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I want to assign a number to an internal variable, depending if a specific option has been given, but it seems to fail. Something like:

paropts={a->1, b->2, c->3}

myFunction[t,  opts:OptionsPattern[paropts]]:=
With[{If[OptionValue[a],x=a/t,x=1/t]},y=x+OptionValue[c]]

I know the syntax is quite wrong here, but I just want to show the concept. If the option "a" is given then use it to calculate x, otherwise use only 1/t.

That means the function should able to work also if at the beginning my opts are:

paropts={b->2, c->3}

I want to do it as easy as possible and other forms I have tried do not work. I also need to keep the structure of the paropts rules as I have them now, because I use them in many other functions.

It would also be helpful to do it with Block or something else.

Thanks,

S

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I am not sure if I completely understood your problem, but maybe the following is what you are looking for:

myFunction[t_, opts : OptionsPattern[paropts]] := 
 Quiet[With[{x = If[OptionValue[a] =!= a, OptionValue[a]/t, 1/t]}, x + OptionValue[c]]]

It works both when the option a is set, either in paropts or in the function call, or not. I also added an underscore to t.

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  • $\begingroup$ Thanks! I think that's the solution, checking if OptionValue does not return the name of the option itself. I think I am safe here, since I use strictly numbers for the options. $\endgroup$ – Santiago Oct 15 '14 at 11:58
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Unfortunately your code is too incomplete to test, but the following should do what you want:

myFunction[t, opts:OptionsPattern[paropts]]:=
  With[{x = If[OptionValue[a], a/Optionvalue[b], 1/OptionValue[b]]},
       y=x+OptionValue[c]]
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  • $\begingroup$ Hi, thanks for your comment. Yes, I will edit and simplify the function so that it works. But I think your solution is still not working for me. $\endgroup$ – Santiago Oct 15 '14 at 10:41
  • $\begingroup$ I edited my question so that it is a bit more clear! Thanks! $\endgroup$ – Santiago Oct 15 '14 at 10:44

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