       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)
• 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:= array = {{1,1},{2,2},{3,3},{4,4},{5,5}};
> In:= array[][]
> Out= 1
> In:= array[][] = 42
> Out= 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:= list = {1,2,3,4,5};
> In[4}:= list[]
> Out= 1
> In:= list[] = 42;
> In:= list
> Out= {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]][] + v*(g+rprev[[n+1]][]) -
>           d*(rprev[[n+2]][]-rprev[[n+1]][])), (rprev[[n+1]][]+
>           a*(rprev[[n]][]-2rprev[[n+1]][]+rprev[[n+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)]][]=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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