Re: ListPlotFit

• To: mathgroup at smc.vnet.net
• Subject: [mg17155] Re: ListPlotFit
• From: BobHanlon at aol.com
• Date: Sun, 18 Apr 1999 00:59:42 -0400
• Sender: owner-wri-mathgroup at wolfram.com

```In a message dated 4/17/99 4:48:55 AM, Maarten.vanderBurgt at icos.be writes:

>With the function ListPlotFit below I combine ListPlot with Fit and Plot
>in
>order to show the data from list l together with the straight line fitted
>through a part of the data points from list l. The list l  is of form
>{{x1,y1},{x2,y2}, ...}.
>
>The part of the data points I want to fit a line through is determined
>by
>start and stop. The default value for the optional argument stop is 10.
>This is not really what I want: I want the default value for stop to be
>stop = Length[l]. How can I achieve this?
>
>I tried ListPlotFit[l_, start_:1, stop_:Hold[Lenght[l]] := Module[{},stop
>=
>ReleaseHold[stop];...], but this did not work (I get an error: "Hold is
>Protected..."). I think I need Hold or something of the sort otherwise
>Lenght[l] gets evaluated to 1 immediately.
>
>I guess I can phrase my question more general: how do I define a function
>F
>with an optional argument v which has as default value a function g of
>a
>non-optional argument u (i.e. F[u_, v_:g[u]])?
>
>
>In[1]:= ListPlotFit[l_, start_:1, stop_:10] :=
>  Module[
>   {lp, fie, fiep},
>   lp = ListPlot[l, DisplayFunction -> Identity, PlotStyle ->
>PointSize[0.015], PlotRange -> All];
>   fie = Fit[l[[Table[j, {j, start, stop}]]], {1, x}, x];
>   fiep = Plot[fie, {x, 1, Length[l]}, DisplayFunction -> Identity];
>   Show[lp, fiep, DisplayFunction -> \$DisplayFunction]
>  ]
>
>In[2]:= dat= Table[{i,3*i^2+5+2*(Random[]-0.5)},{i,1,15}];
>In[3]:= ListPlotFit[dat, 2, 12];
>

Maarten,

Try this.  I take a slightly different approach than yours, but
believe that the behavior is consistent with what you want.

ListPlotFit[l_List, start_Integer:1] :=
ListPlotFit[l, start, Length[l]] ;

ListPlotFit[l_List, start_Integer, stop_Integer] :=
Module[ {lp, fie, fiep},
lp = ListPlot[l, DisplayFunction -> Identity, PlotStyle ->
PointSize[0.015], PlotRange -> All];
fie = Fit[l[[Range[start, stop]]], {1, x}, x];
fiep = Plot[fie, {x, 1, Length[l]}, DisplayFunction -> Identity];
Show[lp, fiep, DisplayFunction -> \$DisplayFunction]
]/; stop > start

dat= Table[{i,3*i^2+5+2*(Random[]-0.5)},{i,1,15}];

ListPlotFit[dat, 2, 12];

ListPlotFit[dat, 2];

ListPlotFit[dat];

Bob Hanlon

```

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