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Re: Modeling and Array Problem
*To*: mathgroup at smc.vnet.net
*Subject*: [mg58746] Re: Modeling and Array Problem
*From*: Bob Hanlon <hanlonr at cox.net>
*Date*: Sun, 17 Jul 2005 03:03:59 -0400 (EDT)
*Reply-to*: hanlonr at cox.net
*Sender*: owner-wri-mathgroup at wolfram.com
array[[m]][[n]] can be written more simply as array[[m,n]]
This latter form must be used to do what you are attempting
array={{1,1},{2,2},{3,3},{4,4},{5,5}};
array[[1,1]]=42;
array
{{42, 1}, {2, 2}, {3, 3}, {4, 4}, {5, 5}}
Here are some equivalent simpler ways of doing things
Table[0 z,{z,0,100}]==Table[0,{z,0,100}]==Table[0,{101}]
True
r=0;s=0;
(Do[rprev[[q]]={r,s^2};(r++);(s++),{q,1,101}];)
rprev==Table[{n,n^2},{n,0,100}]
True
Do[rprevb[[q]]={0,0},{q,1,101}]
rprevb==Table[{0,0},{101}]
True
Bob Hanlon
>
> From: Sycamor at gmail.com
To: mathgroup at smc.vnet.net
> Date: 2005/07/16 Sat AM 01:03:41 EDT
> Subject: [mg58746] Modeling and Array Problem
>
> 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.*)
>
>
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