functions - How do I override options with my own defaults?

I'm calling a function f from within a function g and would like to pass any valid options for f to g, replacing some (say, optA and optB) with defaults different from those defined for f.

What is the right way to do this?

I can get things to (seemingly) work with

Options[g] = Join[FilterRules[Options[f], Except[optA | optB]], {optA -> 0, optB -> 0}];
g[x, opts : OptionsPattern[]] := 
    f[x, opts, OptA -> OptionValue[OptA], optB -> OptionValue[optB]]

but this looks pretty verbose to me. Is there a better, more idiomatic, way?

Answer

Here is what I do in such cases ([edit from 01/10/2020 - apparently, this technique has been exposed on this site earlier, at least partially, in this nice answer by Mr.Wizard - which I definitely read but must have forgotten at the time of posting my answer]) :

ClearAll[g, f];
Options[f] = {optA -> 1, optB -> 1, optC -> 1};
f[x_, opts : OptionsPattern[]] := 
   {OptionValue[optA], OptionValue[optB], OptionValue[optC]}

Options[g] = {optA -> 0, optB -> 0};
g[x_, opts : OptionsPattern[{g, f}]] := f[x, opts, Sequence @@ Options[g]]

In other words, only define options for g which belong to g, and inherit other options from f via the argument of OptionsPattern. Also, pass all options of g explicitly in the call to f, as in Sequence @@ Options[g]. The fact that they come after opts means that they will be over-ridden by opts, as they should be.

Examples:

g[1]

(* {0, 0, 1} *)

g[1, optA -> 2]

(* {2, 0, 1} *)

The added advantage of this scheme is that it is free from the flaw mentioned by Szabolcs, in that subsequent changes of Options[f] will be picked up in this method automatically.

Comments

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