= Set::setps: test in assignment of part is not a symbol.
>
>I suppose the error arises when I attempt to set the numerical contents
>of the array equal to a certain value.  Is there anyway to simply
>overwrite the element at a certain address?  Why does the syntax used
>above seem to work for lists?
>
>In[3]:= list = {1,2,3,4,5};
>In[4}:= list[[1]]
>Out[4]= 1
>In[5]:= list[[1]] = 42;
>In[6]:= list
>Out[6]= {42,2,3,4,5}
>
>I have included what I have written so far at the end of this message
>to put my problem in context.  As I am new to Mathematica, I do not
>know hoe to write elegant, concise programs using the array of powerful
>built in functions, so I appologize for my clunky code.  There is
>probably a much easier way to accomplish my task.
>
>I would appreciate any help on this matter.
>
>
>Thank you,
>
>Peter Hedman
>
>
>
>
>
>
>Bellow is what I have written so far.  The comments are mostly meant
>for myself, and probably do not make much sense.
>
>g = 1;
>	(*sets value of liberated proton constant*)
>
>d=0.1;
>	(*sets value of diffusion constant*)
>
>v= 0.1;
>	(*sets value of boundry velocity*)
>
>a = 0.1;
>	(*sets the value of constant \[Alpha]*)
>
>j = 0;
>
>	(*sets the initial value of j.  This step shouldn't be necessary*)
>
>rprevb=rprev = Table[0z, {z,0,100}];
>
>r = 0; s = 0;
>
>(Do[rprev[([q])] = {r, s^2}; (r++); (s++), {q, 1, 101}];)
>
>	(*These three lines create the initial list of ordered pairs*)
>
>Do[rprevb[[q]]={0,0}, {q,1,101}]
>
>jmax = 100;
>	(*initially sets the maximum horizontal range*)
>
>iter[n_]:= If[
>      n==jmax , (rprev[[n+1]][[2]] + v*(g+rprev[[n+1]][[2]]) -
>          d*(rprev[[n+2]][[2]]-rprev[[n+1]][[2]])), (rprev[[n+1]][[2]]+
>          a*(rprev[[n]][[2]]-2rprev[[n+1]][[2]]+rprev[[n+2]][[2]]))] ;
>
>(*defines transformation function*)
>
>dim =Dimensions[rprev];
>
>dimb = Dimensions[rprevb];
>
>Print["Dimensions of initial list = ",dim]
>
>Print["Dimensions of initial list = ",dimb]
>
>initialplot = ListPlot[rprev]
>
>Do[Do[rprevb[[(j+1)]][[2]]=iter[j]; Print[rprevb]; {j,0,jmax,1}];
>  rprev=rprevb; Print["time = ",t]; Print["jmax = ",jmax ];
>ListPlot[rprev];
>  jmax--,{t,1,10}]
>
>(*With every iteration of the outside loop,
>  jmax is decremented by one.  As iterations progress to flatten the
>curve,
>  the boundry moves.  Should these two processes take place in this
>way?
>      It appears this is not the ideal model, but it will work for
>now.*)
>
>
>  
>

-- 
PGP Key ID: 0x235FDED1
Please avoid sending me Word or PowerPoint attachments.
http://www.gnu.org/philosophy/no-word-attachments.html


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