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Re: passing both arbitary and defined options

  • To: mathgroup at smc.vnet.net
  • Subject: [mg98482] Re: passing both arbitary and defined options
  • From: Bill Rowe <readnews at sbcglobal.net>
  • Date: Fri, 10 Apr 2009 04:58:30 -0400 (EDT)

On 4/9/09 at 5:56 AM, eddie at webassign.net (Eddie) wrote:

>So I'm building a plotting function, and I want it to be able to
>inherit the same options that a ListPlot[] command can inherit, so
>my prototype looks like this:

>NewPlot[data_List,options___Rule]:=Module[{},blah blah blah
>ListPlot[fancy stuff,options]
>]

>Here's the rub, if I additionally want to create a NewPlot-specific
>option, it doesn't mesh.  For example:

>NewPlot[data_List,OptionsPattern[{newOption->newValue}],
>options___Rule] := etc

>Options contains the rule for newOption when NewPlot is called with
>that option utilized, and ListPlot throws an error because it's not
>a recognized option for ListPlot.

>Is there a way I can have my personally declared options and the
>wildcard options both present and working?  Is there a way to "Drop"
>the rule that matches the user-defined option from 'options'?

Use the function FilterRules introduced in version 6. For
earlier versions of Mathematica, the same functionality can be
had by loading the Utilities`FilterOptions`. See the
documentation for examples and syntax.



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