Consider a function myFunc with options

Options[myFunc] = {TellNoOne->True}; (* myFunc has some options *)

myFunc[a_, b_, OptionsPattern[]] := 
    (* Do stuff with OptionValue[TellNoOne] and plot result *)
    ListPlot[{a,b}, moreOptions] (* want to pass some options to ListPlot here *)]

I wish to pass options that will target myFunc, like TellNoOne, and I also want to be able to pass any option to ListPlot that it accepts such as PlotRange->All and others.

Function call example:

myFunc[1, 3, TellNoOne -> False, PlotRange -> All, Joined -> True, Framed -> True]

How can I pass myFunc is own options plus others for ListPlot?

  • 1
    $\begingroup$ Does myFunc[a_, b_, OptionsPattern[Join[Options[myFunc], Options[ListPlot]]]] := (* stuff *) work for you? $\endgroup$ Nov 3, 2017 at 11:22
  • $\begingroup$ You can use FilterRules like in this example from the documentation FilterRules[{PlotRange -> 3, MaxIterations -> 5}, Options[Plot]] $\endgroup$
    – ercegovac
    Nov 3, 2017 at 11:22
  • $\begingroup$ I thinks this is what you ask and more: mathematica.stackexchange.com/q/82626/5478 $\endgroup$
    – Kuba
    Nov 3, 2017 at 11:30
  • $\begingroup$ Does this answer your question? Let us know if it's a suitable duplicate. $\endgroup$
    – Szabolcs
    Nov 3, 2017 at 11:31

1 Answer 1


Here is a very general approach in which we give myFunc default options from ListPlot as well as it own option "Tell", which when given will cause a story :-) to be printed.

Options[myFunc] = {PlotStyle -> Red, Joined -> True, "Tell" -> False};
myFunc[data_List, opts : OptionsPattern[]] :=
  (If[OptionValue[myFunc, FilterRules[{opts}, Options[myFunc]], "Tell"], 
     Print["Once upon a time in a galaxy far, far away ...."]];
   ListPlot[data, FilterRules[{opts, Options[myFunc]}, Options[ListPlot]]])


Test data

data = Table[Sin[2 π t], {t, 0, 1, .02}];

Default options



Giving the "Tell" option

myFunc[data, "Tell" -> True]


Over-riding myFunc's specified graphics options defaults and adding an additional graphics option

myFunc[data, PlotStyle -> Blue, Joined -> False, DataRange -> {0, 1}]


  • $\begingroup$ This works just fine. Thank you. I see by the other answers that there are a few variations on how it can be done. $\endgroup$
    – A. Vieira
    Nov 8, 2017 at 11:10

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